Category: Inter 1st Year

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion

Very Short Answer Questions

Question 1.
Is it necessary that a mass should be present at the centre of mass of any system?
Answer:
No. Any mass need not be present at the centre of mass of a system.
Ex : a hollow sphere, centre of mass lies at its centre.

Question 2.
What is the difference in the positions of a girl carrying a bag in one of her hands and another girl carrying a bag in each of her two hands?
Answer:
When the girl carries a bag in one hand (left) her centre of mass shifts towards other hands (right). In order to bring it in the middle, the girl has to lean towards her other side. When the girl carries a bag in each of her two hands (left and right), the centre of mass does not shift. The girl does not bend any side because of the same hags’ .w in her two hands.

Question 3.
Two rigid bodies have same moment of inertia about their axes of symmetry. Or the two, which body will have greater kinetic energy ?
Answer:
E = \(\frac{1}{2}\) I ω² = \(\frac{1}{2} \frac{\mathrm{L}^2}{1}\) , E oc \(\frac{1}{l}\) (∵ L = constant)
The rigid body having less moment of inertia will have greater kinetic energy.

Question 4.
Why are spokes provided in a bicycle wheel ?
Answer:
By connecting to the rim of wheel to the axle through the spokes the mass of the wheel gets concentrated at its rim. This increases its moment of inertia. This ensures its uniform speed.

Question 5.
We cannot open or close the door by applying force at the hinges, why ?
Answer:
When the force is applied at the hinges, the line of action of the force passes through the axis of rotation i.e, r = 0, so we can not open or close the door by pushing or pulling it at the hinges.

Question 6.
Why do we prefer a spanner of longer arm as compared to the spanner of shorter arm ?
Answer:
The torque applied on the nut by the spanner is equal to the force multiplied by the perpendicular distance from the axis of rotaion.
A spanner with longer arm provides more torque compared to a spanner with shorter arm. Hence longer arm spanner is preferred.

Question 7.
By spinning eggs on a table top, how will you distinguish a hard boiled egg from a raw egg ? [Mar. 13]
Answer:
A raw egg has some fluid in it and a hard boiled egg is solid form inside. Both eggs are spinning on a table top, the fluid is thrown outwards. Therefore (I_r > I_b) That means M. I of raw egg is greater than boiled egg. As I × ω = constant; ∴ ω_r < ω_b. That means Angular Velocity of raw egg is less than angular velocity of boiled egg.

Question 8.
Why should a helicopter necessarily have two propellers ?
Answer:
If there were only one propeller in the hellicopter then, due tio conservation of angular momentum, the helicopter itself would have turned in the opposite direction. Hence, if should have two propellers.

Question 9.
It the polar ice caps of the earth were to melt, what would the effect of the length of the day be ?
Answer:
Earth rotates about its polar axis. When ice of polar caps of earth melts, mass concentrated near the axis or rotation spreads out. Therefore, moment of Inertia I increases.

As no external torgue acts, L = I × ω = \(I\left(\frac{2 \pi}{T}\right)\)= constant. With increase of I, T will increase i.e. length of the day will increase.

Question 10.
Why is it easier to balance a bicycle in motion?
Answer:
When a bicycle is in motion, it is easy to balance because the principle of conservation of angular momentum is involved.

Short Answer Questions

Question 1.
Distinguish between centre of mass and centre of gravity. [A.P. – Mar. ‘18, ‘16, ‘15, ‘14, ‘13; TS – Mar. ‘16, ‘15, ‘14, ‘13]
Answer:
Centre of mass

1. Point at which entire mass of the body is supposed to be concentrated, and the motion of the point represents motion of the body.
2. It refers mass of to body.
3. In a uniform gravitational field centre of mass and centre of gravity coincide
4. Centre of mass of the body is defined to describe the nature of motion of a body as a whole.

Centre of gravity

1. Fixed point through which the weight of the body act.
2. It refers to the weight acting on all particles of the body
3. In a non-uniform gravitational field, centre of gravity and centre of mass do not coincide.
4. Centre of gravity of body is defined to know the amount of stability of the body when supported.

Question 2.
Show that a system of particles moving under the influence of an external force, moves as if the force is applied at its centre of mass.
Answer:
Consider \(\overrightarrow{r_1}, \overrightarrow{r_2}, \overrightarrow{r_3} \ldots \ldots \overrightarrow{r_n}\) be the position vectors of masses m₁, m₂, m₃ …………. m_n of n particle system.
According to Defination of centre of mass.

Differentiating the above equation w.r.t. time, we obtain

Where F_ext represents the sum of all external forces acting on the particles of the system. This equation states that the centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point.

Question 3.
Explain about the centre of mass of earth-moon system and its rotation around the sun.
Answer:
In the solar system the planets have different velocities and have complex two dimensional motion. But the motion of the centre of mass of the planet is simple and translational. consider the earth and moon system. We consider that the earth is moving around the sun in an elliptical path. But actually the centre of mass of earth and moon moves in an elliptical path round the sun. But the motion of either earth or moon is complicated when considered separately, more over we say that moon goes round the earth.

But actually earth and moon are revolving round their centre of mass such that they are always on opposite sides of the centre of mass. Here the forces of attraction between earth and moon are internal forces.

Question 4.
Define vector product. Explain the properties of a vector product with two examples. [T.S. AP – Mar. 15]
Answer:
The cross product of two vectors is given by \(\vec{C}=\vec{A} \times \vec{B}\). The magnitude of the vector defined from cross product of two vectors is equal to product of magnitudes of the vectors and sine of angle between the vectors.

Direction of the vectors is given by right hand corkscrew rule and is perpendicular to the plane containing the vectors.
∴ \(|\vec{C}|\) = AB sin θ. and \(\vec{C}\) = AB sin θ \(\hat{n}\). Where, \(\hat{n}\) is the unit vector perpendicular to the plane containing the vectors \(\vec{A}\) and \(\vec{B}\)
Example: 1) Torque is cross product of position vector and Force. i.e., \(\overrightarrow{\mathrm{T}}=\vec{r} \times \vec{F}\)
2) Angular momentum is cross product of position vector and momentum
i.e., \(\vec{L}=\vec{r} \times \vec{p}\)

Properties:

1. Cross product does not obey commutative law. But its magnitude obeys commutative low.
   \(\vec{A} \times \vec{B} \neq \vec{B} \times \vec{A} \Rightarrow(\vec{A} \times \vec{B})=-(\vec{B} \times \vec{A}),|\vec{A} \times \vec{B}|=|\vec{B} \times \vec{A}|\)
2. It obeys distributive law \(\vec{A} \times(\vec{B} \times \vec{C})=\vec{A} \times \vec{B}+\vec{A} \times \vec{C}\)
3. The magnitude of cross product of two vectors which are parallel is zero.
   Since θ = 0; \(|\vec{A} \times \vec{B}|\) = AB sin 0° = 0
4. For perpendicular vectors, θ = 90°, \(|\vec{A} \times \vec{B}|\) = AB sin 90° \(|\hat{n}|\)= AB

Question 5.
Define angular velocity(u). Derive v = r ω. [T.S. Mar. 16]
Answer:
Angular velocity (ω):
The rate of change of angular displacement of a body is called angular velocity, i.e., ω = \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\)
Derivation of v = rω
consider a rigid body be moving with uniform speed (v)along the circumference of a circle of radius r. Let the body be displaced from A to B in a small interval of time At making an angle ∆θ at the cantre. Let the linear displacement be ∆x from A to B.
From the property of circle, length of arc = radius × angle
∆x = r ∆θ
This equation is divided by ∆t, and taking

Question 6.
Define angular acceleration and torque. Establish the relation between angular acceleration and torque. [T.S. Mar. 18]
Answer:
Angular acceleration :
The rate of change of angular velocity is called angular acceleration
i.e., α = \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\)
Torque : The rate of change of angular momentum is called torque or The moment of Force is called Torque.

Relation between angular acceleration and Torque: Consider a rigid body of mass ‘M’ rotating in a circular path of radius ‘R’ with angular velocity ‘ωω’ about fixed axis.

By definition, τ = \(\frac{\mathrm{dL}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{I} \omega)}{\mathrm{dt}}\)
Where I = MR² = moment of inertia of a body
τ = I \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\) [∵ I = constant]
But \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\) α
∴ τ = Iα

Question 7.
Write the equations of motion for a particle rotating about a fixed axis?
Answer:
Equations of motion for a particle rotating about a fixed axis:
1. ω_f = ω_f + αt [∵ like v = u + at]
2. θ = \(\left(\frac{\omega_f+\omega_t}{2}\right) t\) [∵ like v = \(\left(\frac{v_1+v_2}{2}\right) t\)]
3. θ = ω_it + \(\frac{1}{2}\) α t² [∵ like s = ut + \(\frac{1}{2}\) at²]
4. ω_f² – ω_i² = 2 α θ [∵ like v² – u² = 2as]
Where ω_i = Initial angular velocity
ω_f = Final angular velocity
α = Angular acceleration
θ = Angujar displacement
t = time

Question 8.
Derive expressions for the final velocity and total energy of a body rolling without slipping.
Answer:
Expression of velocity of a body Rolling down an inclined plane:
Consider a rigid body of mass M and radius R rolling down an inclined plane from a height h . Let v the linear speed acquired by the body when it reaches the bottom of the plane and k is its radius of gyration.

According to law of conservation of Energy, we have P.E of body on top of inclined plane = K.E of a body at the bottom of inclimed plane
When a body rolls down an incline of height h, we apply the principle of conservation of energy.
P.E at the top – K.E of translation + K.E of rotation

Expression of Total energy of a body Rolling down on an inclined plane:
Suppose a body (Sphere) is rolling on a surface. Its motion can be treated as a combination of the translation of the centre of mass and rotation about an axis passing through the centre of mass. The total kinetic energy E can written as
E = E_T + E_R
Where E_T = Translational kinetic energy
and E_R = Rotational kinetic energy
E = \(\frac{1}{2}\) Mv² + \(\frac{1}{2}\) Iω²
= \(\frac{1}{2}\) MR² ω² + \(\frac{1}{2}\) Mk²ω²
E = \(\frac{1}{2}\) M ω² (R² + K²)
where k is radius of gyration
E = \(\frac{1}{2}\) Mv² (1 + \(\frac{K^2}{R^2}\)) [∵ ω = \(\frac{V}{R}\)]

Long Answer Questions

Question 1.
a) State and prove parallel axis theorem.
Answer:
Statement: The moment of inertia of a rigid body about any axis is equals to the sum of moment of inertia about a parallel axis passing through centre of mass (lg) and product of mass of body and square of perpendicular distance between two parallel axis.
∴ I = I_G + mr²
Proof : Consider a rigid body of mass M. Let I and I_G are the moment of inertia of a body about parallel axes X and Y respectively.

Let ‘r’ be the perpendicular distance between two axes.
Moment of inertia about ‘X’ axes is
I = Σm(OP)² ………………. (1)
Moment of inertia about Y axes is
I_G = Σm(GP)² …………….. (2)
from ∆⁴ OQP, OP² = (OQ)² + (PQ)²
= (OG + GQ)² + (PQ)²
= OG² + (GQ)² + 2.OG.GQ + (PQ)²
= OG² + (GP)² + 2(OG) (GQ) (∴ (GP)² = (GQ)² + (PQ)²)
Multiplying both sides by Σm,
⇒ Σm(OP)² = Σm(OG)² + Σm(GP)² + 2Σm(OG)(GQ)
from (1) and (2), I = mr² + IG + 2Σm (OG) (GQ)
But sum of moments of particles about centre of mass is zero.
i.e., Σm (OG) (GQ) = 0
∴ I = I_G + mr²
Hence proves.

b) For a thin flat circular disk, the radius of gyration about a diameter as axis is k. If the disk is cut along a diameter AB as shown in to two equal pieces, then find the radius of gyration of each piece about AB.

Answer:
For thin circular disk, the radius of gyration about a diameter
AB is, K = \(\sqrt{\frac{I}{M}}\)
Where M = mass of disk
I = M.I of disk
The disk is cut into two halves about AB, when each,
Mass M’ = \(\frac{\mathrm{M}}{\mathrm{2}}\) and each M.I, I = \(\frac{1}{2}\)
Therefore Radius of gyration of each piece is
K’ = \(\sqrt{\frac{I^{\prime}}{M^{\prime}}}=\sqrt{\frac{\left(\frac{I}{2}\right)}{\left(\frac{M}{2}\right)}}=\sqrt{\frac{I}{M}}\) = k

Question 2.
a) State and prove perpendicular axes theorem.
Answer:
Statement: The moment of inertia of a plane lamina about an axes perpendicular to its plane is equal to sum of moment of inertia of lamina about the perpendicular axes in its plane intersecting each other at a point, where the perpendicular axes passes.
i.e., I_z = I_x + I_y
Proof’ Consider a particle of mass’m’ at p. Let it be at a distance Y from Z-axis. Here ‘X’ and Y axes are in plane Ramina and Z-axes perpendicular to plane lamina.

Now,
Moment of inertia about X-axix, I_x = Σmx²
Moment of inertia about Y-axis, I_y = Σmy²
Moment of inertia about Z-axis, I_z = Σmr²
From ∆⁴ OQP, r² = x² + y²
multiplying both side with Σm,
⇒ Σmr² = Σmx² + Σmy²
⇒ I_z = I_x + I_y
Hence proved.

b) If a thin circular ring and a thin flat circular disk of same mass have same moment of inertia about their respective diameters as axes. Then find the ratio of their radii.
Answer:
For a thin flat circular dist, M.I., I_r = \(\frac{\mathrm{MR}_{\mathrm{r}}^2}{2}\)
For a thin flat circular dist, M.I., I_d = \(\frac{\mathrm{MR}_{\mathrm{d}}^2}{2}\)
Given that I_r = I_d
\(\frac{M R_r^2}{2}=\frac{M R_d^2}{4} \Rightarrow \frac{R_r^2}{R_d^2}=\frac{2}{4}=\frac{1}{2}\)
∴ \(\frac{R_r}{R_d}=\frac{1}{\sqrt{2}} \text { or } R_r: R_d=1: \sqrt{2}\)

Question 3.
State and prove the principle of conservation of angular momentum. Explain the principle of conservation of angular momentum with examples. [A.P. Mar 16]
Answer:
Statement: Angular momentum of a body remains constant when the external torque is zero.
L = I ω = constant K.
or I₁ ω₁ = I₂ ω₂.
If the moment of Inertia of a body is lowered, the angular velocity of the body co increases.
Proof:
By derfination, the rate of change of angular momentum is called Torque.
i.e., τ = \(\frac{\mathrm{dL}}{\mathrm{dt}}\)
If τ = 0 ⇒ \(\frac{\mathrm{dL}}{\mathrm{dt}}\) = 0
or L = constant k
⇒ L₁ = L₂
∴ I₁ ω₁ = I₂ ω₂
Example 1): When a man with stretched out arms stands on a turn table which is revolving then his moment of inertia is high. If he folded his hands, the moment of inertia decreases and hence the angular velocity, linear velocity increase, but the period decreases. In both cases angular momentum remains constant.

Example 2): An acrobat from a swing in a circus, leaves the swing with certain angular momentum, with his arms and legs stretched. As soon as he leaves the swing he pulls his hands and legs together thus lowering his M.I. and increasing his angular velocity. He then quickly makes somersaults in air and finally lands on a net or ground.

Problems

Question 1.
Show that a(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.
Solution:
Let a parallele piped be formed on three vectors.

Which is equal in magnitude to the volume of the parallel piped.

Question 2.
A rope of negligible mass is wound 4. round a hollow cylinder of mass 3kg and redius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N ? What is the linear acceleration of the rope ? Assume that there is no slipping.
Solution:
Here M = 3kg, R = 40 cm = 0.4 m
Moment of intertia of the hollow cylinder about its axis
I = MR² = 3(0.4)² = 0.48 kgm²
Force applied F = 30 N
∴ Torque, τ = F × R= 30 × 0.4 = 12 N – m.
If α is the angular acceleration, produced, then from τ = Iα
α = \(\frac{\tau}{\mathrm{l}}=\frac{12}{0.48}\) = 25 rads^-2
Linear acceleration, a = Rα = 0.4 × 25 = 10 m/s²

Question 3.
A coin is kept a distance of 10 cm from the centre of a circular turntable. If the coefficient of static friction between the table and the coin is 0.8 find the frequency of rotation of the disc at which the coin will just begin to slip.
Solution:
Here, radius, r = 10cm = 0.1 m; μ_s = 0.8
F = μmg
mrω² = μmg
rω² = μg
ω = \(\frac{\mu \mathrm{g}}{\mathrm{r}}=\sqrt{\frac{0.8 \times 9.8}{0.1}}=\sqrt{8 \times 9.8}\)
= 8.854 rad/s
w = 2πn
∴ frequency n = \(\frac{\omega}{2 \pi}=\frac{8.854}{2 \times 3.14}\) = 1 .409rps
n = 1.409 × 60 = 84.54 rpm

Question 4.
Particles of masses 1g, 2g, 3g…. 100g are kept at the marks 1 cm. 2cm, 3cm ….100cm respectively on 2 meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale. –
Solution:
By adding all the masses, we get M = 5050g
= 5.050 kg
= 5.1 kg
and L = 1 m
M.I. of the meter scale = \(\frac{\mathrm{mL}^2}{12}=\frac{5.1 \times 1^2}{12}\)
= 0.425kg m²
= 0.43kg -m²

Question 5.
Three particles each of mass 100g are placed at the vertices of an equilateral triangle of side length 10 cm. Find the moment of inertia of the system about an axis passing through the centroid Of the triangle and perpendicular to its plane.
Solution:
m = 100 g = 100 × 10^-3kg
Side a = 10 cm

Question 6.
Four particles each of mass 100g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system.
Solution:
mass m = 100g
= 100 × 10^-3kg
M = 4m = 400 × 10^-3kg
radius r = 10cm
= 10 × 10^-2m
Moment of Inertia I = Mr²

Question 7.
Two uniform circular discs, each of mass 1 kg and radius 20 cm, are kept in contact about the tangent passing through the point of contact. Find the moment of inertia of the system about the tangent passing through the point of contact.
Solution:
Mass m = 1kg; r = 20cm = 20 × 10^-2m
I = I₁ + I<sub2

I₁ = \(\frac{\mathrm{MR}^2}{4}\) + MR²
I₁ = \(\frac{5\mathrm{MR}^2}{4}\)
Similarly I₂ = \(\frac{5\mathrm{MR}^2}{4}\)
∴ I = \(\frac{10 \mathrm{MR}^2}{4}=\frac{10 \times 1 \times\left(20 \times 10^{-2}\right)^2}{4}\) = 0.1 kg-m²

Question 8.
Four spheres, each diameter 2a and mass ‘m’ are placed with their centres on the four corners of a spuare of the side b. Calculate the moment of inertia of the system about any side of the square.
Solution:
I₁ = mb²
I₂ = \(\frac{2}{5}\) ma²
I₃ = \(\frac{2}{5}\) ma²
I₄ = mb²
∴ Moment of inertia of the system
I = I₁ + I₂ + I₃ + I₄
= mb² + \(\frac{2}{5}\) ma² + \(\frac{2}{5}\) ma² + mb²
I = \(\frac{4}{5}\) ma² + 2mb²

Question 9.
To maintain a rotor at a uniform angular speed or 200 rads^-1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine ? (Note : uniform angular velocity in the absence of friction implies zero torque. In practice applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Solution:
Here, ω = 200 rad/s, torque τ = 180 N – m, power p = ?
As p = τ ω
∴ p = 180 × 200 = 3600.watt = 36 kw.

Question 10.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick ?
Solution:
Let M be the mass of stick concentrated at L, the 50 cm, mark.
For equilibrium about G’ the 45 cm mark,
10g (45 – 12) = mg(50 – 45)
10 g × 33 = mg × 5
M = \(\frac{10 \times 33}{5}\) = 66 gram.

Question 11.
Determine the kinetic energy of a circular disc rotating with a speed of 60 rpm about an axis passing through a point on its circumference and perpendicular to its plane. The circular disc has a mass of 5kg and radius 1m.
Solution:
Here M = 5kg; R = 1 m;
ω = 2π × \(\frac{\mathrm{N}}{\mathrm{t}}\) = 2n × \(\frac{60}{60}\) rad /s = 2π rad/s
The M.I of disc about parallel axis passing through a point on its circumferance
I = \(\frac{\mathrm{MR}^2}{2}\) + MR² = \(\frac{3}{2}\) MR²
∴ Kinetic energy = \(\frac{1}{2}\) I ω²
= \(\frac{1}{2}\) × \(\frac{3}{2}\) MR² ω² = \(\frac{3}{4}\) × 5 × (1)² × (2π)²
= \(\frac{3}{4}\) × 5 × 4π² = 15 × (\(\frac{22}{7}\)²
∴ K.E = 148.16 J.

Question 12.
Two particles each of mass m and speed v Travel in opposite directions along parallel lines separated by a distance. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.
Solution:
Given that m₁ = m₂ = m;
\(\vec{V}_1=\vec{V}_2=\vec{V}\)
momentum of 1^st particle p₁ = \(\mathrm{m} \vec{\mathrm{V}}\)
∴ p₁ = \(\mathrm{m} \vec{\mathrm{V}}\)
momentum of 2^nd particle \(\vec{\mathrm{P}}_2=\mathrm{mV}\)

If two particles moves oppositely on
Circumference of circle, distance d = 2r
Angular momentum of 1^st particle w.r.t
Centre ‘O’ is \(\vec{L_1}=\vec{r} \times \vec{P}_1=\vec{r} \times m \vec{V}\)
Angular momentum of 2^nd particle w.r.t
centre ‘O’ is \(\vec{\mathrm{L}_2}=\vec{\mathrm{r}} \times \vec{\mathrm{P}_2}=\vec{\mathrm{r}} \times \mathrm{m} \vec{\mathrm{V}}\)
∴ \(\vec{L_1}=\vec{L_2}\)

Question 13.
The moment of inertia of a fly wheel making 300 revolutions per minute is 0.3 kgm ² Find the torque required to bring it to rest in 20s.
Solution:

Question 14.
When 100 J of work is done on a fly wheel, its angular velocity is increased from 60 rpm to 180 rpm. What is the moment of inertia of the wheel ?
Solution:
Here, Initial frequency
n₁ = \(\frac{60}{60}\) = 1Hz
Initial angular velocity
ω₁ = 2π n₁ = 2π rad /sec
Final frequency n₂ = \(\frac{180}{60}\) = 3Hz
Final angular velocity
ω₂ = 2π n₂ = 2π × 3 = 6π rad/sec
Work done = 100 J from work – energy therorem,
Workdone = change in K.E.
W = \(\frac{1}{2}\) I ω₂² – \(\frac{1}{2}\) I ω₁²
100 = \(\frac{1}{2}\) I [(6π)² – (2π)²]
100 = \(\frac{1}{2}\) I (32π)²
I = \(\frac{200}{32 \pi^2}\) = 0.634 kg – m²
∴ I = 0.634 kg- m²

Additional Problems

Question 1.
Give the location of the centre of mass of a (i) sphere (ii) cylinder (iii) ring and (iv) cube. each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
Solution:
In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres.
No, it is not necessary that the center of mass of a body should lie on the body for example in caSe of a circular ring. Center of mass is at the centre of the ring, where there is no mass.

Question 2.
In the HCl molecule the separation between the nuclei of the two atoms Is about 1.27 Å(1 Å = 10^-10m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom Is concentrated in its nucleus.
Solution:
Let the mass of the H atom = m unit, mass of the Cl atom = 35.5 m units
Let cm be at the distance xÅ from Htom
∴Distance of cm from CL atom = (1.27 – x) Å
It cm is taken at the origin,
then mx + (1.27 – x) 55.5m = 0
mx = (1.27 – x) 35.5 m.
Negative sign indicates that if chlorine atom is on the right side of cm (+), the hydrogen atom is on the left side of cm, so leavning negative sign, we get
x + 35.5 x = 1.27 × 35.5
36.5 x = 45.085
x = \(\frac{45.085}{36.5}\)
= 1.235
x = 1.235 Å
Hence cm is located on the line joining centres of H and d atoms at a distance 1.235 Å from H.

Question 3.
A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?
Solution:
The speed of the centre of mass of the system , (trolley + child) shall remain unchanged, when the child gets up and runs about on the trolley in any manner. This is because forces involved in the exercise are purely internal i.e., from within the system. No external force acts on the system and hence there is no change in velocity of the system.

Question 4.
Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b
Solution:
Let \(\vec{\mathrm{a}}\) represented by \(\vec{\mathrm{OP}}\) and \(\vec{\mathrm{b}}\) be represented by \(\vec{\mathrm{OQ}}\).
Let ∠POQ = θ

∴ area of ∆ OPQ = \(\frac{1}{2}|\vec{a} \times \vec{b}|\), which was to be proved.

Question 5.
Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.
Solution:
Let a parallel piped be formed on three vectors.

Where \(\hat{n}\) is unit vector along \(\overrightarrow{\mathrm{OA}}\) perpendicular to the plane containing \(\) and \(\vec{\mathrm{b}}\). Now \(\vec{\mathrm{b}}\) = (a) (be) cos 0°
= abc
Which is equal in magnitude to the volume of the parallelepiped.

Question 6.
Find the components along the x. y. z axes of the angular momentum 1 of a particle whose position vector is r with components x, y, z and momentum is p with components p_x, p_y, and p_z,. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Solution:
For motion in 3D, the position vector \(\vec{\mathrm{r}}\) and linear momentum vector \(\vec{\mathrm{p}}\) can be written in terms of their rectangular components as follows.

Comparing the coefficient of k on both sides, we have L_z = xp_y – yp_x.
Therefore the particle moves only in x-y plane. The angular momentum has only z component.

Question 7.
Two particles, each of mass m and speed v. travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same what ever be the point about which the angular momentum is taken.
Solution:
From fig, vector angular momentum of the two particle system any point A on x₁ y₁ is.

Similary, vector angular momentum of the tw.o particle system, a bout any pt. B on x₂y₂ is
\(\vec{L_B}=m \vec{v} \times d+m \vec{v} \times 0=m \vec{v} d\)
Let us consider any other point (on AB, where AC = x)
∴ Vector angular momentum of the two particle system about c is
\(\vec{L_c}=m \vec{v}(x)+m \vec{v}(d-x)=m \vec{v} d\)
Clearly, \(\vec{\mathrm{L}_{\mathrm{A}}}=\vec{\mathrm{L}_{\mathrm{B}}}=\vec{\mathrm{L}_C}\)
Which has to be proved.

Question 8.
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in figure. The angles made by the strings with the vertical are 36.9 and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
Solution:

From fig (b)
θ₁ = 36.9°, θ₂ = 53.1°
If T₁, T₂ are the tensions in two strings, then for equilibrium along the horizontal,
T₁ sin θ₁ = T₂ sin θ₂
(or) \(\frac{T_1}{T_2}=\frac{\sin \theta_2}{\sin \theta_1}=\frac{\sin 53.1^{\circ}}{\sin 36.9^{\circ}}\)
= \(\frac{0.7407}{0.5477}\) = 1.3523
Let d be the distance of center of gravity c of the bar from the left end.
For rotational equilibrium about c,
T₁ cos θ₁ Xd = T₂ cos θ₂ (2 – d)
T₁ cos 36.9° × d = T₂ cos 53.1° (2 – d)
T₁ × 0.8366 d = T₂ × 0.6718 (2 – d)
Put T₁ = 1.3523 T₂ and solve to get
d = 0.745 m.

Question 9.
A car weight 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Solution:
Here, m = 1800 kg
Distance between front and .back axles = 1.8 m
Distance of center of gravity (c) behind the front axle = 1.05 m
Let R₁ and R₂ be the forces exerted by the level ground on each front wheel and each back wheel. As it is dear from fig.
R₁ + R₂ = mg = 1800 × 9.8

For rotational equilibrium about C,
R₁ × 1.05 = R₂ (1.8 – 1.05) = R₂ × 0.75 ………….. (i)
\(\frac{R_1}{R_2}=\frac{0.75}{1.05}=\frac{5}{7}\)
Putting in (i)
\(\frac{5}{7}\) R₂ + R₂ = 1800 × 9.8
R₂ = \(\frac{7 \times 1800 \times 9.8}{12}\)
= 10290 N
R₁ = \(\frac{5}{7}\) R₂
= \(\frac{5}{7}\) × 10290
= 7350 N.

Question 10.
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 MR²/5. where M is the mass of the sphere and R is the radius of the sphere.
Solution:
a) Moment of inertia of sphere about any diameter = \(\frac{2}{5}\)MR²
Applying theorem of parallel axes,
Moment of inertia of sphere about a tangent to the sphere = \(\frac{2}{5}\)MR² + M(R)²
= \(\frac{7}{5}\)MR².

(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR²/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Solution:
We are given, moment of inertia of the disc about any of its diameters = \(\frac{1}{4}\)MR².
i) Using theorem of perpindicular axes, moment of inertia of the disc about an axis passing through its center and normal to the disc = 2 × \(\frac{1}{4}\)MR² = \(\frac{1}{2}\)MR²
ii) Using theorem of parallel axes, moment of inertia of the disc passing through a point on its edge and normal to the disc
= \(\frac{1}{2}\)R² + MR² = \(\frac{3}{2}\) MR².

Question 11.
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.
Solution:
If M is mass and R is radius of the hollow cylinder and the solid sphere, then
M.I of hollow cylinder about its axis of symmetry I₁ = MR² and
M.I of solid sphere about an axis through its 2 , centre, I₁ = \(\frac{2}{5}\) MR²
Torque applied, I = I₁ α₁ = I₂ α₂
\(\frac{\alpha_2}{\alpha_1}=\frac{l_1}{l_2}=\frac{M^2}{\frac{2}{5} M R^2}=\frac{5}{2}\)
α₂ > α₁
From ω = ω₀ + αt, we find that for given ω₀ and t, ω₂ > ω₁ to, i.e. angular speed of solid sphere will be greater than angular speed of hallow sphere.

Question 12.
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s^-1. The radius of the cylinder is 0.25m. What is the kinetic energy associated with the rotation of the cylinder ? What is the magnitude of angular momentum of the cylinder about its axis ?
Solution:
Here, M = 20 kg, R = 0.25 m, w – 100 g^-1
Moment of inertia of solid cylinder
= \(\frac{\mathrm{MR}^2}{2}=\frac{20 \times(0.25)^2}{2}\) = 0.625 kg/m²
K.E of rotation = \(\frac{1}{2}\) Iω² = \(\frac{1}{2}\) × 0.625 × (100)² = 3125 J
Angular momentum, L = Iω
= 0.625 × 100
= 62.5 Js.

Question 13.
a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/miri. How much is the angular spped of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction.
Solution:
Here, intial angular speed ω₁ = 40 rev/min, ω₂ = ?
Final moment of inertia, I₂ = \(\frac{2}{5}\)I, Intial moment of Inertia
As no external torque acts in the process, therefore
Iω = constant
i.e. I₂ω₂ = I₁ω₁
ω₂ = \(\frac{I_1}{I_2}\) ω₁ = \(\frac{5}{2}\) × 40
= 100 rpm

b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy ?
Solution:
Final K.E of rotation, E₂ = \(\frac{1}{2}\)I₂ω₂²
Intial K.E of rotation, E₁ = \(\frac{1}{2}\)I₂ω₁²
\(\frac{E_2}{E_1}=\frac{\frac{1}{2} l_2 \omega_2^2}{\frac{1}{2} l_1 \omega_1^2}=\left(\frac{l_2}{l_1}\right)\left(\frac{\omega_2}{\omega_1}\right)^2\)
= \(\frac{2}{5} \times\left(\frac{100}{40}\right)^2=\frac{5}{2}\) = 2.5
∴ K.E of rotation increase. This is because child spends internal energy folding back his hands.

Question 14.
A rope of negligible mass is wound roung a hollow cylinder of mass 3kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? what is the linear acceleration of the rope ? Assume that there is no slipping.
Solution:
Here M = 3kg, R = 40 cm = 0.4 m
Moment of Inertia of the hollow cylinder about its axis
I = MR² = 3(0.4)² = 0.48 kgm²
Force applied F = 30 N
∴ Torque, τ = F × R = 30 × 0.4 = 12 N – m.
If α is the angular acceleration, produced, then from τ = Iα
α = \(\frac{\tau}{\mathrm{I}}=\frac{12}{0.48}\) = 25 rads^-2
Linear acceleration, a = Rα = 0.4 × 25 = 10 m/s².

Question 15.
To maintain a rotor at a uniform angular speed of 200 rad s^-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note : uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Solution:
Here, ω = 200 rad/s, torque τ = 180 N – m, power p = ?
As p = τω
∴ p = 180 × 200 = 3600 watt = 36 kw.

Question 16.
From a uniform disc of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is a R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Solution:
Suppose mass per unit area of the disc = M
∴ Mass of original disc

M = πR² × m
Mass of portion removed from the disc
M’ = π\(\left(\frac{R}{2}\right)^2 \times m=\frac{\pi R^2}{4} m=\frac{M}{4}\)
In fig, mass M is connected at O and mass
M’ is concentrated at O’, where OO’ = \(\frac{\mathrm{R}}{\mathrm{2}}\).
After the circular disc of mass M’ is removed, the remaining portion can be considered as a system of two masses M at 0 and – M’ = \(\frac{\mathrm{-M}}{\mathrm{4}}\) at O’. If x is the distance of centre of 4 mass (p) of the remaining part, then
x = \(\frac{M \times O-M^{\prime} \times \frac{R}{2}}{M-M^{\prime}}\)
= \(\frac{\frac{-M}{4} \times \frac{R}{2}}{M-\frac{M}{4}}=\frac{-M R}{8} \times \frac{4}{3 M}=\frac{-R}{6}\)
Negative sign shows that p is the left O.

Question 17.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick ?

Solution:
Let m be the mass of stick concentrated at c, the 50 cm, mark,
For equilibrium about c’ the 45 cm mark,
10g (45 – 12) = mg(50 – 45)
10 g × 33 = mg × 5
m = \(\frac{10 \times 33}{5}\)
= 66 gram.

Question 18.
A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination, (a) Will it reach the bottom with the same speed in each case ? (b) Will jt take longer to roll down one plane than the other ? (c) If so, which one and why ?
Solution:
Let v be the speed of the solid sphere at the bottom of the incine. Applying principle of conservation of energy, we get
\(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω² = mgh
As I = \(\frac{2}{5}\) mr²; \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (\(\frac{2}{5}\) mr²)ω² = mgh
as rω = v, \(\frac{1}{2}\) mv² + \(\frac{1}{5}\) mv² = mgh
v = \(\sqrt{\frac{10}{7} \mathrm{gh}} .\)
As h is same in two cases, v must be same i.e., It will reach the bottom with the same speed. Time taken to roll down the two planes will also be the same, as their height is the same.

Question 19.
A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it ?
Solution:
Here, R = 2m, M = 100 kg v = 20 cm/s = 0.2 m/s.
Total energy of the hoop = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω²
= \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (MR)²ω² = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) mv²
= mv².
Work required to stop the hoop = total energy of the hoop
w = mv² = 100(0.2)² = 4 joule.

Question 20.
The oxygen molecule has a mass of 5.30 × 10^-26 kg and a moment of inertia of 1.94 × 10^-46 kg m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of trans-lation. Find the average angular velocity of the molecule.
Solution:
Here, m = 5.30 × 10^-26 kg
I = 1.94 × 10-^-46 . kgm²
v = 500 m/s
If \(\frac{\mathrm{m}}{\mathrm{2}}\) is mass of each atom of oxygen and 2r is the distance between the two atoms as shown in fig. then

= 6.7 × 10-¹² rod/s.

Question 21.
A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
a) How far will the cylinder go up the plane ?
b) How long will is take to return to the bottom ?
Solution:
Here θ = 30°, v = θm/s
Let the cylinder go up the plane upto a height h
From \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω² = mgh
\(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (\(\frac{1}{2}\) mr²) ω² = mgh
\(\frac{3}{4}\) mv² = mgh
h = \(\frac{3 v^2}{4 g}=\frac{3 \times 5^2}{4 \times 9.8}\) = 1.913 m
If s is the distance up the inclined plane, then as sin θ = \(\frac{\mathrm{h}}{\mathrm{s}^2}\)
s = \(\frac{\mathrm{h}}{\sin \theta}=\frac{1.913}{\sin 30^{\circ}}\) = 3.826 m
Time taken to return to the bottom
t = \(\sqrt{\frac{2 s\left(1+\frac{k^2}{r^2}\right)}{g \sin \theta}}=t \sqrt{\frac{2 \times 3.826\left(1+\frac{1}{2}\right)}{9.8 \sin 30^{\circ}}}\) = 1.53 s.

Question 22.
As shown in Fig. the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F.1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder.
(Take g = 9.8 m/s²)

(Hint: Consider the equilibrium of each side of the ladder separately.)
Solution:
Data seems to be insufficient

Question 23.
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand . The angular speed of the platform is 30 revolutions per minute. The man then brings his arms dose to his body with the distance of each weight from the axis changing from 90cm to 20 cm. Thd moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m²
(a) What is his new angular speed ? (Neglect friction.)
b) Is kinetic energy conserved in the process ? If not, form where does the change come about ?
Solution:
Here l₁ = 7.6 × 2 × 5(0.9)² = 15.7 kgm²
ω₁ = 30 rpm
l₂ = 7.6 + 2 × 5(0.2)² = 8.0 kgm²
ω₂ = ?
According to the principle of conservation of angular momentum
l₂ω₂ = l₁ω₁
ω₂ = \(\frac{l_1}{l_2} \omega_1=\frac{15.7 \times 30}{8.0}\) = 58.88 rpm

Question 24.
A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.
(Hint : The moment of inertia of the door about the vertical axis at one end is ML²/3.)
Solution:
Angular momentum imparted by the bullet
I_ = mv × r = (10 × 10^-3) × 500 × \(\frac{1}{2}\) = 2.5
Also, L = \(\frac{M L^2}{3}=\frac{12 \times 1.0^2}{3}\) = 4kgm
As L = Lω
∴ ω = \(\frac{\mathrm{L}}{1}=\frac{2.5}{4}\) = 0.625 rad/sec.

Question 25.
Two discs of moments of inertia I, and I2 about their respective axes (normal to the disc and passing through the centre), and rotawing with angular speeds ω₁, and ω₂ are brought into contact face to face with their axes of rotation coincident,
(a) .What is the angular speed of the two-disc system ?
(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy ? Take ω₁ ≠ ω₁.
Solution:
Here, total intial angular momentum of the two discs L₁ = I₁ω₁ + I₂ω₂
Under, the given conditions, moment of intertia of the two disc system = (I₁ + I₂)
If ω is angular speed of the combined system, the final angular momentum of the system
L₂ = (I₁ + I₂)ω
As no external torque is involved in this excercise, therefore, L₂ = L₁
(I₁ + I₂)ω = I₁ω₁ + I₂ω₂
ω = \(\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}\)

b) Initial K.E of two disc E₁ = \(\frac{1}{2}\) I₁ω₁² + \(\frac{1}{2}\) I₂ω₂²
Final K.E of the system E₂ = \(\frac{1}{2}\) (I₁ + I₂)ω²

∴ E₁ – E₂ > 0 or E₁ > E₂ or E₂ < E₁
Hence there occurs a loss of K.E in the process. Loss of energy = E₁ – E₂. This loss must be due to friction in the contact of the two discs.

Question 26.
a) Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point (x, y) in the x – y plane from an axis through the origin and perpendicular to the plane is x² + y²).
Answer:
Statement: The sum of moments of inertia of a plane lamina about any two perpendicular axes in its plane is equal to its moment of inertia about an axis perpendicular to the plane and passing through the point of intersection of the first two axes.

Proof : Consider a plane lamina revolving about the Z axis. Let ‘O’ be the origin of the axis. Imagine a particle of mass ‘m’ lying at a distance ‘r’ from point ‘o’ on the plane. Let x, y be the coordinates of the point P.
Thus r² = x² + y²
Then the moment of the body about x-aixs
I_x = Σm y²
The moment of inertia of the body about y-axis.
I_y = Σm x²
Then the moment of inertia of the body about Z-axis

I_z = Σm r²
I_z = Σm(x² + y²)
I_z = Σm x² + Σm y² = I_y + I_x
∴ I_z = I_x + I_y
Hence perpendicular axes theorem is proved.

b) Prove the theorem of parallel axes.
(Hint: If the centre of mass is chosen to be the origin Σm_ir_i = 0).
Answer:
Statement: The moment of inertia of a plane lamina about an-axis is equal to the sum of the moment of inertia about a parallel axis passing through the centre of mass and the product of its mass and square of the distance between the two axes i.e., I₀ = I_G + Mr²
Let I_G is the moment of inertia of the plane lamina about the axis Z₂ passing through the centre of mass.
I₀ is the moment of inertia of the plane lamina about an axis Z₁
Let M be the mass of the lamina and r be the distance between the two axes . Then
I₀ = I_G + mr².
Proof : Let a particle of mass m is situated at P. Moment of inertia about the axis passing through 0 is
dl = m op² or I = Σm op².
Join the lines PO and PG and draw the line PQ and Join with the line extending from OG.
From the trainagle POQ, OP² = OQ² + PQ²
OP² – (OG + GQ)² + PQ² + OQ = OG + GQ
OP² = OG² + 2OG . GQ + (GQ² + PQ²)
OP² – OG² + 2OG. GQ + GP²
OP² = OG² + GP² + 2OG.GQ
[∵ From the ∆le PGQ, GP² = PQ² + GQ²]

Multiplying with Σm on both side
Σm OP² = Σm OG² + Σm GP² + Σm OG. GQ
But Σm OG² = Mr²
(∵ OG is constant and Σm = M, total mass of the body)
Σm GP² = I_G
Σm OP² = I₀
∴ I₀ = Mr² – I_G + 2r ΣmGQ
Σm.GQ = 0
[∵ The moment of all the particles about the centre of mass is always zero]
I₀ = I_G + Mr²
Thus the theorem is proved.

Question 27.
Prove the result that the velocity u of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by υ² = \(\frac{2 g h}{\left(1+k^2 / R^2\right.}\) using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
Solution:
When a body rolls down an incline of height h, we apply the principle of conservation of energy.

K.E of translation + K.E of rotation = P.E at the top
i.e . \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω² = mgh
\(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (mk²) ω² = mgh
As w = \(\frac{\mathrm{V}}{\mathrm{R}}\)
∴ \(\frac{1}{2}\)mv² + \(\frac{1}{2}\)m \(\frac{k^2}{R^2}\)v² = mgh
or mv² (1 + \(\frac{k^2}{R^2}\)) = mgh
v² = \(\frac{2 g h}{\left(1+\frac{K^2}{R^2}\right)}\)

Question 28.
A disc rotating about its axis with angular speed ω₀ is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. ? Will the disc roll in the direction indicated ?

Solution:
Using the reaction v = rω,
we get
for point A, V_A = Rω₀, along AX
for point B, V_B = Rω₀, along BX
for point C, V_c = (\(\frac{\mathrm{R}}{\mathrm{2}}\) ω₀ parallel to AX,
The disc will not rotate, because it is placed on a perfectly frictionless table, without, rolling is not possible.

Question 29.
Explain why friction is necessary to make the disc in Figure roll in the direction indicated.

(a) Give the direction of frictional force at B and the sense of frictional torque before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins ?
Solution:
To roll a disc, we require a torque, which can be proved only by a tangential force. As force of friction is the only tangential force in this case, it is necessary.
(a) As frictional force at B opposes the velocity of point B, which is to the left, the frictional force must be to the right. The sense of frictional torque will be perpendicular to the plane of the disc and outwards.

(b) As frictional force at B decreases the velocity of the point of contact B with the surface, the perfect rolling begins only when velocity of point B becomes zero. Also, force of friction would become zero at this stage.

Question 30.
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s^-1. Which of the two will start to roll earlier ? The co-efficient of kinetic friction is μ_k = 0.2.
Solution:
Here, intial velocity of centre of mass is zero i.e, u = 0.
Frictional force causes the GM to accelerate
μ_k mg = ma ∴ a = μ_k g
As v = u + at ∴ v = 0 + μ_k gt
Torque due to friction causes retardation in the intial angular speed ω₀.
i.e. μ_k mg × R = – Iα
α = \(\frac{\mu_k \mathrm{mgR}}{\mathrm{I}}\)
ω = ω₀ + αt
∴ ω = ω₀ – \(\frac{\mu_k \mathrm{mgRt}}{\mathrm{I}}\)
Rolling begins, when v = Rω from (ii) and (iv)

Comparing (vi) and (vii) we find that the disc would begin to roll earlier than the ring We can calculate the values of t from (vi) and (vii) using known values of μ_k, g, R and ω₀.

Question 31.
A cylinder of mass 10kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction μ_s = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?
Solution:
Here, m = 10 kg, r = 15 cm = 0.15 m
θ = 30°, μ_s = 0.25
Accelaration of the cylinder down the incline,
α = \(\frac{2}{3}\)g sin θ = \(\frac{2}{3}\) × 9.8 sin 30° = \(\frac{9.8}{3}\) m/s²
a) Force of friction, f = mg sin θ – ma = m(g sin θ – α) = 10(9.8 sin 30° – \(\frac{9.8}{3}\)) = 16.4 N
b) During rolling the point of contact is at rest. Therefore work done against friction is zero
c) For rolling without slipping/skidding μ = \(\frac{1}{3}\) tan θ
tan θ = 3μ
= 3 × 0.25 = 0.75
θ = 37°.

Question 32.
Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly italic inclined plane will undergo slipping (not roiling) motion.
Solution:
a) The statement is false.

b) True. This is because rolling body can be imagined to be rotating about an axis passing through the point of contact of the body with the ground.
Hence its instantaneous speed is zero.

c) This is not true. This is because when the body is rotating its instantaneous acc is not zero.

d) It is true. This is because once the perfect rolling begins, force of friction becomes zero. Hence work done against friction is zero.

e) The statement is true. This is because rolling occurs only on account of friction which is tangential force capable of providing torque. When the inclined plane is perfectly smooth it will simply slip under the effect of its own weights.

Question 33.
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass :
(a) Show P = \(\mathrm{P}_{\mathrm{i}}^{-}+\mathrm{m}_{\mathrm{i}} \mathrm{V}\)
Where P_i is the momentum of the i_i the particle (of mass m_i) and p’_i + m_iv’_i. Note relative velocity of the ith particle relative to the centre of mass. Also, prove using the definition of the centre of mass Σp’_i = 0
(b) Show K = K’ + 1/2MV²
Where K is the total kinetic energy of the system of particles. K’ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV²/2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.
(c) Show L = L’ + R × MV
Where L’ = Σr_i P_i is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember r’_i = r_i – R rest of the notation is the standard notation used in the chapter. Note L1 and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
(d) Show \(\frac{d l^{\prime}}{d t}=\sum r_i^{\prime} \frac{d p^{\prime}}{d t}\)
Further, show that \(\frac{\mathrm{dL}^{\mathrm{l}}}{\mathrm{dt}}\) = τ’_ext
Whereτ’_ext is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)
Solution:
a) Let m₁, m₂, … m_i mass points have the position vectors \(\overrightarrow{r_1}, \overrightarrow{r_2}, \overrightarrow{r_i}\) w.r.t origin ‘O’. The position vector of C.M. say \(\overrightarrow{\mathrm{OP}}\)
i.e., \(\overrightarrow{\mathrm{OP}}=\frac{m_1 \overrightarrow{r_1}+m_2 \overrightarrow{r_2}+\ldots}{m_1+m_2+\ldots}=\frac{m_i r_i}{M}\)
where i = 1, 2, 3 .
Now, let us change the origin to O’ and assume that the C.M is now at p’ with

Multiplying eqn. (1) by m_i and differentiating w.r.t to time, we get

even if we change the origin, the position of
centre of mass will not change i.e., Σp’_i = 0

b) In rotational kinematics, K.E_T of the system of particles = K.E_T of the system when the particle velocities are taken w.r.t to CM + K.E of the translation of the system as a whole (i.e. of the C.M motion of the system)
i.e. \(\frac{1}{2}\) m₁v₁² + \(\frac{1}{2}\) m₂v₂² + …..
= [\(\frac{1}{2}\) m₁v₁² + \(\frac{1}{2}\)m₂v₂ +……] + \(\frac{1}{2}\)Mv²
\(\frac{1}{2}\) m_iv_i² = \(\frac{1}{2}\)m_iv_i² + \(\frac{1}{2}\) mv²
∴ k = k’ + \(\frac{1}{2}\)
where M = total mass of particles
and y = velocity of C.M motion of the system.

c) From eq. (1), we have

Textual Examples

Question 1.
Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150 g, and 200 g respectively. Each side of the equilateral triangle is 0.5 long. [A.P. & T.S. Mar.18]
Solution:

With the x – and y-axes chosen as shown in Fig. 7.9, the coordinates of points O, A and B forming the equilateral triangle are
respectively (0, 0), (0, 5, 0), (0.25. \(\sqrt{3}\) ). Let the masses 100 g, 150g and 200g be located at O, A and B be respectively. Then.

Question 2.
Find the centre of mass of a triangular lamina.
Solution:
The lamina (∆LMN) may be subdivided into narrow strips each parallel to the base (MN) as shown in Fig.

By symmetry each strip has its centre of mass at its midpoint. If we joint the midpoint of all the strips we get the median LP. The centre of mass of the triangle as a whole therefore, has to lie on the median LP. Similarly, we can argue that it lies on the median MQ and NR. This means the centre of mass lies on the point of concurrence of the medians, i.e. on the centroid G of the triangle.

Question 3.
The density of the atmosphere at sea level is 1.29 kg/m³. Assume that it does not change with altitude. Then how high would the atmosphere extend ?
Solution:
The centres of mass C₁, C₂ and C₃ of the squares are, by symmetry, their geometric centres and have coordinates (1/2,1/2), (3/ 2,1/2), (1/2,3/2) respectively.

Question 4.
A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet reston the floor 1 m from the wall as shown in Fig. Find the reaction forcesof the wall and the floor.
Solution:

Let AB is 3 m long, A is at distance AC = 1 m from the wall. Pythagoras theorem,
BC = 2\(\sqrt{2}\) m.
For translational equilibrium, taking the forces in the vertical direction, N – W = 0 ……….. (i)
Taking the forces in the horizontal direction,
F – F₁ = 0 ……………. (ii)
taking the moments of the forces about A,
2\(\sqrt{2}\) F1 – (1/2) W = 0 ………….. (iii)
Now W = 20 g = 20 × 9.8 N = 1960.0 N
From (i) N = 196.0
From (iii) = F₁ = W/4 \(\sqrt{2}\) = 196.0/4 \(\sqrt{2}\)
= 34.6 N
From (ii) F = F₁ = 34.6 N
F₂ = \(\sqrt{F^2+N^2}\) = 199.0 N

Question 5.
Find the scalar and vector products of two vectors
a = \((3 \vec{i}-4 \vec{j}+5 \vec{k})\) and b = \((-2 \vec{i}+\vec{j}-3 \vec{k})\)
Solution:
a.b = (\((3 \vec{i}-4 \vec{j}+5 \vec{k})\)) . (\((-2 \vec{i}+\vec{j}-3 \vec{k})\))
= -6 – 4 – 15 = -25
a × b = \(\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
3 & -4 & 5 \\
-2 & 1 & -3
\end{array}\right|=7 \vec{i}-\vec{j}-5 \vec{k}\)
Note b × a = \(7 \vec{i}-\vec{j}-5 \vec{k}\)

Question 6.
Obtain Equation from first principles.
Solution:
The angular acceleration is uniform,
ω = αt + c (as α is comstant)
At t = 0, ω = ω₀ (given)
From (i) we get at t = 0, ω = c = ω₀
Thus, ω = αt + ω₀ as required

Question 7.
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm is 16 seconds, (i) What is its angular acceleration, assuming the acceleration to be uniform ? (ii) How many revolutions does the engine make during this time ?
Solution:
(i) We shall use ω = ω₀ + αt
ω₀ = initial angular speed in rad/s
= 2π × angular speed in rev/s
= \(\frac{2 \pi \times \text { angular speed in rev } / \mathrm{min}}{60 \mathrm{~s} / \mathrm{min}}\)
= \(\frac{2 \pi \times 1200}{60} \mathrm{rad} / \mathrm{s}\)
= 40π rad/ s
Similarly ω = final angular speed in rad /s
= \(\frac{2 \pi \times 3120}{60} \mathrm{rad} / \mathrm{s}\)
= 2π × 52 rad/s = 104 π rad/s
∴ Angular acceleration
α = \(\frac{\omega-\omega_0}{t}\) = 4π rad/s²
The angular acceleration of the engine = 4π rad/s²

(ii) The angular displacement in time t is given by
θ = ω₀t + \(\frac{1}{2}\) αt²
= (40π × 16 + \(\frac{1}{2}\) × 4π × 16²) rad
= (640 π+ 512 π )rad
= 1150π rad
Number of revolutions = \(\frac{1152 \pi}{2 \pi}\) = 576

Question 8.
Find the torque of a force \((\bar{i}-\bar{j}+\bar{k})\) about the origin. The force acts on a particle whose position vector is \((\bar{i}-\bar{j}+\bar{k})\) (Mar.’14, 13)
Answer:
Here r = \(\vec{i}-\vec{j}+\vec{k}\)
and F = \(7 \vec{i}-3 \vec{j}+5 \vec{k}\)
We shall use the determinant rule to find the torque τ = r × F
τ = \(\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & -1 & 1 \\
7 & 3 & -5
\end{array}\right|=(5-3) \vec{i}-(-5-7) \vec{j}\) + \((3-(-7)) \vec{k}\)
or τ = \(2 \vec{i}+12 \vec{j}+10 \vec{k}\)

Question 9.
Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.
Solution:
Let the particle with velocity v be at point P at some instant t. We want to calculate the angular momentum of the particle about an arbitrary point O.

The angular momentum is 1 = r × mv. Its magnitude is mvr. sine, where θ is the angle between r and v as shown in Fig. Although the particle changes position with time, the line of direction of v remains the same and hence OM = r sin θ ; a constant.

Question 10.
Show that moment of a couple does not depend on the point about which you take the moments.
Solution:

The moment of the couple = sum of the moments of the two forces making the couple
= r₁ × (-F) + r₂ × F
= r₂ × F + r₁ × F
= (r₂ – r₁) × F
But r₂ + AB = r₂, and hence AB = r₂ – r₁.
The moment of the couple, therefore, is AB × F.

Question 11.
What is the moment of intertia of a disc about one of its diameters ?
Solution:

x and y-axes lie in the plane of the disc and z is perpendicular I_z = I_x = I_y
Now, x and y axes are along two diameters of the disc, and by symmetry the moment of inertia of the disc is the same about any diameter. Hence
I_x = I_y
and I_z = 2I_y
But I_x = I_y
So finally, I_x = I_z/2 = MR²/4
Thus the moment of inertia of a disc about any of its diameter is MR²/4.
Find similarly the moment of inertia of a ring about any of its diameter. Will the theorem be applicable to a solid cylinder?

Question 12.
What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end ?
Solution:
For the rod of mass M and length l, I = Ml²/ 12. Using the parallel axes theorem, I’ = I + Ma² with a = l/2 we get.
I’ = \(\mathrm{M} \frac{l^2}{12}+\mathrm{M}\left(\frac{l}{2}\right)^2=\frac{\mathrm{M} l^2}{3}\)
We can check this independently since I is half the moment of inertia of a rod of mass 2M and length 21 about its midpoint,
I’ = \(2 \mathrm{M} \frac{4 l^2}{12} \times \frac{1}{2}=\frac{\mathrm{M} l^2}{3}\)

Question 13.
What is the moment of inertia of a ring about a tangent to the circle of the ring ?
Solution:
The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring. The distance between these two parallel axes is R, the radius of the ring. Using the parallel axes theorem.

= 0.4kg m²
α = angular acceleration
= 5.0 Nm/0.4 kg m² = 12.5 S^-2

b)Work done by the pull unwinding 2m of the cord
= 25 N × 2 m = 50 J

(c) Let ω be the final angular velocity. The kinetic energy gained = \(\frac{1}{2}\) Iω²
since the wheel starts from rest. Now,
ω₂ = ω²₀ +2αθ, ω₀ = 0
The angular displacement θ = length of unwound string / radius of wheel = 2m / 0.2 m = 10 rad
ω² = 2 × \(\frac{1}{2}\) × 12.5 × 10.0 = 250 (rad/s)²
∴ K.E. gained = \(\frac{1}{2}\) × 0.4 × 250 = 50 J

(d) The answers are the same, i.e. the kinetic energy gained by the wheel work done by the force. There is no loss of energy due to friction.

Question 14.
A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. The flywheel is mounted on a horizontal axle -with frictionless bearings.
(a) Compute the angular acceleration of the wheel.
(b) Find the work done by the pull, when 2m of the cord is unwound
(c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest.
(d) Compare answers to parts (b) and (c).
Solution:

a) We use Iα = τ
the toque τ = FR
= 25 × 0.20 Nm (as R = 0.20m)
= 5.0 Nm

Question 15.
Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity ?
Solution:
We assume conservation of energy of the rolling body, i.e. there is no loss of energy due to friction etc. The protential energy lost by the body in rolling down the inclined plane (= mgh) must, therefore, be equal to kinetic energy gained. (See fig.) Since the bodies start from rest the kinetic energy gained is equal tot he final kinetic energy of the odies. From K = \(\frac{1}{2} m v_{\mathrm{cm}}^2\left[1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right]\)
K = \(\frac{1}{2} m v^2 1+\left(\frac{K^2}{R^2}\right)\) Where v is the final velocity of (the centre of mass of) the body.
Equating K and mgh

Note v² is independent of the mass of the rolling body:
For a ring, k² = R²
v_ring = \(\sqrt{\frac{2 g h}{1+1}}=\sqrt{g h}\)
For a solid cylinder K² = R²/0.2
v_disc = \(\sqrt{\frac{2 g h}{1+1 / 2}}=\sqrt{\frac{4 g h}{3}}\)
For a solid sphere K² = 2R²/5
v_sphere = \(\sqrt{\frac{2 g h}{1+1 / 52}}=\sqrt{\frac{10 g h}{7}}\)

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 6th Lesson Work, Energy and Power Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 6th Lesson Work, Energy and Power

Very Short Answer Questions

Question 1.
State the conditions under which a force does no work. [T.S. Mar. 15]
Answer:

1. When the displacement is zero,
2. When the displacement is perpendicular to the direction of the force,
3. When the body moves under the action of a conservative force over a closed path.

Question 2.
Define : Work, Power and Energy. State their SI units.
Answer:
Work : The product of magnitude of displacement (S) and component of force (F cos θ) along the direction of displacement is called work, ie., W = \(\vec{F} \cdot \vec{S}\) = F S cos θ.
Unit: Joule

Power : The rate of doing work by a force is called power, i.e., P = \(\frac{\mathrm{W}}{\mathrm{t}}\)
Unit: watt or J/S

Energy : The ability or the capacity to do work is called energy.
Unit: Joule

Question 3.
State the relation between the kinetic energy and momentum of a body.
Answer:
Kinetic energy (E_K) = \(\frac{\mathrm{P}^2}{2 \mathrm{~m}}\); where P = momentum of a body, m = mass of the body

Question 4.
State the sign of work done by a force in the following.
a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
b) Work done gravitational force in the above case.
Answer:
a) To lift a bucket out of a well, force equal to the weight of the bucket has to be applied by the man along the vertical in upward direction. Since the displacement is also in upward direction θ = 0 and hence the workdone is positive.

b) The angle between the gravitational force and the displacement is 180°. Therefore workdone by the gravitational force is negative.

Question 5.
State the sign of work done by a force in the following.
a) Work done by friction on a body sliding down an inclined plane.
b) Work done by gravitational force in the above case.
Answer:
a) Friction always acts in a direction, opposite to the direction of motion. Hence work done is negative.
b) The work done is positive.

Question 6.
State the sign of work done by a force in the following.
a) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
b) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer:
a) The applied force and the displacement are in same direction. Work is positive.
b) The direction of resistive force is opposite to the direction of motion of the pendulum. Hence work done is negative.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer.
a) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
b) The work done by earth’s gravitational force in keeping the moon in its orbit for its one revolution is zero.*
Answer:

1. False
2. True. Because gravitational force is conservative force.

Question 8.
Which physical quantity remains constant

1. in an elastic collision
2. in am. inelastic collision ?

Answer:

1. In elastic collision : Both momentum and kinetic energy is constant.
2. In an inelastic collision : Momentum remains constant.

Question 9.
A body freely falling from a certain height h, after striking a smooth floor rebounds and h rises to a height W2. What is the coefficient of restitution between the floor and the body? [T.S. – Mar.’18]
Answer:
h₁ = h, h₂ = \(\frac{\mathrm{h}}{2}\)
e = \(\sqrt{\frac{\mathrm{h}_2}{\mathrm{~h}_1}}=\sqrt{\frac{\frac{\mathrm{h}}{2}}{\mathrm{~h}}}=\frac{1}{\sqrt{2}}\)

Question 10.
What is the total displacement of a freely falling body, after successive rebounds from the same place of ground, before it comes to stop ? Assume that ‘e’ is the coefficient of restitution between the body and the ground.
Answer:
Total distance = \(\frac{h\left(1+e^2\right)}{\left(1-e^2\right)}\), where h is the height through it has fallen and e is the coefficient of restitution.

Short Answer Questions

Question 1.
What is potential energy ? Derive an expression for the gravitational potential energy.
Answer:
Potential energy: Definition : The energy possessed by a body by virtue of its position or state is called potential energy.
e.g. : 1) Energy possessed by water stored in a dam.
2) A stretched rubber card.
Formula for P.E = mgh : The potential energy is measured by the work done in lifting a body through a height ‘h’ against gravitational force. Consider a body of mass m on the body against gravitational force.

∴ Gravitational force F = Weight of the body
Minimum force needed to lift the body = F in upwards direction
F = mg
Height lifted = h
Work done W = gravitational force × height lifted
W = mgh
This work done is stored as potential energy.
∴ Potential energy P.E = mgh.

Question 2.
A lorry and a car moving with the same momentum are brought to rest by the application of brakes, which provide equal retarding forces. Which of them will come to rest in shorter time ? Which will come to rest in less distance ?
Answer:
F₁ = F₂ and P₁ = P₂; W = \(\frac{1}{2}\) mv² – \(\frac{1}{2}\) mu²; \(\frac{1}{2}\) mu² = \(\frac{\mathrm{P}^2}{2 \mathrm{~m}}\)
∴ W = F.S = \(\frac{\mathrm{P}^2}{2 \mathrm{~m}}\)
∴ S ∝ \(\frac{1}{m}\)
The heavier body (lorry) comes to rest in shorter time. The lorry will come to rest in less distance.

Question 3.
Distinguish between conservative and non-conservative forces with one example each.
Answer:
Forces can be classified into two types.

1. Conservative force,
2. Non conservative force.

1) Conservative force : Conservative forces are the forces under the action of which a body returns to its starting point with same K.F. with which it is projected work done by conservative force is path independent.

Explanation : If work done along path I, path II, path III are w₁, w₂, w₃ respectively and w₁ = w₂ = w₃ then the force involved in the work is conservative force.
Workdone by a conservative force along a closed path is zero.
Ex. : Gravitational force, electrostatic force, spring force etc.

2) Non-conservative force : Non-conservative force are the forces under the action of which the work done depends upon the path followed and the K.E changes work done by non¬conservative force is path dependent.

Explanation : If workdone along path I, path II, path III are w₁, w₂, w₃ respectively and w₁ ≠ w₂ ≠ w₃, then the force involved in the work is non conservative. Workdone by a non conservative force along a closed path is not zero.
Ex. : Frictional force, viscous force etc.

Question 4.
Show that in the case of one dimensional elastic collision, the relative velocity of approach of two colliding bodies before collision is equal to the relative velocity of – separation after collision. [T.S. Mar. 18]
Answer:
Consider two spheres (bodies) which have smooth, non-rotating of masses m₁ and m₂ (m₁ > m₂) are moving along the straight line joining the centres of mass with initial velocities u₁ and u₂ (u₁ > u₂). They undergo head on collision and move along the same line after collision with final velocities of v, and v2. These two bodies exert forces on each other during collision.
Let the collision be elastic, then both momentum and K.E are conserved.
According to law of conservation of linear momentum, Momentum of the system before collision = Momentum of the system after collision.
m₁u₁ + m₂u₂ = m₁v₁ + m₁v₂
m₁(u₁ – v₁) = m₂ (v₂ – u₂) ……………….. (1)
According to law of conservation of kinetic energy, K.E of the system before collision = K.E of the system after collision.
\(\frac{1}{2}\)m₁u₁² + \(\frac{1}{2}\)m₂u₂² = \(\frac{1}{2}\) m₁v₁² + \(\frac{1}{2}\)m₂v₂²
m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
m₁(u₁² – v₁²) = m₂(v₂² – u₂²) ……………….. (2)
\(\frac{\text { (2) }}{(1)} \Rightarrow \frac{m_1\left(u_1^2-v_1^2\right)}{m_1\left(u_1-v_1\right)}=\frac{m_2\left(v_2^2-u_2^2\right)}{m_2\left(v_2-u_2\right)}\)
\(\frac{\left(u_1-v_1\right)\left(u_1+v_1\right)}{\left(u_1-v_1\right)}=\frac{\left(v_2-u_2\right)\left(v_2+u_2\right)}{\left(v_2-u_2\right)}\)
(u₁ – v₁) = u₂ + v₂
u₁ – u₂ = v₂ – v₁
Relative velocity of approach from the above equation before collision = Relative velocity of separation after collision.

Question 5.
Show that two equal masses undergo oblique elastic collision will move at right angles after collision, if the second body initially at rest.
Answer:
Oblique elastic collision : If the centres of mass of the colliding bodies are not initially moving along the line of impact, then the impact is called oblique collision.

Two equal masses undergo oblique elastic collision will move at right angles after collision, if the second body initially at rest:

Consider two smooth and perfectly elastic spheres of masses m₁ and m₂. Let u₁ and u₂ be their initial velocities before collission. Let v₁ and v₂ be their final velocities after collision, (α, β) and (θ, Φ) are the angles, the directions of motion make with the line of impact before collision and after collision.

Since the spheres are smooth, there is no (impulse) change of velocities perpendicular to the line of impact. Hence the resolved parts of velocity of the two spheres in the direction perpendicular to the line of impact remain unchanged.
∴ v₁ sinθ = u₁ sinα ……………………. (1)
and v₂ sinΦ = u₂ sin β …………………. (2)
According to the principle of conservation of momentum, the sum of the moment of the two spheres along the line of impact before collision and after collision are equal.
∴ m₁u₁ cosα + m₂u₂ cosβ = m₁v₁ cosθ + m₂v₂ cosΦ ……………… (3)
⇒ m₁(u₁ cosα – v, cosθ) = m₂ (v₂ cosΦ – u2 cosβ) ……………….. (4)
According to the principle of conservation of kinetic energy, the sum of the K.E along the line of impart before and after are equal.
⇒ \(\frac{1}{2}\) m₁ (u₁cosα)² + \(\frac{1}{2}\)m₂ (u₂cosβ)² = \(\frac{1}{2}\) m₁ (v₁cosθ)² + \(\frac{1}{2}\)m₂(v₂cos126,Φ)²
⇒ m₁ [(u₁cosα)² – (v₁cosθ)² ] = m₂ [(v₂cosΦ)² – (u₂cosβ)²] …………….. (5)
\(\frac{(5)}{(4)}\) ⇒ u₁ cosα + v₁ cosθ = v₂ cosΦ + u₂ cosβ
v₁cosθ = v₂ cosΦ + u₂ cosβ – u₁ cosα ……………… (6)
and v₂ cosΦ = v₁ cosθ + u₁ cosα – u₂ cosβ ……………….. (7)
sub. equation (7) in equation (3), we get
v₁ cosθ = \(\left(\frac{m_1-m_2}{m_1+m_2}\right)\) u₁ cosα + \(\left(\frac{2 m_u u_2}{m_1+m_2}\right)\) cosβ ……………… (8)
Sub-equation (6) in equation (3), we get
and v₂ cosΦ = \(\left(\frac{m_2-m_1}{m_1+m_2}\right)\) u₂ cosβ + \(\left(\frac{2 m_1 u_1}{m_1+m_2}\right)\) cosα ………………. (9)
If u₂ = 0 and m₁ = m₂ then equation (2), we get Φ = 0 and from equation (8), θ = 90°. This means that if a sphere of mass ‘m’ collides obliquely on another perfectly elastic sphere of the same mass at rest, the directions of motions of the spheres after impact will be at right angles.

Question 6.
Derive an expression for the height attained by a freely falling body after ‘n’ number of rebounds from the floor.
Answer:
Consider a small sphere fall freely at a height ‘h’ onto the floor. It strikes the floor with a velocity of ‘u’
So that u₁ = \(\sqrt{2 g h}\) …………….. (1)
At the time of collision between the sphere and the floor, the initial and final velocities of floor are zero i.e., u₂ = 0 and v₂ = 0

Let ‘v₁‘ be the final velocity of the sphere after the first collision.
∴ e = \(\frac{v_2-v_1}{u_1-u_2}=\frac{0-v_1}{\sqrt{2 g h}-0}=\frac{-v_1}{\sqrt{2 g h}}\)
∴ v₁ = -e\(\sqrt{2 g h}\) ……………… (2)
(-) indicates that the sphere rebounds
∴ The height (h₁) attained by the sphere after first rebound is
h₁ = \(\frac{v_1^2}{2 g}=\frac{(e \sqrt{2 g h})^2}{2 g}\) = e²h
h₁ = (e²)¹ h ………………… (3)
Similarly we can that velocity attained by the sphere after second rebound.
v₂ = -e² \(\sqrt{2 g h}\) …………………. (4)
and maximum height attained by the sphere after second rebound
h₂ = (e²)² h ………………… (5)
From the (2) and (4) equations
velocity of the sphere rebounds after ‘n’ collisions
v_n = eⁿ \(\sqrt{2 g h}\)
From the (3) and (5) equations, maximum height attained by the sphere after n rebounds h_n = (e²)ⁿ h

Question 7.
Explain the law of conservation of energy.
Answer:
The total mechanical energy of the system is conserved if the forces doing work on it are conservative. If some of the forces involved are non-conservative, part of the mechanical energy may get transformed into other forms such as heat, light and sound. However, the total energy of an isolated system does not change, as long as one accounts for all forms of energy. Energy may be transformed from one form to another but the total energy of an isolated system remains constant. Energy can neither be created, nor destroyed.

Since the universe as a whole may be viewed as an isolated system, the total energy of the universe is constant. If one part of the universe looses energy, another part must gain an equal amount of energy.

Long Answer Questions

Question 1.
Develop the notions of work and kinetic energy and show that it leads to work energy theorem. [A.P. Mar. 17; T.S. Mar. 15, 14]
Answer:
Statement : The change in kinetic energy of a particle is equal to the workdone on it by the net force, i.e., k_f – k_i = W
Proof : Consider a particle of mass ‘m’ is moving with initial speed ‘u’ to final speed ‘v’. Let ‘a’ be its constant acceleration and S be its distance traversed. The kinematic relation is given by
v² – u² = 2as ………………. (1)
Multiplying both sides by \(\frac{\mathrm{m}}{2}\), we have 11
\(\frac{1}{2}\) mv² – \(\frac{1}{2}\) mu² mas = FS …………….. (2)
Where the last step follows from Newton’s second law.
We can generalise Equation (1) to three dimensions by employing vectors
v² – u² = 2 \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}\)
Once again multiplying both sides by \(\frac{\mathrm{m}}{2}\), we obtain
\(\frac{1}{2}\) mv² – \(\frac{1}{2}\) mu² = m \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}\) – \(\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{d}}\) …………………. (3)
The above equation provides a motivation for the definitions of work and kinetec energy.
In equation (3), \(\frac{1}{2}\) mv² – \(\frac{1}{2}\) mu² = k_f – k_i
Where k_i, k_f are initial and final kinetic energies and \(\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{d}}\) = W, where W refers workdone by a force on the body over a certain displacement.
k_f – k_i = W ……………….. (4)
Equation (4) is also a special case of the work-energy (WE) theorem.

Question 2.
What are collisions ? Explain the possible types of collisions ? Develop the theory of one dimensional elastic collision. [T.S. Mar. 18]
Answer:
Collisions : A strong interactions between bodies that occurs for a very short interval during which redistribution of momenta occur ignoring the effect of other forces are called collisions.
Collisions are of two types :
i) Elastic collision : The collision in which both momentum and kinetic energy is constant is called elastic collision.
ii) Inelastic collision: The collision in which momentum remains constant but not kinetic energy is called Inelastic collision.
One dimensional elastic collision : Consider two spheres (bodies) which have smooth, non-rotating of masses m₁ and m₂ (m₁ > m₂) are moving along the straight line joining their centres of mass with initial velocities u₁and u₂ (u₁ > u₂). They undergo head on collision and move along the same line after collision with final velocities v₁ and v₂. These two bodies exert forces on each other during collision.
Hence both momentum and kinetic energy are conserved. According to law of conservation of linear momentum.

From the above equation, in one dimensional elastic collision, the relative velocity of approach before collision is equal to the relative velocity of separation after collision.
From equation (3) u₁ – u₂ = v₂ – v₁, v₂ = u₁ – u₂ + v₁ ………………… (4)
Sub equation (4) in equation (1) we get
m₁ (u₁ – v₁) = m₂ (u₁ – u₂ + v₁ – u₂), m₁ u₁ – m₁v₁ = m₂ u₁ – m₂ u₂ + m₂ v₁ – m₂ u₂
m₁ u₁ – m₂ u₁ + 2 m₂ u₂ = m₁ v₁ + m₂ v₁ (m₁ – m₂) u₁ + 2 m₂ u₂ = (m₁ + m₂) u₁
v_i = \(\left(\frac{m_1-m_2}{m_1+m_2}\right) u_1+\left(\frac{2 m_2}{m_1+m_2}\right) u_2\) ………………… (5)
From the equation (4), v₁ = v₂ – u₁ + u₂
Sub. this value in equation (1) we get
m₁ (u₁ – v₂ + u₁ – u₂) = m₂ (v₂ – u₂), m₁ u₁ – m₁ v₂ + m₁ u₁ – m₁ u₂ = m₂ v₂ – m₂ u₂
2m₁ u₁ + u₂ (m₂ – m₁) = (m₂ + m₂)v₂
v₂ = \(\frac{2 m_1 u_1}{\left(m_1+m_2\right)}+\frac{\left(m_2-m_1\right)}{\left(m_1+m_2\right)}\) u₂

Question 3.
State and prove law of conservation of energy in case of a freely falling body. [AP – Mar. ’18, ’16, ’15, ’13; TS – Mar. ’17, ’16]
Answer:
Statement: “Energy can neither be created nor destroyed. But it can be converted from one form to another form. The total energy of a closed system always remains constant”. This law is called law of conservation of energy.

Proof : In case of freely falling body : A body of mass’m’ is dropped or freely falling at a height ‘H‘ from the ground. The total mechanical energy of the body E = K + U.

Where k = kinetic energy; U = potential energy.
Suppose A, B and C are the points at heights H, h and ground respectively.
At ‘A’ : At the highest point, velocity u = 0
Kinetic energy K.E. = \(\frac{1}{2}\) mu² = 0
Potential energy P.E. = mgH
Total mechanical energy of the body = E_A = P.E. + K.E. = mgH + 0
∴ E_A = mgH ………………. (1)

At ‘B’: When the body falls from A to B, which is at a height ‘h’ from the ground. The velocity of the body at ‘B’ is V_B.
At ‘B’ the body possesses both P.E and K.E
S = H – h
u = 0
a = g
v = v_B = ?
v_B² – u² = 2as
a = g
v = v_B = ?
v_A² – 0 = 2g (H – h)
V_B = 2g(H – h)
Kinetic energy of the body (K) = \(\frac{1}{2}\) mv_B² = \(\frac{1}{2}\) m 2g(H – h)
K = mg (H – h)
Potential energy (u) = mgh
∴ Total mechanical energy at B = k + u
E_B = mg (H – h) + mgH
E_B = mgh ……………… (2)

At ‘C’ : When the body reaches the point ‘c’ on the ground. The velocity of the body at ‘c is ‘v_c’ can be found by using v² – u² = 2as equation.
S = H
u = 0
a = g
v = v_c = ?
v_c² – 0 = 2gH
v_c² = 2gH
v_c = \(\sqrt{2 \mathrm{gH}}\)
Kinetic energy of the body (k) = \(\frac{1}{2}\) mv_c²
= \(\frac{1}{2}\) m 2gH = mgH
Potential energy (u) = 0
Total energy at c = k + u
∴ E_C = mgH ………………. (3)
From the above three equations (1), (2) and (3), the mechanical energy of the body remains constant under the action of gravitational force. Hence, law of conservation of energy is proved incase of freely falling body.

Problems

Question 1.
A test tube of mass 10 grams closed with a cork of mass 1 gram contains some ether. When the test tube is heated the cork flies out under the pressure of the ether gas. The test tube is suspended horizontally by a weight less rigid bar of length 5cm. What is the minimum velocity with which the cork should fly out of the tube, so that test tube describing a full vertical circle about the point O. Neglect the mass of ether.
Answer:
Mass of test tube M = 10g; Mass of ether m = 1g ;
Length of rigid bar = radius of circle (r) = 5cm

∴ r = 5 × 10^-2 m, g = 10 ms²
[∴ action = – reaction]
Given the minimum velocity of cork flies out (= -v) (v) = The minimum velocity of test tube right
v = \(\sqrt{5 r g}=\sqrt{5 \times 5 \times 10^{-2} \times 10}\)
= 5 × \(\sqrt{10^{-1}}\) m/s

Question 2.
A machine gun fires 360 bullets per minute and each bullet travels with a velocity of 600 ms^-1. If the mass of each bullet is 5 gm, find the power of the machine gun ? [A.P. – Mar. ’18, ’16; Mar. ’13]
Answer:
Here n = 360; t = 60 sec; V = 600 ms^-1; m = 5g = 5 × 10^-3 kg; Power,
P = \(\frac{\text { K.E of n} \text { bullets }}{t}=\frac{\frac{1}{2} \mathrm{mnV}^2}{t}\)
= \(\frac{\frac{1}{2} \times 5 \times 10^{-3} \times 360 \times 600 \times 600}{60}\)
∴ P = 5400, W = 5.4 KW.

Question 3.
Find the useful power used in pumping 3425 m³ of water per hour from a well 8 m deep to the surface, supposing 40% of the horse power during pumping is wasted. What is the horse power of the engine ?
Answer:
Here V = 3425 m³; d = 10³ kg m^-3; h = 8m;
g = 9.8ms^-2; t = 1 hour = 60 × 60s

166.6 HP [1 hp = 746 Watt]

Question 4.
A pump is required to lift 600 kg of water per minute from a well 25 m deep and to eject it with a speed of 50 ms^-1. Calculate the power required to perform the above task ? [Mar. 16 (TS), A.P (Mar. ’15)
Answer:
Here m = 600kg; h = 25m; V = 50ms^-1; t = 60s;
Power of the motor,
P = \(\frac{\mathrm{mgh}+\frac{1}{2} m \mathrm{~V}^2}{\mathrm{t}}=\frac{\mathrm{m}\left(\mathrm{gh}+\frac{\mathrm{V}^2}{2}\right)}{\mathrm{t}}\)
= \(\frac{600}{60}\left(9.8 \times 25+\frac{50 \times 50}{2}\right)\)
= 10 (245 + 1250) = 14950 = 14.95 KW.

Question 5.
A block of mass 5 kg initially at rest at the origin is acted on by a force along the X-positive direction represented by F = (20 + 5x)N. Calculate the work done by the force during the displacement of the block from x = 0 to x = 4 m.
Answer:
Given m = 5kg;
F = (20 + 5x)
W = \(\int_{x=0}^{x=4} F d x\)
= \(\int_0^4(20+5 x) d x\)
= \([20 x]_0^4+\left[5 \frac{x^2}{2}\right]_0^4\)
= 20 × 4 + 5 × \(\frac{16}{2}\)
= 120 J.

Question 6.
A block of mass 2.5 kg is sliding down a smooth inclined plane as shown. The spring arranged near the bottom of the inclined plane has a force constant 600 N/m. Find the compression in the spring at the moment the velocity of the block is maximum?

Answer:
Here m = 2.5 kg
µ = 0; K = 600 N/m
Let ‘x’ be the compression of the spring.
According to Newton’s third law, we have
Force due to block on spring
F_B = – Restoring force of spring (F_R)
In magnitude F_B = F_R = mg sinθ = Kx
2.5 × 10 × \(\frac{3}{5}\) = 600 × x
⇒ x = \(\frac{15}{600}\)
= 0.025 cm.

Question 7.
A force F = –\(\frac{K}{x^2}\)(x ≠ 0)acts on a particle along the X-axis. Find the work done by the force in displacing the particle from x = C +a to x = +2a. Take K as a positive constant.
Answer:

Question 8.
A force F acting on a particle varies with the position x as shown in the graph. Find the work done by the force in displacing the particle from x = -a to x = +2a.
Answer:
Average force, F = \(\frac{+2 b-b}{2}=\frac{b}{2}\)
⇒ x = -a to x = 2a
W = \(\int_{-a}^{+2 a} F d x=F[x]_a^{+2}=\frac{b}{2}[2 a-(-a)]=\frac{3 a b}{2}\)

Question 9.
From a height of 20m above a horizontal floor, a ball is thrown down with initial velocity 20 m/s. After striking the floor, the ball bounces to the same height from which it was thrown. Find the coefficient of restitution for the collision between the ball and the floor.
Answer:
Here u = 20 mIs; s = h = 20m; g = 10m/s²;
v = u₁
using, v² – u² = 2as
u₁² = 20² = 2 × 10 × 20
u₁² = 20² + 400 = 800
∴ u₁ = \(\sqrt{800}\)
And v₁ = \(\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 20}=\sqrt{400}\)
Since floor is at rest u₂ = 0; v₂ = 0
e = \(\frac{v_2-v_1}{u_1-u_2}=\frac{0-\sqrt{400}}{800-0}=-\frac{1}{\sqrt{2}}\)
∴ e = \(\frac{1}{\sqrt{2}}\) […. -ve sign indicates, if rebounds]

Question 10.
A Ball falls from a height of 10 m on to a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is \(\frac{1}{\sqrt{2}}\). What is the total distance travelled by the ball before it ceases to rebound ?
Solution:
Given h = 10m; e = \(\frac{1}{\sqrt{2}}\)
d = h\(\left[\frac{1+\mathrm{e}^2}{1-\mathrm{e}^2}\right]=10\left[\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right]=10\left[\frac{\left(\frac{3}{2}\right)}{1 / 2}\right]\)
∴ d = 30 m

Additional Problems

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
a) Work done by a man ¡n lifting a bucket out of a well by means of a rope tied to the bucket.
b) Work done by gravitational force in the above case,
c) Work done by friction on a body sliding down an inclined plane,
d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer:
Workdone W = \(\vec{F} \cdot \vec{S}\) = FS cosθ where θ is smaller angle between force j and displacement \(\vec{S}\).
a) To lift the bucket, force equal to weight of b

b) The bucket moves in a direction opposite to the gravitational force which acts vertically downwards.
θ = 180°
W = FS cos 180° = -FS. It is negative

c) Friction always opposite the relative motion.
∴ θ = 180°, W = FS cos 180° = -FS. It is negative.

d) As the body moves along the direction of applied force θ = 0, W = FS cos 0° = FS. It is positive.

e) The direction of resistive face is opposite to the direction of motion of the bob i.e. θ = 180°. Hence workdone, in this case, is negative.

Question 2.
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
a) Work done by the applied force in 10s,
b) Work done by friction in 10s,
c) Work done by the net force on the body in 10s.
d) Change in kinetic energy of the body in 10s. and interpret your results.
Answer:
Here, m = 2kg, u = 0, F = 7N; µ = 0.1,
W = 2, t = 10s
Acceleration produced by applied force;
a = –\(\frac{F}{m}=\frac{7}{2}\) = 3.5 m/s²
force of friction, f = µR = µmg = 0.1 × 2 × 9.8 = 1.96 N
Retardation produced by friction
a₂ = –\(\frac{F}{m}=\frac{-1.96}{2}\) = 0.98 m/s²
Net acceleration with which body moves,
a = a₁ + a₂
= 3.5 – 0.98 = 2.52m/s²
Distance moved by the body in 10 second from
S = Ut + \(\frac{1}{2}\) at²
= 0 + \(\frac{1}{2}\) × 2.52 × (10)² = 126m.
a) Workdone by the applied force = F × S
W₁ = 7 × 126 = 882J

b) Workdone by force of friction W₂
= -f × s = -1.96 × 126 = 246.9J

c) Workdone by the net force W₃ = Net force × distance
= (F – f)s = (7 – 1.96)126 = 635 J.

d) From v = u + at
v = 0 + 2.52 × 10 = 25.2 m^S-1
Final K.E = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) × 2 × (25.2)²
= 635J
Initial K.E = \(\frac{1}{2}\)mu² = 0
∴ Change in K.E = 635 – 0 = 635 J.
This shows that change in K.E of the body is equal to workdone by the net force on the body.

Question 3.
Given in Fig. are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.



Answer:
We know that total energy E = P.E + K.E (or) K.E = E – P.E and K.E can never be negative. The object cannot exist in the region where its K.E would become negative.

1. For x > a, P.E (v₀) > E
   ∴ K.E becomes negative. Hence the object cannot exist in the region x > a.
2. For x < a and x > b, P.E (v₀) > E
   ∴ K.E becomes negative. Hence the object cannot be present in the region x < a and x > b.
3. Object cannot exist in any region because P.E (v₀) > E in every region.
4. On the same basis, the object cannot exist in the region -b/2 < x < – a/2 and a/2 < x < b/2.

Question 4.
To potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx²/ 2, where k is the force constant of the oscillator. For k= 0.5 N m^-1, the graph of V(x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2m

Answer:
At any instant, the total energy of an oscillator is the sum of K.E and P.E i.e.
E = K.E + P.E, E = \(\frac{1}{2}\) mu² + \(\frac{1}{2}\) kx²
The particles turns back at the instant, when its velocity becomes zero i.e. u = 0.
∴ E = 0 + \(\frac{1}{2}\) kx² as E = 1 Joule and K = \(\frac{1}{2}\) N/m
∴ 1 = \(\frac{1}{2}\) × \(\frac{1}{2}\) x2 (or) x² = 4, x = ± 2m.

Question 5.
Answer the following :
a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained ? The rocket or the atmosphere ?
b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ?
c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ?

d) In fig. (i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ?
Answer:
a) The total energy of a rocket in a flight depends on its mass i.e. P.E + K.E = mgh + \(\frac{1}{2}\) mv². When the casing burns up, its mass decreases. The total energy of the rocket decreases Hence heat energy required for burning is obtained from the rocket itself and not from the atmosphere.

b) This is because gravitational force is a conservative force. Work done by the gravitational force of the sun over a closed path in every complete orbit of the comet is zero.

c) When the artificial satellite orbiting the earth comes closer and closer to the earth, its potential energy decreases as sum of potential energy and kinetic energy is constant, therefore K.E of satellite and hence its velocity goes on increasing. However total energy of satellite decreases a little on account of dissipation against atmospheric resistance.

d) In fig. (a), force is applied on the mass, by the man in vertically upward direction but distance is moved along the horizontal.
∴ θ = 90°, W = FS cos 90° = zero.
In fig. (b), force is applied along the horizontal and the distance moved is also along the horizontal. Therefore v = 0°.
W = FS cosθ = mg × S cos0°
W = 15 × 9.8 × 2 × 1 = 294 Joule.
workdone in 2^nd case is greater.

Question 6.
Underline the correct alternative :
a) When a conservative force does positive work on a body, the potential energy of the body increases / decreases / remains unaltered.
b) Work done by a body against friction always .results in a loss of its kinetic / potential energy.
c) The rate of change of total momentum of a many-particle system is proportional to the external force / sum of the internal forces on the system.
d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear mometum. total energy of the system of two bodies.
Answer:
a) Potential energy of the body decreases. The conservative force does positive work on a body, when it displaces the body in the direction of force. The body, therefore, approaches the center of force, decreasing x. Hence P.E decreases.

b) Workdone by a body against friction at the expense of its kinetic energy. Hence K.E of the body decreases.

c) Internal forces cannot change the total (or) net momentum of a system. Hence the rate of change of total momentum of many particle system is proportional to the external force on the system.

d) In an inelastic collision of two bodies, the quantities which do not change after the collision are total linear momentum and total energy of the system of two bodies. The total K.E changes as some energy appears in other forms.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer.
a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
c) Work done in the motion of a body over a closed loop is zero for every force in nature.
d) In an inelastic collision/the final kinetic energy is always less than the initial kinetic energy of the system.
Answer:
a) False. The total momentum and total energy of the system are conserved and not of each body.
b) False. The external forces on the body may change the total energy of the body.
c) False. Work done in the motion of a body over a closed loop is zero only the body is moving under the action of conservative forces (like gravitational (or) electrostatic faces).
It is not zero when the forces are non conservative, e.g. frictional forces etc.
d) True, because in an inelastic collision, some kinetic energy usually changes into some other form of energy.

Question 8.
Answer carefully, with reasons ;
a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ?
b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ?
c) What are the answer to (a) and (b) for an inelastic collision ?
d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Answer:
a) No, K.E is not conserved during the given elastic collision. K.E. before and after collision is the same. Infact during collision, K.E of the balls gets converted into potential energy.
b) Yes, the total linear momentum is conserved during the short time of an elastic collision of two balls.
c) In an in elastic collision, total K.E is not conserved during collision.
d) The collision is elastic, because the forces involved are conservative.

Question 9.
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
i) t^1/2
ii) t
iii) t^3/2
iv) t²
Answer:
From v – u + at, v = 0 + at = at
As power, p = F × v = (ma) × at = ma²t
As m and a are constants, therefore, p ∝ t.

Question 10.
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to
i) t^1/2
ii) t
iii) t^3/2
iv) t²
Answer:
As power, P = force × velocity
∴ P = [MLT^-2] [LT^-1] = [mL² T^-3]
As P = [mL^-2 T^-3] = constant
∴ L² T^-3 = constant
∴ L² ∝ T³ (or) L ∝ T^3/2
(or) \(\frac{\mathrm{L}^2}{\mathrm{~T}^3}\) = constant

Question 11.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = \(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) N where \(\hat{i}, \hat{j}, \hat{k}\) are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ?
Answer:
Here, F = \(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)N
\(\overrightarrow{\mathrm{S}}=4 \hat{\mathrm{k}}\) (∵ 4m distance is along z-axis)
W = 2; As W = \(\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{S}}\)
∴ W = \((-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{k}})=12 \hat{\mathrm{k}} \cdot \hat{\mathrm{k}}\) = 12J.

Question 12.
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV and the second With 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds, (electron mass = 9.11 × 10^-31 kg, proton mass = 1.67 × 10^-27 kg, 1 eV = 1.60 × 10^-19 J).
Answer:
E₁ = \(\frac{1}{2}\) m_ev_e² = 10 × 1.6 × 10^-16 ……………… (i)
E₂ = \(\frac{1}{2}\) m_pv_p² = 100 × 1.6 × 10^-16 ……………… (ii)
Divide (i) by (ii) \(\frac{m_v v_v^2}{m_p v_p^2}=\frac{1}{10}\)
\(\frac{v_{\mathrm{e}}}{v_\rho}=\sqrt{\frac{1}{10}} \frac{m \rho}{m_e}\)
= \(\sqrt{\frac{1.67 \times 10^{-27}}{10 \times 9.11 \times 10^{-31}}}=10^2 \sqrt{\frac{1.67}{91.1}}\) = 13.53

Question 13.
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it .attains its maximum (terminal) speed and moves with uniform speed thereafter. What is the work done by the gravita-tional force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms^-1 ?
Answer:
Here, r = 2mm = 2 × 10^-3m
Distance moved in each half of journey,
s = \(\frac{500}{2}\) = 250 m
Density of water, ρ = 10³ kg/m³
Mass of rain drop
= Volume of drop × Density
m = \(\frac{4}{3}\)πr³ × ρ = \(\frac{4}{3}\) × \(\frac{22}{7}\) (2 × 10^-3)³ × 10³
= 3.35 × 10^-5 kg
w = mg × s = 3.35 × 10^-5 × 9.8 × 250 = 0.082J
It should be clearly understood that whether the drop moves with decreasing acceleration (or) with uniform speed, work done by the gravitational force on the drop remains the same.
If there were no resistive forces, energy of drop on reaching the ground.
E₁ = mgh = 3.35 × 10^-5 × 9.8 × 500 = 0.164J
Actual energy, E₂ = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) × 3.35 × 10^-5 × (10)² = 1.675 × 10^-3 J.
∴ Workdone by the resistive forces,
W = E₁ – E₂ = 0.164 – 1.675 × 10^-3
W = 0.1623 Joule.

Question 14.
A molecule in a gas container hits a horizontal wall with speed 200 m s^-1 and angle 30° with the normal and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?
Answer:
We know that momentum is conserved in all collisions, elastic as well as inelastic. Let us now check if K. E is conserved (or) not.

If the wall is too heavy, the recoiling molecule produces no velocity in the wall.
If m is mass of the gas molecule and M is mass of wall, then total K.E after collision
E₂ = – m(200)² + \(\frac{1}{2}\) M(0)²
E₂ = 2 × 10⁴ mJ
Which is the K.E. of the molecule before collision.
[E₁ = \(\frac{1}{2}\)m(200)² = 2 × 10⁴ mJ]
Hence the collision is elastic.

Question 15.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 min. If the tank is 40 m above the ground and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?
Answer:
Here, volume of water = 30 m², t = 15 min = 15 × 60 = 900s.
height h = 40m, efficiency η = 30%
As density of water = ρ = 10³ kg /m³
∴ Mass of water pumped, m = volume x density = 30 × 10³ kg
Actual power consumed (or) out power
P₀ = \(\frac{\mathrm{w}}{\mathrm{t}}=\frac{\mathrm{mgh}}{\mathrm{t}}\)
= \(\frac{30 \times 10^3 \times 9.8 \times 40}{900}\) = 10370 watt.
If P_i is input power (required), then as η = \(\frac{P_0}{P_i}\)
∴ P_i =\(\frac{P_0}{\eta}=\frac{10370}{\frac{30}{100}}\) = 43567 watt = 43.567 KW.

Question 16.
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision ¡s elastic, which of the following (Fig.) is a possible result after collision ?

Answer:
Let m be the mass of each ball bearing
Before collision, total K.E of the system =
\(\frac{1}{2}\)mV² + 0 = \(\frac{1}{2}\) mV²
After collision K.E of the system is
Case I, E₁ = \(\frac{1}{2}\) (2m) (V/2)2 = \(\frac{1}{4}\) mV²
Case II, E₂ = \(\frac{1}{2}\) mV²
Case III, E₃ = – (3m) (V/3)² = \(\frac{1}{6}\)mV²
We observe that is conserved only in Case II. Hence the Case II is the only possibility.

Question 17.
The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a tables as shown in Fig. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic.

Answer:
The bob A shall not rise. This is because when two bodies of same mass undergo an elastic collision, their velocities are interchanged. After collision, ball a will come to rest and the ball B would move with the velocity of A.

Question 18.
The bob of a pendulum is released from a horizontal position, If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ?
Answer:
Here, h = 1.5m, V = ?
Energy dissipated = 5%
Taking B as the lowest position of the bob, its potential energy at B is zero. At the horizontal position, A total potential energy of the bob is mgh.
In going from A to B, P.E of the bob is converted into K.E
Energy conserved = 95% (mgh)
If V is velocity acquired at B, then K.E = \(\frac{1}{2}\) mv²
= \(\frac{95}{100}\)mgh
V = \(\sqrt{\frac{95}{100} \times 2 \mathrm{gh}}=\sqrt{\frac{19}{20} \times 2 \times 9.8 \times 1.5}\)
= 5.285ms^-1.

Question 19.
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/ h oh a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s^-1. What is the speed of the trolley after the entire sand bag is empty ?
Answer:
As the trolley carrying the sane bag is moving uniformly, therefore, external force on the system = zero.
When the sand leaks out, it does not lead to the application of any external force on the trolley. Hence the speed of the trolley shall not change.

Question 20.
A body of mass 0.5 kg travels in a straight line with velocity v = ax^3/2 where a = 5m^-1/2 s^-1. What is the workdone by the net force during its displacement from x = 0 to x = 2 m ?
Answer:
Here, m = 0.5 kg; v = ax^3/2, a = 5m^-1/2 s^-1,
w = ?
Intial vel. at x = 0, v₁ = a × 0 =0
Final vel, at x = 2, v₂ = a2^3/2 = 5 × 2^3/2
Workdone = increase in K.E. = \(\frac{1}{2}\) m
(v₂² – v₁²), W = \(\frac{1}{2}\) × 0.5 [(5 × 2^3/2)²) – 0]
= 50 J

Question 21.
The blades of a windmill sweep out a circle of area A.
(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ?
(b) What is the kinetic energy of the air ?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy and that A = 30 m², v = 36 km/h and the density of air is 1.2 kg m^-3. What is the electrical power produced ?
Answer:
a) Volume of wind flowing /sec = AV
Mass of wind flowing / sec = AVρ
Mass of air passing in t sec = AVρt

b) K.E of air = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) (AVρt)
v² = \(\frac{1}{2}\) av³ρt.

c) Electrical energy produced = \(\frac{25}{100}\)
K.E of air = \(\frac{1}{4}\) × \(\frac{1}{2}\) × Av³ρt
Electrical power = \(\frac{1}{8} \frac{A v^3 \rho t}{t}=\frac{1}{8} A v^3 \rho t\)
P = \(\frac{1}{8}\) × 30 × (10)³ × 1.2 = 4500 watt
= 4.5 Kw.

Question 22.
A person trying to lose weight (dieter) lifts a 10kg mass, one thousand times, to a height of 0.5 m ech time. Assume that the potential energy lost each time she lowers the mass is dissipated,
(a) How much work does she do against the gravitational force ?
(b) Fat supplies 3.8 × 10⁷J of energy per kilo-gram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up ?
Answer:
Here, m = 10 kg, h = 0.5 m, n = 1000
a) Workdone against gravitational force W = n(mgh)
= 1000 × (10 × 9.8 × 0.5 = 49000 J

b) Mechanical energy supplied by 1 kg of fat
= 3.8 × 10 × \(\frac{20}{100}\) = 0.76 × 10⁷ J/kg
∴ Fat used up by the dieter
= \(\frac{1 \mathrm{~kg}}{0.76 \times 10^7}\) = 4000 = 6.45 × 10^-3 kg.

Question 23.
A family uses 8 kW of power,
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW ?
(b) Compare this area to that of the roof of a typical house.
Answer:
Let area be A’ square meter
∴ Total power = 200A
Useful electrical energy produced/sec = \(\frac{20}{100}\)
(200A) = 8KW = 40A = 8000 (watt)
Therefore, A = \(\frac{8000}{40}\) = 200 sq.m
This area is comparable to the roof of a large house of 250 sq.mt.

Question 24.
A bullet of mass 0.012 kg and horizontal speed 70 ms^-1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Answer:
Here m₁ = 0.012kg. u₁ = 70 m/s m₂ = 0.4 kg, u₂ = 0
As the bullet comes to rest w.r. to the block, the two behave as one body. Let v be the velocity acquired by the combination. Applying principle of conservation of linear momentum, (m₁ + m₂)v
= m₁u₁ + m₂u₂ = m₁u₁
V = \(\frac{m_1 u_1}{m_1+m_2}=\frac{0.012 \times 70}{0.012+0.4}=\frac{0.84}{0.412}\)
= 2.04 m/s.
Let the block rise to a height h.
RE of combination = K.E. of the combination
(m₁ + m₂)gh = \(\frac{1}{2}\) (m₁ + m₂)v²
h = \(\frac{v^2}{2 g}=\frac{2.04 \times 2.04}{2 \times 9.8}\) = 0.212m.
For calculating heat produced. We calculate energy lost (W). where
W = intial K.E of bullet – final K.E of combination
= \(\frac{1}{2}\) m₁u₁² \(\frac{1}{2}\) (m₁ + m₂)v²
= \(\frac{1}{2}\) × 0.012 (70)² – \(\frac{1}{2}\) (0.412) (2.04)²
W = 29.4 – 0.86 = 28.54 Joule.
∴ heat produd. H = \(\frac{W}{j}=\frac{28.54}{4.2}\) = 6.8 cal.

Question 25.
A 1kg block situated on a rough incline is connected to a spring of spring constant 100 N m^-1 as shown in Fig. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Answer:
From Fig.
R = mg cosθ
F = μR = μmg cosθ
Net force on the block down the incline
= mg sin θ – F = mg sin θ – μ mg cos θ
= mg (sin – μ cos θ)
Distanced moved, x = 10cm = 0.1m.
In equilibrium, work done = P.E of streched spring
mg (sin θ – μ cos θ) x = \(\frac{1}{2}\) Kx²
2mg (sin θ – μ cos θ) = Kx
2 × 1 × 10 (sin 37° – μ cos 37°) = 100 × 0.1
20(0.601 – μ 0.798) = 10
∴ μ = 0.126

Question 26.
A bob of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 ms^-1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact ? Would your answer be different if the elevator were stationary ?
Answer:
Here, m 0.3kg, v = 7 m/s h = length of elevator = 3m
As relative of the bob w.r.t elevator is zero, therefore, in the impact, only potential energy of the fall is converted into heat energy.
Amount of heat produced = P.E lost by the bob = mgh = 0.3 × 9.8 × 3 = 8.82 J.
The answer shall not be different, if the elevator were stationary as the bob too in that cause would start from stationary position, i.e. relative velocity of the ball w.r.t elevator would continue to be zero.

Question 27.
A trolley of mass 200 kg moves with a uniform speed of 36 km / h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of p m s^-1 relative to the trolley in a direction opposite to the its motion and jumps out of the trolley. What is the final speed of the trolley ? Flow much has been trolley moved from the time the child begins to run ?
Answer:
Here, mass of trolley, m₁ = 200 kg
speed of the trolley v = 36 km/h = 10m/s
mass of the child, m₂ = 20 kg
Before the child starts running, momentum of the system.
P₁ = (m₁ + m₂)v = (200 + 20)10
= 2200kg ms^-1.
∴ Momentum of the system when the child is running
P₂ = 200v¹ + 20 (v¹ – 4) = 220v¹ – 80
As no external force is applied on the system.
∴ P₂ = P₁
220v¹ – 80 = 2200
220v¹ = 2200 + 80 = 2280
v¹ = \(\frac{2280}{220}\) = 10.36 ms^-1
Time taken by the child to run a distance of 10m over the trolley, t = \(\frac{10 \mathrm{~m}}{4 \mathrm{~ms}^{-1}}\) = 2.5 s
Distance moved by the trolley in this time = Velocity of trolley × time = 10.36 × 2.5 = 25.9 m

Question 28.
Which of the following potential energy curves in Fig. cannot possibly describe the elastic collision of two billiard balls ? Here r is the distance between centres of the balls.

Answer:
The potential energy of a system of two masses varies inversely as the distance (r) between them i.e,
v(r) ∝ \(\frac{\mathrm{1}}{\mathrm{r}}\).
When the two billiard balls touch each other, P.E become zero i.e. at r = R + R = 2R;
v(r) = 0. Out of the given graphs, curve (v) only satisfies these two conditions.

Question 29.
Consider the decay of a free neutron at rest : n → p + e^– Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus (Fig.).

(Note : The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin 1/2 (like e-, p or n), but is neutral, and either masslessor having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n → p + e- + v
Answer:
In the decay process, n → p + e
energy of electrons is equal to (∆m)c²
Where ∆m = mass defect = mass of neutron – mass of proton and electron;
which is fixed. Therefore, two body decay of this type cannot explain the observed continuous energy distribution in the β-decay of a neutron or a nucleus.

Textual Examples

Question 1.
Find the angle between force F = \((3 \bar{i}+4 \bar{J}-5 \bar{k})\) unit and displacement d = \((3 \bar{i}+4 \bar{J}-5 \bar{k})\) unit. Also find the projection of F on d.
Answer:
F.d = F_xd_x + F_yd_y + F_zd_z = 3(5) + 4(4) + (-5)
(3) = 16 unit
Hence F.d = F d cos θ = 16 unit
Now F.F = F² – F_x² + F_y² + F_z²
= 9 + 16 + 25 = 50 unit
and d.d = d² = d_x² + d_y² + d_z²
= 25 + 16 + 9 = 50 unit
∴ cosθ = \(\frac{16}{\sqrt{50} \sqrt{50}}=\frac{16}{50}\) = 0.32
= θ = cos^-1 0.32.

Question 2.
It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height 1.00 km. It hits the ground with a speed of 50.0 ms^-1. (a) What is the work done by the gravitational force? What is the work done by the unknown resistive force?
Answer:
(a) The change in kinetic energy of the drop is
∆k = \(\frac{1}{2}\)mv² – 0
= \(\frac{1}{2}\) × 10^-3 × 50 × 50 = 1.25 J
where we have assumed that the drop is initially at rest.
Assuming that g is a constant with a value 10 m/s², the work done by the gravitational force is,
W_g = mgh = 10^-3 × 10 × 10^-3 = 10.0 J

(b) From the work-energy theorem
∆K = W_g + W_r
Where W_r is the work done by the resistive force on the raindrop. Thus
W_r = ∆K – W_g = 1.25 – 10 = -8.75 J in negative.

Question 3.
A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion,
(a) How much work does the road do on the cycle ?
(b) How much work does the cycle do on the road ?
Answer:
Work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road.
(a) The stopping force and the displacement make an angle of 180° (π rad) with each other.
Thus, work done by the road,
W_r = Fd cosθ = 200 × 10 × cos π
= -2000 J
It is this negative work that brings the cycle to a halt in accordance with WE theorem.

(b) From Newton’s Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude is 200 N. However, the road, undergoes no displacement. Thus work done by cycle on the road is zero.

Question 4.
In a ballistics demonstration a police officer fires a bullet of mass 50.0g with speed 200 ms^-1 (Bullet of mass = 5 × 10^-2, Bullet of speed = 200; Bullet of K(J) = 10³; on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet ?
Answer:
The initial kinetic energy of the bullet is mv² / 2 = 1000 J. It has a final kinetic energy of 0.1 × 1000 = 100 J. If v_f is the emergent speed of the bullet,
\(\frac{1}{2}\) mv²_f = 100 J
v_f = \(\frac{\sqrt{2 \times 100 \mathrm{~J}}}{0.05 \mathrm{~kg}}\) = 63.2 ms^-1
The speed is reduced by approximately 68% (not 90%).

Question 5.
A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 50 N. The total distance through which the trunk has been moved is 20 m. Plot the force applied by the woman and the frictional force, which is 50 N versus displacement. Calculate the work done by the two forces over 20m.
Answer:

Plot of the force F applied by the woman and the opposing frictional force f versus displacement.
The plot of the applied force is shown in Fig. At x = 20 m, F = 50 N (≠ 0). We are given that the frictional force f is |f| = 50N. It opposes motion and acts in a direction opposite to F. It is therefore, shown on the negative side of the force axis.
The work done by the woman is
W_F → area of the rectangle ABCD + area of the trapezium CEID
W_F = 1000 × 10 + \(\frac{1}{2}\)(100 + 50) ×10
= 1000 + 750 = 1750 J
The work done by the frictional force is
W_f → area of the rectangle AGHI
W_f → (-50) × 20 = -1000 J
The area on the negative side of the x- axis has a negative sign.

Question 6.
A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v₀ at the lowest point A such that it completes a semi circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. Obtain an expression for (i) v₀ (ii) the speeds at points B and C; (iii) the ratio of the kinetic energies (K_B/K) at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C.

Answer:
Thus, at A:
E = \(\frac{1}{2} {\mathrm{mv}_0^2}\) ………………… (1)
T_A – mg = \(\frac{\mathrm{mv}_0^2}{\mathrm{~L}}\) [Newtons Second Law]
where T_A is the tension in the string at A. At the highest point C, the string slackens, as the tension in the string (T_c) becomes zero.
E = – \(\frac{1}{2} {\mathrm{mv}_c^2}\) + 2mgL ………………… (2) .
mg = \(\frac{\mathrm{mv}_c^2}{\mathrm{~L}}\) [Newtons Second law] ……………… (3)
where v_c is the speed at C. From equations (2) & (3)
E = \(\frac{5}{2}\)mgL
Equating this to the energy at A
\(\frac{5}{2}\) mgL = \(\frac{m}{2} v_0^2\) or, v₀ = \(\sqrt{5 \mathrm{gL}}\)

(ii) It is clear from Equation (3)
v_c = \(\sqrt{ \mathrm{gL}}\)
At B, the energy is
E = \(\frac{1}{2} m v_B^2\) + mgL
Equating this to the energy at A and empLoying the result from (i),
namely v₀² = 5gL
\(\frac{1}{2} m v_B^2+m g L=\frac{1}{2} m v_0^2\)
= \(\frac{5}{2}\)mgl ∴ v_B = \(\sqrt{3\mathrm{gL}}\)

(iii) The ratio of the kinetic energies at Band C is :
\(\frac{\mathrm{K}_{\mathrm{B}}}{\mathrm{K}_{\mathrm{C}}}=\frac{\frac{1}{2} m v_{\mathrm{B}}^2}{\frac{1}{2} m v_{\mathrm{c}}^2}=\frac{3}{1}\)
At point C, the string becomes slack and the velocity of the bob is horizontal and to the left.

Question 7.
To simulate car accidents, auto manufactures study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 km / h on a smooth road and colliding with a horizontally mounted spring of spring constant 6.25 × 10³ N m¹. What is the maximum compression of the spring ?
Answer:
At maximum compression the kinetic energy of the car is converted entirely into the potential energy of the spring.
The kinetic energy of the moving car is 11
K = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × 10³ × 5 × 5
K = 1.25 × 10⁴ J
where we have converted 18 km h^-1 to 5ms^-1 [It is useful to remember that 36 km h^-1 = 10 ms^-1]. At maximum compression x_m, the potential energy. V of the spring is equal to the kinetic energy K of the moving car from the principle of conservation of mechanical energy.
V = \(\frac{1}{2}\) K x_m² = 1.25 × 10⁴ J
We obtain x_m = 2.00 m

Question 8.
Consider example 8 taking the coefficient of friction, μ, to be 0.5 and calculate the maximum compression of the spring.
Answer:
In presence of friction, both the spring force and the frictional force act so as to oppose the compression of the spring as shown in Fig.
We invoke the work-energy theorem, rather than the conservation of mechanical energy.
The change in kinetic energy is

∆K = K_f – K_i = 0 – \(\frac{1}{2}\)mv²
The work done by the net force is
W = \(\frac{1}{2}\)Kx_m² – μm g x_m
Equating we have
\(\frac{1}{2}\)mv² = \(\frac{1}{2}\)K x_m² + μm g x_m
Now μmg = 0.5 × 103 × 10 = 5 × 10³ N
(taking g = 10.0 ms^-2]. After rearranging the above equation we obtain the following quadratic equation in the unknown x_m.
K x_m² + 2μm gx_m – mv² = 0
x_m = \(\frac{-\mu m g+\left[\mu^2 m^2 g^2+m k v^2\right]^{1 / 2}}{k}\)
where we take the positive square root since x_m is positive. Putting in numerical values we obtain
x_m = 1.35 m
which, as expected, is less than the result in example 7
If the two forces on the body consist of a conservative force F_c and a non- conservative force F_nc, the conservation of mechanical energy formula will have to be modified. By the WE theorem
But Hence, (F_c + F_nc) ∆x = ∆K
F_c ∆x = ∆V
∆ (K + V) = F_nc ∆x
∆E = F_nc ∆x
where E is the total mechanical energy. Over the path this assumes the form E_f – E_i = W_nc
Where W_nc is the total work done by the non-conservative forces over the path. Note that unlike the conservative force. W_nc depends on the particular path i to f.

Question 9.
Examine express
(a) The energy required to break one bond in DNA in eV(1.6 × 10^-19J);
(b) The kinetic energy of an air molecule (10^-21J) in eV;
(c) The daily in take of a human adult in kilocalories (10⁷).
Answer:
(a) Energy required to break one bond of DNA is \(\frac{10^{-20} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) ; 0.06 eV
Note 0.1 eV = 100 meV (100 millielectron volt).

(b) The kinetic energy of an air molecule is
\(\frac{10^{-21} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\); 0.0062 eV
This is the same as 6.2 meV.

(c) The average human consumption in a day
is \(\frac{10^7 \mathrm{~J}}{4.2 \times 10^3 \mathrm{~J} / \mathrm{kcal}}\); 2400 kcaL

Question 10.
An elevator can carry a maximum load of 1800 kg (elavator + passengers) is moving up with a constant speed of 2ms^-1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.
Answer:
The downward force on the elevator is F = mg + F_f = (1800 × 10) + 4000 = 22000 N
The motor must supply enough power to balance this force. Hence,
P = F.v = 22000 × 2 = 44000 W = 59 hp

Question 11.
Slowing down of neutrons : In a nuclear reactor of high speed (typically 107ms^-1) must be slowed to 10³ ms^-1 so that it can have a high probability of interacting with isotope \({ }_{92}^{235} \mathrm{U}\) and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water (D₂O) or graphite, is called a moderator.
Answer:
The initial kinetic energy of the neutron is
K_1i = \(\frac{1}{2}\) m₁ v_1i²
While its final kinetic energy from
V_1f = \(\frac{\left(m_1-m_2\right)}{m_1+m_2}\) v_1i²
K_1f = \(\frac{1}{2}\)m₁ v₁²_f = \(\frac{1}{2}\)m₁ \(\left(\frac{m_1-m_2}{m_1+m_2}\right)^2\) v_1i²
The fractional kinetic energy lost is
f₁ = \(\frac{K_{1 f}}{k_{1i}}\left(\frac{m_1-m_2}{m_1+m_2}\right)^2\)
while the fractional kinetic energy gained by the moderating nuclei K_2f/ K_1i is f₂ = 1 – f₁
elastic collision = \(\frac{4 m_1 m_2}{\left(m_1+m_2\right)^2}\)
One canalso verify this result by substituting from
v_2f = \(\frac{2 m_1 v_1i}{m_1+m_2}\)
For deuterium m₂ = 2m₁ and we obtain f₁ = 1/9 while f₂ = 8/9. Almost 90% of the neutron’s energy is transferred to deuterium. For carbon f₁ = 71.6% and f₂ = 28.4%. In practice, however, this number is smaller since head-on collisions are rare.

Question 12.
Consider the collision depicted in Fig. to be between two billiard balls with equal masses m₁ = m₂. The first bail is called the cue while the second bail is called the target. The billiard player wants to ‘sink’ the target ball in a corner pocket, which is at an angle θ₂ = 37°. Assume that the collision is elastic and that friction and rotational motion are not important. Obtain θ₁.
Answer:
From momentum conservation, since the masses are equal
v_1i = v_1f + v_2f
(or) v_1i² = (v_1f + v_2f). (v_1f + v_2f)
= v_1f² + V_2f² + 2v_1fv_2fcos (θ₁ + 37°) …………….. (1)
Since the collision is elastic and m1 = m2 it follows from conservation of kinetic energy that
v_1i²= v_1f² + v_2f² ………………. (2)
Comparing equations (1) and (2), we get
cos (θ₁ + 37°) = 0
or θ₁ +37° = 90°
Thus, θ₁ = 53°
This proves the following results : When two equal masses undergo a glancing elastic collision with one of them at rest, after the collision, they will move at right angles to each other.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 5th Lesson Laws of Motion Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 5th Lesson Laws of Motion

Very Short Answer Questions

Question 1.
What is inertia? What gives the measure of inertia? [T.S. Mar. 17]
Answer:
The resistance of the body to change its state of rest the state of uniform motion in a straight line is called inertia of the body.
Acceleration = \(\frac{\text { Force }}{\text { Mass }}\)
∴ The more the mass, the less acceleration and the more inertia. The mass of a body is a quantitative measure of its inertia.

Question 2.
According to Newton’s third law, every force is accompanied by an equal and opposite force. How can a movement ever take place?
Answer:
Because both action and reaction are taking place on different bodies.

Question 3.
When a bullet is fired from a gun, the gun gives a kick in the backward direction. Explain. [A.P. Mar. 15]
Answer:
Due to law of conservation of momentum, Recoil of the gun V = \(\frac{\mathrm{mu}}{\mathrm{M}}\)
Where M – mass of the gun; m = mass of the bullet; u = velocity of the bullet

Question 4.
Why does a heavy rifle not recoil as strongly as a light rifle using the same cartridges ?
Answer:
Recoil of the Gun V = \(\frac{\mathrm{mu}}{\mathrm{M}}\)
Due to heavy mass of rifle the recoil is less.

Question 5.
If a bomb at . _st explodes into two pieces, the pieces must travel in opposite directions. Explain. [T.S. Mar. 16, 15]
Answer:
According to law of conservation of momentum,
Mu = m₁v₁ + m₂v₂
Initially the bomb is at rest u = 0
m₁v₁ + m₂v₂ = 0
or m₁v₁ – m₂v₂
(Negative sign indicates that the pieces must travel in opposite direction)

Question 6.
Define force. What are the basic forces in nature ?
Answer:
The force is on which changes or tends to change the state of rest or motion of a body. Basic forces :

1. Gravitational force
2. Electromagnetic force
3. Nuclear force
4. Weak interaction force

Question 7.
Can the coefficient of friction be greater than one ?
Answer:
Yes, coefficient of friction may be greater than one. In some particular cases it is possible. They are

1. Due to increase the inner molecular attractive forces between surfaces when the contact surfaces are highly polished.
2. When the contact surfaces of the bodies are inter locking the coefficient friction may be greater than one.

Question 8.
Why does the car with a flattened tyre stop sooner than the one with inflated tyres ?
Answer:
Flattened deforms more than the inflated tyre. Due to greater deformation of the type rolling friction is large nence it stops soon.

Question 9.
A horse has to pull harder during the start of the motion than later. Explain. [Mar. 13]
Answer:
We know the limiting frictional force is greater than kinetic frictional force. For starting motion of the cart, the limiting friction is to be overcome. Once motion is set, frictional force reduces. Therefore, the horse has to pull harder during starting of the cart.

Question 10.
What happens to the coefficient of friction if the weight of the body is doubled? [A.P. – Mar. 16]
Answer:
If weight of a body is doubled, coefficient of friction does not change. Coefficient of friction is independent of normal reaction. If weight is doubled, normal reaction doubled and correspondingly frictional force doubled. So, coefficient of friction does not change.

Short Answer Questions

Question 1.
A stone of mass 0.1 kg is thrown vertically upwards. Give the magnitude and direction of the net force on the stone
(a) during its upward motion.
(b) during the downward motion.
(c) at the highest point, where it momentarily comes to rest.
Answer:
Given that, mass of stone, m = 0.1 kg, g = 9.8 ms^-2.
a) During upward motion: Magnitude of net force on the stone,
F = |-mg|; F = 0.1 × 9.8 = 0.98N.
Direction of net force is in upward direction.

b) During downward motion: Magnitude of net force on the stone,
F = ma = 0.1 × 9.8 = 0.98N.
Direction of net force is in downward direction.

c) At the heighest point : Magnitude of net force, F = mg = 0.1 × 9.8 = 0.98N..
At highest point of stone, direction is indeterminate.

Question 2.
Define the terms momentum and impulse. State and explain the law of conservation of linear momentum. Give examples.
Answer:
Momentum : The product of mass and velocity of a body is called momentum momentum p = mv .
Impulse : The product of force and time that produces finite change in momentum of the body is called impulse.
Impulse (I) = Force × time duration = mat = \(m \frac{(v-u)}{t} t\)
= (mv – mu)
Law of conservation of linear momentum : The total momentum of an isolated system of interacting particles remains constant it there is no resultant external force acting on it.

Explanation : Consider two smooth, non-rotating spheres of masses m₁ and m₂ (m₁ > m₂). Let u₁ and u₂ be their initial velocities. Let v₁ and v₂ be final velocities after head on collision. According to law of conservation of linear momentum, we have
Momentum of the system before collision = Momentum of the system after collision.
t.e., m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Examples :

1. Motion of a Rocket
2. Bullet-Gun motion

Question 3.
Why are shock absorbers used in motor cycles and cars ?
Answer:
When a scooter or a car moves on a rough road it receives an impulse due to the Jerkey motion. In case the shockers are used in the vehicle, the time of impact increases, and decreases the impulsive force, due to increased value of the time of impact the force of impact is reduced. So it saves the vehicle and its occupants from experiencing reverse jerks.

Question 4.
Explain the terms limiting friction, dynamic friction and rolling friction.
Answer:
Limiting friction : The maximum value of static friction is called “Limiting friction”.
It is denoted by F_L = F_s(max) [∵ F_s ≤ μ_sN]
Dynamic friction (Kinetic friction) : The resistance encountered by a sliding body on a surface is called kinetic or dynamic friction F_k.
If the applied force overcomes the limiting friction and sets the body into motion. Then motion of the body is resisted by another friction called “Dynamic friction” or “Kinetic friction”.

Rolling friction : ‘The resistance encountered by a rolling body on a surface is called “Rolling friction”.
If a wheel or a cylinder or a spherical body like a marble rolls on horizontal surface, the speed of rolling gradually decreases and it finally stops.

Question 5.
Explain advantages and disadvantages of friction. [T.S., A.P. Mar. 15]
Answer:
Advantages of friction :

1. Safe walking on the floor, motion of vehicles etc., are possible only due to friction.
2. Nails, screws are driven into walls (or) wooden surfaces due to friction.
3. Friction helps the fingures hold the things (or) objects like pen, pencil and water tumbler etc.
4. Speed running vehicles etc. can be stopped suddenly when friction is present, otherwise accidents become large. Due to friction vehicles move on the roads without slipping and they can be stopped.
5. The mechanical power transmission of belt drive is possible due to friction.

Disadvantages of friction:

1. Due to friction there is large amount of power loss in machines and engines.
2. Due to friction wear and tear of the machines increases and reducing their life.
3. Due to friction some energy gets converted into heat which goes as waste.

Question 6.
Mention the methods used in decrease friction [A.P. Mar. 18; T.S. Mar. 16, Mar. 14]
Answer:

1. Polishing : By polishing the surfaces of contact, friction can be reduced.
2. Bearings : The rolling friction.is less than the sliding friction hence free wheels of a cycle, motor
   car, dynamos etc., are provided with ball bearings to reduce friction. Bearings convert sliding motion into rolling motion.
3. Lubricants: The lubricant forms a thin layer between surfaces of contact. It reduces the friction. In light vehicles or machines, oils like “three in one” are used as lubricants. In heavy machines greasure is used. In addition to this they guard the mechanical parts from over heating.
4. Streamlining : Automobiles and aeroplanes are streamlined to reduce the friction due to air.

Question 7.
State the laws of rolling friction.
Answer:
Laws of friction – rolling friction :

1. The smaller the area of contact, the lesser will be the rolling friction.
2. The larger the radius of the rolling body, the lesser will be the rolling friction.
3. The rolling friction is directly proportional to the normal reaction.
   If F_R is the rolling friction and ‘N’ is the normal reaction at the contact, then F_R ∝ N
   F_R = μ_RN; where μ_R is the coefficient of rolling friction.

Question 8.
Why is pulling the lawn roller preferred in pushing it ?
Answer:
Pulling of lawn roller : Let a lawn roller be pulled on a horizontal road by a force ‘F, which makes an angle θ with the horizontal, to the right as shown in the figure. The weight of the body “mg” acts vertically downwards.

Let the force ‘F’ be resolved into two mutually perpendicular components F sin θ, vertically upwards and F cos θ horizontally along the road.
∴ The normal reaction N = mg – F sin θ
Then the frictional force acting towards left is F_R = μ_RN, where μ_R is the coefficient of rolling friction between he roller and the road, or F_R= μ_R (mg – F sin θ)
∴ The net pulling force on roller is P = F cos θ – F_R = F cos θ – (μ_R mg – F sin θ)
or P = F(cos θ + μ_R sin θ) – μ_R mg ………………(1)

Pushing of lawn roller : When a lawn roller is pushed by a force ‘F1, which makes an angle ‘θ’ with the horizontal, the component of force acting vertically downwards is F sin θ. The horizontal component F cos θ pushes the roller to the right as shown in figure.
The weight ‘mg’ of the lawn roller acts vertically downwards. Therefore the normal reaction N of the surface on the roller
N = mg + F sin θ
Then the frictional force acting towards left.
F_R = μ_RN = μ_R (mg + F sin θ)
The net pushing force on roller is
P’ = F cos θ – F_R = F cos θ – μ_R (mg + F sin θ)
or P’ = F(cos θ – μ_R sin θ) – μ_R mg ……………… (2)
From equations (1) and (2) that it is easier to pull than push a lawn roller.

Long Answer Questions

Question 1.
a) State Newton’s second law of motion. Hence derive the equation of motion F = ma from it. [T.S. Mar. 18; A.P. Mar. 16, Mar. 13]
b) A body is moving along a circular path such that its speed always remains constant. Should there be a force acting on the body ?
Answer:
a) Newton’s Second of motion : “The rate of change of momentum of a body is directly proportional to the external force applied and takes place in the same direction”.
To show F = ma : Let a body of mass ‘m’ moving with a velocity ‘v’ under the action of an external force F in the direction of velocity.
Momentum ‘P’ of a body is the product of the mass and velocity V.
∴ P = mv ……………… (1)
According to Newton’s second law of motion, we have
\(\frac{\mathrm{dp}}{\mathrm{dx}}\) ∝ F, where F = external force
(or)F = K\(\frac{\mathrm{dp}}{\mathrm{dx}}\) ………………. (2)
From equations (1) and (2) we have
F = \(K \frac{d(m v)}{d t}=K \cdot m \frac{d v}{d t}\)
= Kma …………………. (3)
Since the rate of change of velocity \(\frac{\mathrm{dv}}{\mathrm{dx}}\) is the acceleration ‘a’ of the body.
In SI system the unit of force is Newton and is defined as that force which when acting on a body of mass 1 kg produces in it an acceleration of 1 ms^-2.
i.e., from equation (3),
If F = 1, m = 1 and a = 1
we get K = 1
Hence F = \(\frac{\mathrm{dp}}{\mathrm{dx}}\) = ma .
∴ F = ma

b) Suppose a body is moving along a circular part though its speed always remains constant its velocity changes at every point and resultant force acts on the body.

Question 2.
a) Define angle of friction and angle of repose. Show that angle of friction is equal to angle of repose for a rough inclined place.
Answer:
Angle of friction : The angle of friction is defined as the angle made by the resultant of the normal reaction and the limiting friction with the normal reaction is called angle of friction.

Angle of repose : The angle of repose is defined as the angle of inclination of a plane with respect to horizontal for which the body will be in equilibrium on the inclined plane is called angle of repose.

Angle of friction is equal to angle of repose for a rough inclined plane : Let us consider a body of mass’m1 on a rough inclined plane. The angle of inclination of the rough surface is ‘θ’. By increasing the angle of inclination at one end, the body tends to slide on the surface. Then the angle of inclination ’0′ is called angle of repose.

The weight (mg) of the body resolved into two components, the component mg cos 0 acts perpendicular to the inclined surface, which is equal to normal reaction ‘N‘.
i.e., N = mg cos θ ……………. (1)
The other component mg sin θ acts parallel to inclined plane and opposite to the frictional force ‘f’.
f_s = my sin θ ……………… (2)
From \(\frac{(2)}{(1)} \Rightarrow \frac{f_s}{N}=\frac{m g \sin \theta}{m g{con} \theta}\)
= \(\frac{f_s}{N}\) = tan θ = µ_s
∴ µ_s = tan θ …………………. (3)
when ‘α’ is angle of friction then from the definition of co-efficient of static friction,
µ_s = tan α ………………… (4)
from (3) and (4)
⇒ tan θ = tan α
θ = α
Hence angle of friction is equal to angle of repose.

b) A block of mass 4 kg is resting on a rough horizontal plane and is about to move when a horizontal force of 30 N is applied on it. If g = 10 m/s². Find the total contact force exerted by the plane on the block.
Solution:
Given m = 4kg
F = 30 N
g = 10 ms^-2
a = \(\frac{F}{m}=\frac{30}{4}\) = 7.5 ms^-2
µ = \(\frac{\mathrm{a}}{\mathrm{g}}=\frac{7.5}{10}=\frac{3}{4}\)
∴ Contact force = Frictional force
= µ mg
= \(\frac{3}{4}\) × 4 × 10 = 30 N.

Problems

Question 1.
The linear momentum of a particle as a function of time t is given by p = a + bt, where a and b are positive constants. What is the force acting on the particle ?
Answer:
Linear momentum of a particle p = a + bt
Force F = \(\frac{\mathrm{dp}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(a + bt) = 0 + b
∴ F = b

Question 2.
Calculate the time needed for a net force of 5 N to change the velocity of a 10 kg mass by 2 m/s.
Answer:
F = 5N, m = 10kg; (v-u) = 2m s^-1, t = ?
F = \(m \frac{(v-u)}{t} \Rightarrow 5=\frac{10 \times 2}{t}\)
∴ t = 4s.

Question 3.
A ball of mass m is thrown vertically upward from the ground and reaches a height h before momentarily coming to rest. If g is acceleration due to gravity. What is the impulse received by the ball due to gravity force during its flight ? (neglect air resistance).
Answer:
mass = m
Initial velocity of a ball to reach h is,
u = \(\sqrt{2 g h}\)

On return journey, velocity of a ball to reach the ground, v = –\(\sqrt{2 g h}\)
Impluse I = m (u – v) = m [\(\sqrt{2 g h}\) – (-\(\sqrt{2 g h}\))]
= m 2 \(\sqrt{2 g h}\) = 2m \(\sqrt{2 g h}\)
∴ I = \(\sqrt{8 m^2 g h}\)

Question 4.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 in m s^-1 to 3.5 m s^-1 in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force ?
Answer:
m = 3.0 kg; u = 2.0 ms^-1,
v = 3.5ms^-1, t = 25 s;
F = m \(\left(\frac{v-u}{t}\right)\) = 3\(\left(\frac{3.5-2}{25}\right)\)
= \(\frac{3 \times 15}{25}=\frac{0.9}{5}\) = 0.18 N
This force acts in the direction of change in velocity.

Question 5.
A man in a lift feels an apparent weight W when the lift is moving up with a uniform acceleration of 1/3^rd of the acceleration due to gravity. If the same man were in the same lift now moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity, then what is his apparent weight ?
Answer:
When the lift is moving up, a = \(\frac{g}{3}\)
Apparent weight W = m(g + a)
= m(g + \(\frac{g}{3}\)) = \(\frac{4mg}{3}\)
⇒ mg = \(\frac{3 W}{4}\) …………….. (1)
When the lift is moving down, a = \(\frac{g}{2}\)
Apparent weight W¹ = m(g – a)
= m(g – \(\frac{g}{2}\)) = \(\frac{mg}{2}\)
∴ |W¹| = \(\frac{\left[\frac{3 W}{4}\right]}{2}=\frac{3 W}{8}\)

Question 6.
A container of mass 200 kg rests on the back of an open truck, if the truck accelerates at 1.5 m/s², what is the minimum coefficient of static friction between the container and the bed of the truck required to prevent the container from sliding off the back of the truck ?
Answer:
m = 200kg, a = 1.5 ms^-2, g = 9.8 ms^-2
ma = μ_smg
μ_s = \(\frac{a}{g}=\frac{1.5}{9.8}\) = 0.153

Question 7.
A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments. One of them starts moving vertically down- wards with an initial speed of 10 m/s. If acceleration due to gravity is 10 m/s². What is the separa-tion between the fragments 2s after the explosion ?
Answer:
Bomb explodes into two fragments say 1 and 2.
For 1^st fragment, u₁ = 10m/s, t = 2 sec; g = 10 m/s^-2, s₁ = ?
Now displacement of 1^st fragment,
s₁ = u₁t + \(\frac{1}{2}\) gt²
= 10 × 2 × \(\frac{1}{2}\) × 10 × 2² = 40m

2^nd fragment is move opposite to 1^st fragment like an object from a tower.
For 2^nd fragment u₂ = – u₁ = 10m/s
t = 2 sec; g = 10 m/s²
Now displacement of 2^nd fragment
s₂ = u₂t + \(\frac{1}{2}\) gt²
= 10 × 2 × \(\frac{1}{2}\) × 10 × 2² = 0
∴ The seperation of two fragments
= s₁ + s₂ = 40 + 0 = 40 m

Question 8.
A fixed pulley with a smooth grove has a light string passing over it with a 4 kg attached on one side and a 3 kg on the other side. Another 3 kg is hung from the other 3 kg as shown with another light string. If the system is released from rest, find the common acceleration ? (g = 10 m/s²).

Answer:
From fig,
m₁ = 3 + 3
= 6 kg
m₂ = 4 kg
g = 10ms^-2
Acceleration of the system,
a = \(\left(\frac{\dot{m}_1-m_2}{m_1+m_2}\right) g=\left(\frac{6-4}{6+4}\right) \times 10\) = 2 ms^-2

Question 9.
A block of mass of 2 kg slides on an inclined plane that makes an angle of 30% with the horizontal. The coefficient of friction between the block and the surface is \(\frac{\sqrt{3}}{2}\).
(a) What force should be applied to the block so that it moves up without any acceleration ?
(b) What force should be applied to the block so that it moves up without any acceleration ?
Answer:
m = 2kg; θ = 30°; µ = \(\frac{\sqrt{3}}{2}\)
i) F_down = mg (sinθ – µ cosθ)
= 2 × 9.8 (sin30° – \(\frac{\sqrt{3}}{2}\)cos30°)
= 2 × 98 [\(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) × \(\frac{\sqrt{3}}{2}\)] = 49 N

ii) F_up = mg (sinθ + µ cosθ)
= 2 × 9.8 [sin30° + \(\frac{\sqrt{3}}{2}\) × cos30°]
= 2 × 9.8 [\(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) × \(\frac{\sqrt{3}}{2}\)] = 24.5 N

Question 10.
A block is placed on a ramp of parabolic shape given by the equation g = \(\frac{x^2}{20}\), see If i_s = 0.5, what is the maximum height above the ground at which the block can be placed without slipping ? (tan θ = µ_s = \(\frac{\mathrm{d} y}{\mathrm{~d} x}\))

Answer:

Question 11.
A block of metal of mass 2 kg on a horizontal table is attached to a mass of 0.45 kg by a light string passing over a frictionless pulley at the edge of the table. The block is subjected to a horizontal force by allowing the 0.45 kg mass to fall. The coefficient of sliding friction between the block and table is 0.2.

Calculate (a) the initial acceleration, (b) the tension in the string, (c) the distance the block would continue to move if. after 2 s of motion, the string should break.
Answer:
Here, m₁ = 0.45kg
m₂ = 2kg
µ = 0.2
a) Initial acceleration.
a = \(\left(\frac{m_1-\mu m_2}{m_1+m_2}\right) g=\left[\frac{0.45-0.2 \times 2}{0.45+2}\right] \times 9.8\)
a = 0.2ms^-2

b) From fig, we have
T – f = m₂ a
T – 3.92 = 2 × 0.2
[∵ f = µm₂g = 0.2 × 2 × 9.8] = 3.92 N
⇒ N = 0.4 + 3.92 = 4.32 N

c) Velocity of string after 2 sec
= u in this case: µ’ = 0
Stoping distance s = \(\frac{\mu^2}{2 \mu g}=\frac{0.4 \times 0.4}{2 \times 0.2 \times 9.8}\) = 0.0408 m

Question 12.
On a smooth horizontal . surface a block A of mass 10 kg is kept. On this block a second block B of mass 5 kg is kept. The coefficient of friction between the two blocks is 0.4. A horizontal force of 30 N is applied on the lower block as shown.

The force of friction between the blocks is (take g = 10 m/s²).
Answer:
Here m_A = 10kg; m_B = 5kg;
F = 30N; µ = 0.4
F = (m_A + m_B)a
⇒ a = \(\frac{F}{\left(m_A+m_B\right)}\)
= \(\frac{30}{10+5}\) = 2ms^-2
f = m_Ba = 5 × 2 = 10 N

Additional Problems

(For simplicity in numerical calculations, take g = 10 ms^-2)

Question 1.
Give the magnitude and direction of the net force acting on
a) a drop of rain falling down with a constant speed,
b) a cork of mass 10 g floating on water,
c) a kite skillfully held stationary in the sky,
d) a car moving with a constant velocity of 30 km/h on a rough road,
e) a high-speed electron in space far from all material objects and free of electric and magnetic fields.
Answer:
a) As the rain drop is falling with a constant speed, its acceleration a = 0. Hence net force F = ma = 0.
b) As the cork is floating on water, its weight is being balanced by the upthrust (equal-to weight of water displaced). Hence net force on the cork is zero.
c) As the kite is held stationary, net force on the kite is zero, in accordance with Newton’s first law.
d) Force is being applied to overcome the force of friction. But as velocity of the car is constant, its acceleration, a = 0. Hence net force on the car F = ma = 0.
e) As no field (gravitational / electric / magnetic) is acting on the electron, net force on it is zero.

Question 2.
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble.
a) during its upward motion,
b) during its downward motion
c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at ah angle of 45° with the horizontal direction ?
Ignore are resistance.
Answer:
When a body is thrown vertically upwards (or) it moves vertically downwards, gravitational pull of earth gives it a uniform acceleration a = + g = + 9.8 ms^-2 in the downward direction. Therefore, the net force on the pebble in all the three cases is vertically downwards.
As m = 0.05 kg and a = + 9.8 m/s²
In all the three cases,
∴ F = ma
= 0.05 × 9.8 = 0.49 N, vertically downwards.
If the pebble were thrown at an angle of 45° with the horizontal direction, it will have horizontal and vertical components of velocity. These components do not affect the force on the pebble. Hence our answers do not alter in any case. However in each case (C), the pebble will not be at rest. It will have horizontal component of velocity at highest point.

Question 3.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg.
a) just after it is dropped from the window of stationary train,
b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
c) just after it is dropped from the window of a train accelerating with 1 ms^-2,
d) lying on the floor of a train which is accelerating with 1 ms-2, the stone being at rest relative to the train.
Answer:
a) Here, m – 0.1 kg, a = +g = 9.8 m/s²
Net force, F = ma = 0.1 × 9.8 = 0.98 N This force acts vertically downwards.

b) When the train is running at a constant velocity its acc. = 0. No force acts on the stone due to this motion.
Therefore, force on the stone F = weight of stone mg = 0.1 × 9.8 = 0.98 N
This force also acts vertically downwards.

c) When the train is accelerating with 1m/s², an additional force F¹ = ma = 0.1 × 1 = 0.1 N acts on the stone in the horizontal direction. But once the stone is dropped from the train, F¹ becomes zero and the net force on the stone is F = mg = 0.1 × 9.8 = 0.98 N, acting vertically downwards.

d) As the stone is lying on the horizontal direction of motion of the train. Note the weight of the stone in this case is being balanced by the normal reaction.

Question 4.
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is ;
i) T,
ii) T – \(\frac{\mathrm{m} v^2}{l}\),
iii) T + \(\frac{\mathrm{m} v^2}{l}\),
iv) 0
T is the tension in the string. (Choose the correct Answer)
Answer:
The net force on the particle directed towards the center is T. This provides the necessary centripetal force to the particle moving in the circle.

Question 5.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop ?
Answer:
Here, F = -50 N, m = 20 kg
μ = 15 m/s, v = 0, t = ?
From F = ma,
a = \(\frac{F}{m}=\frac{-50}{20}\) = -2.5 m/s²
From v = u + at
0 = 15 – 2.5t
t = \(\frac{15}{2.5}\) = 6s.

Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms^-1 to 3.5 ms^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ?
Answer:
Here m = 3.0 kg
μ = 2.0 m/s
v = 3.5 m/s,
t = 25s, F = ?
F = ma = \(\frac{m(v-u)}{t}=\frac{3.0(3.5-2.0)}{25}\)
= 0.18 N.
The force is along the direction of motion.

Question 7.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body ?
Answer:

θ = 36° 52
This the direction of resultant force and hence the direction of acceleration of the body, fig.
Also a = \(\frac{F}{m}=\frac{10}{5}\) = 2ms^-2

Question 8.
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three wheeler is 400 kg and the mass of the driver is 65 kg.
Answer:
Here, u = 36km/h = 10 m/s, v = 0, t = s
m = 400 + 65 = 465 kg
Retarding force
F = ma = \(\frac{m(v-u)}{t}=\frac{465(0-10)}{4}\) = -1162 N.

Question 9.
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Answer:
Here m = 20,000 kg = 2 × 10⁴ kg
Initial ace = 5 m/s²
Thrust, F = ?
Clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 m/s². Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 m/s² As thrust = force = mass × acceleration
∴ F = 2 × 10⁴ × 14.8 = 2.96 × 10⁵N

Question 10.
A body of mass 0.40 kg moving initially with a constant speed of 10 ms-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at the time to be x = 0 and predict its position at t = -5 s, 25 s, 100 s.
Answer:
Here, m = 0.40 kg, µ = 10m/s due N
F = – 8.0 N
a = \(\frac{F}{m}=\frac{-8.0}{0.40}\) = -20 m/s²
for 0 ≤ t ≤ 30s.
i) At t = -5s, x = Ut = 10 × (-5) = -50 m

ii) At t = 25s, x = Ut + \(\frac{1}{2}\) at²
= 10 × 25 + \(\frac{1}{2}\) (-20) (25)² = – 6000m

iii) At t = 100s, The problem is divided into two parts, upto 30s, there is force/acc.
∴ from x₁ = Ut + \(\frac{1}{2}\) at²
= 10 × 30 + \(\frac{1}{2}\) (-20) (30)²
= -8700
At t = 30s, v = U + at = 10 – 20 × 30 = – 590 m/s,
∴ for motion from 30s to 100s
x₂ = vt = – 590 × 70 = – 41300 m
x = x₁ + x₂ = -8700 – 41300
= -50,000 m = – 50km.

Question 11.
A truck starts from rest and accelerates uniformly at 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity and (b) acceleration of the stone at t = 11 s ? (Neglect air resistance).
Answer:
Here, u = 0, a = 2 m/s², t = 10s

Let v be the velocity of the truck when the stone is dropped.
From v = u + at
v = 0 + 2 × 10 = 20m/s
a) Horizontal velocity of stone, when it is dropped, v_x = v = 20 m/s.
As air Resistance is neglected, v_x remains constant.
In the vertical direction, initial velocity of stone, µ = 0, a = g = 9.8 m/s²,
time t = 11 – 10 = 1s
From v = u + at
v_y = 0 + 9.8 × 1 = 9.8 ms^-1
Resultant velocity of stone, OC is given by
v = \(\sqrt{v_x^2+v_y^2}=\sqrt{20^2+(9.8)^2}\)
v = 22.3 m/s.
Let θ is the angle with the resultant velocity OC of stone makes with the horizontal direction OA, then from fig.
tan θ = \(\frac{v_y}{v_x}=\frac{9.8}{20}\) = 0.49
∴ θ = 29°

b) The moment the stone is dropped from the car, horizontal force on the stone = 0. The only acceleration the path followed by the stone is, however parabolic.

Question 12.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms^-1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
Answer:
a) We shall study in unit x that at each extreme position, velocity of the bob is zero. If the string is cut at the extreme position, it is only under the action of ‘g’. Hence the bob will fall vertically downwards.

b) At the mean position, velocity of the bob is 1m/s. along the tangent to the arc, which is in the horizontal direction. If the string is let at mean position, the bob will be have as a horizontal projectile. Hence it will follow a parabolic path.

Question 13.
A man of mass 70 kg stands on a weighing scale in a lift which is moving
a) upwards with a uniform speed of 10 ms^-1
b) downwards with a uniform acceleration of 5 ms^-2
c) upwards with a uniform acceleration of 5 ms^-2
What would be the readings on the scale in each case ?
d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ?
Answer:
Here, m = 70 kg, g = 9.8 m/s²
The weighing machine in each case measures the reaction R i.e. the apparent weight.
a) When lift moves upwards with a uniform speed, its accelerations is zero.
R = mg = 70 × 9.8 = 686N

b) When the lift moves downwards with a = 5 m/s²
R = m(g – a) = 70 (9.8 – 5) = 336 N

c) When the lift moves upwards with a = 5 m/s²
R = m(g + a) = 70 (9.8 + 5) = 1036 N
If the lift were to come down freely under gravity, downward acc. a = g
∴ R = m (g – a) = m(g – g) = zero

Question 14.
Figure shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t < 4 s, 0 < t < 4s ? (b) impulse at t = 0 and t = 4 s ?
(Consider one-dimensional motion only)

Answer:
i) For t < 0, the position time graph is OA which means displacement of the particle is zero.
i.e. particle is at rest at the origin. Hence force on the particle must be zero.

ii) For 0 < t < 4s, s, the position time graph OB has a constant slope. Therefore, velocity of the particle is constant in this interval i.e. particle has zero acceleration. Hence force on the particle must be zero. iii) For t > 4s, the position time graph BC is parallel to time axis. Therefore, the particle remains at a distance 3m from the origin, i.e. it is at rest. Hence force on the particle is zero.

iv) Impulse at t = 0 :
We know. Impulse = change in linear momentum, Before t = 0 particle is at rest i.e. u = 0. After t = 0, particle has a constant velocity v = \(\frac{3}{4}\)
= 0.75 m/s

∴ Impulse = m(v – u)
= u (0.75-0)
= 3kg m/s
∴ Impulse at t = 4s
Before t = 4s, particle has a constant velocity u = 0.75 m/s
After t = 4s, particle is at rest i.e. v = 0
Impulse = m(v – u) = 4 (0 – 0.75) = 3kg m/s.

Question 15.
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string, a horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of String. What is the tension in the string in each case ?
Answer:
Here, F = 500 N
m₁ = 10kg, m₂ = 20kg
Let T be the tension in the string and a be the acceleration of the system, in the direction of the force applied.
a = \(\frac{F}{m_1+m_2}=\frac{500}{10+20}=\frac{50}{3}\) m/s²
a) When force is applied on heavier block,
T = m₁ a = 10 × \(\frac{50}{3}\) N
T = 166.66 N

b) When force is applied on lighter block,
T = m₂a = 20 × \(\frac{50}{3}\) N
= 333.33 N
Which is different from value T in case (a) Hence our answer depends on which mass end, the force is applied.

Question 16.
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Answer:
Here, m₂ = 8 kg, m₁ = 12kg
as a = \(\frac{\left(m_1-m_2\right) g}{m_1+m_2}\)
a = \(\frac{(12-8) 9.8}{12+8}=\frac{39.2}{20}\)
= 1.96 m/s²
Again
T = \(\frac{2 m_1 m_2 g}{m_1+m_2}=\frac{2 \times 12 \times 8 \times 9.8}{(12+8)}\) = 94.1 N

Question 17.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Answer:
Let m₁m₂ be the masses of products and \(\vec{v}_1, \vec{v}_2\) be their respective velocities. Therefore total linear momentum after disintegration = \(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}\). Before disintegration, the nucleus is at rest. There, its linear momentum before dis-integration is zero. According to the principle of conservation of linear momentum.
\(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}\) (Or) \(\vec{v}_2=\frac{-m_1 \vec{v}_1}{m_2}\)
Negative sign shows that \(\overrightarrow{v_1}\) and \(\overrightarrow{v_2}\) are in opposite direction.

Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other ?
Answer:
Here, initial momentum of the ball,
A = 0.05 (6) = 0.3 kg m/s
As the speed is reversed on collision, final momentum of the ball A = 0.05 (-6)
= – 0.3 kg ms^-1
Impulse imparted to ball A = Change in momentum of ball A = Final momentum – Intial momentum = – 0.3 – 0.3 = – 0.6 kg m/s

Question 19.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun ?
Answer:
Here, mass of shell m = 0.02 kg
Mass of gun M = 100 kg
Muzzle speed of shell v = 80 m/s
Recoil speed of gun V = ?
According to the principle of conservation of linear momentum mv + MV = 0
(Or) V = \(\frac{-\mathrm{mv}}{\mathrm{M}}=\frac{-0.02 \times 80}{100}\) = 0.016 m/s

Question 20.
A batsman deflects a ball by the angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball ? (Mass of the balls is 0.15 kg).
Answer:
In fig. the ball hits the bat KL along AO and is deflected by the bat along OB. Where LAOB = 45°. ON is normal to the portion of the bat KL deflecting the ball.

∴ θ = LNOA
= 45°/2 = 22.5°
Intial vel along AO = u = 54 km/h
= 15 m/S₁ and mass of ball m = 0.15 kg Intial velocity along AO has the two rectangular components : u cos θ along NO produced and u sin 6 along the horizontal OL.
Final velocity along OB has the same magnitude = u
It is resolved into two rectangular components u cos 6 along ON and u sin θ along OL. We observe that there is no change in velocity along the horizontal, but velocity along vertical is just reserved.
∴ Impulse imparted to the ball
= Change in linear momentum of the ball
= m u cos θ – (- m u cos θ)
= 2 m u cos θ
= 2 × 0.15 × 15 cos 22.5°
= 4.5 × 0.9239
= 4.16 kg m/s

Question 21.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./ min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Answer:
Here, m = 0.25 kg, r = 15m
n = 40 rpm = \(\frac{40}{60}\), rps = \(\frac{2}{3}\), T = ?
T = mrw² = mr(2 π)² = 4 π²m²
T = 4 × \(\left(\frac{22}{7}\right)^2\) × 0.25 × 1.5 × \(\left(\frac{2}{3}\right)^2\) = 6.6N
If T_max = 200 N. then from
T_max = \(\frac{\mathrm{mv}_{\max }^2}{r}\)
\(v_{\max }^2=\frac{T_{\max } \times r}{m}=\frac{200 \times 1.5}{0.25}\) = 1200
v_max = \(\sqrt{1200}\) = 34.6 m/s

Question 22.
If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :
a) the stone moves radially outwards,
b) the stone flies off tangentially from the instant the string breaks,
c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?
Answer:
The instant the string, breaks, the stone flies off tangentialy, as per Newton’s first law of motion.

Question 23.
Explain why
a) a horse cannot pull a cart and run in empty space,
b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
c) it is easier to pull a lawn mower than to push it.
d) a cricketer moves his hands back-wards while holding a catch.
Answer:
a) While trying to pull a cart, a horses pushes the ground backwards with a certain force at an angle. The ground offers an equal reaction in the opposite direction on the feet of the horse. The forward component of this reaction is responsible for the motion of the cart. In empty space, there is no reaction and hence a horse cannot pull the cart and run.

b) This is due to inertia of motion. When the speeding bus stops suddenly, lower part of the body in contact with the seats stops. The upper part of the body of the passengers tend to maintain its uniform motion. Hence the passengers are thrown forward.

c) While pulling a lawn roller, force is applied upwards along the handle. The vertical component of this force is upwards and reduces the effective weight of the roller, fig. (a) while pushing a lawn roller. Face is applied downwards along the handle. The vertical component of this force is downwards and increases the effective weight of the roller, fig. (b). As the effective weight is lesser in case of pulling than in a case of pushing, therefore pulling is easier than pushing.

d) While holding a catch, the impulse receives by the hands F × t = Change in linear momentum of the ball is constant. By moving his hands, backwards, the cricketer increases the time of impact (t) to complete the catch. As t increases, F decreases and as a reaction, his hands are not hurt severly.

Question 24.
Figure shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body ? What is the magnitude of. each impulse ?

Answer:
Here, m = 0.04 kg. position time graph shows that the particle moves from x = 0 at 0 to x = 2 cm at A in 2 sec.
As x – t graph is a st. line, the motion is with a constant velocity.
μ = \(\frac{(2-0) \mathrm{cm}}{(2-0) \mathrm{s}}\) = 1 cm s^-1
= 10^-2 ms^-1

From x = 2 cm at A, particle goes to x = 0 at B in 2 sec.
As AB is a stline, motion is with constant velocity v = 1 cm/s = 10^-2 m/s
Negative sign indicates the reversal of direction of motion. This is being repeated. We can visualise a ball moving between two walls located at x = 0 and x = 2 cm, getting rebounded repeatedly on striking against each wall on every collision with a wall, linear momentum of the ball changes. Therefore, the ball receives impulse after every two seconds. Magnitude of impulse = Total change in linear momentum.
= mu -(mv) = mu – mv = m(u – v)
= 0.04(10^-2 + 10^-2)
= 0.08 × 10^-2
= 8 × 10^-4 kg m/s

Question 25.
Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms-2. What is the net force on the man ? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt ? (Mass of the man = 65 kg).

Answer:
Here, acceleration of conveyer belt, a = 1 m/s²
As the man is standing stationary w.r.t the belt, acceleration of the man = Acceleration of belt = a = 1 m/s²
As m = 65 kg
∴ Net force on the man, F = ma = 65 × 1 = 65 N
Now, µ = 0.2
Force of limiting friction- F = µR = µmg
It the man remains stationary upto max.acc. a of the belt, then F = ma¹ = g mg
a¹ = mg = 0.2 × 9.8 = 1.96 ms^-2

Question 26.
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : (Choose the correct alternative)
Lowest Point – Highest Point
a) mg – T₁ – mg + T₂
b) mg + T₁ – mg – T₂
c) mg + T₁ – (mυ₁²)/R – mg – T₂ + (mυ₁²)/R
d) mg – T₁ – (mυ₁²/R – mg + T₂ + (mυ₁²)/R
T₁ and υ₁ denote the tension and speed at the lowest point T₂ and υ₂ denote corresponding values at the highest point.
Answer:
The net force at the lowest point α is
F_L = (mg – T₁) and the net force at the highest point H is F_H = mg + T₂. Therefore, alternative (a) is correct.

Question 27.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms^-2. The crew and the passengers weight 300 kg. Give the magnitude and direction of the
a) force on the floor by the crew and passengers,
b) action of the rotor of the helicopter on the surrounding air,
c) force on the helicopter due to the surrounding air.
Answer:
Here, mass of the helicopter, m₁ = 100 kg
Mass of the crew and passengers m₂ = 300 kg
Upward acceleration a = 15 ms^-2 and g = 10 ms^-2
a) Force on the floor of helicopter by the crew and passengers = appeard weight of crew and passengers = m₂(g + a)
= 300(10 + 15) = 7500 N

b) Action of rotor of helicopter on surrounding air is obviously vertically downwards because helicopter rises on account reaction to this force. Thus, force of action F = (m₁ + m₂) (g + a)
= (1000 + 300) (10 + 15)
= 1300 × 25 = 32500 N

c) Force on the helicopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore, force of reaction F¹ = 32500 N, vertically upwards.

Question 28.
A stream of water flowing horizontally with a speed of 15 ms^-1 gushes out of a tube of cross-sectional area 10^-2m² and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?
Answer:
Here v = 15 ms^-1
Area of cross section a = 10² m^-2, F = ?
Volume of water pushing out/sec
= a × v = 10^-12 × m³ s^-1
As density of water is 10³ kg/m², therefore, mass of water striking the wall per sec.
m = (15 × 10^-2) × 10³ = 150 kg/s
Change in linear momentum
As F = \(\frac{\text { Change in linear momentum }}{\text { Time }}\)
∴ F = \(\frac{m \times v}{t}=\frac{150 \times 15}{1}\) = 2250 N

Question 29.
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of
a) the force on the 7^th coin (counted from the bottom) due to all the coins on its top
b) the force on the 7^th coin by the eighth coin,
c) the reaction of the 6^th coin on the 7^th coin.
Answer:
a) The force on 7^th coin is due to weight of the three coins lying above it. Therefore,
F (3m) kgf = (3mg)N
Where g is acceleration due to gravity. This force acts vertically downwards.

b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e.,
F = 2m + m = 3(m) kgf = (3mg)N
The force acts vertically downwards.

c) The sixth coin is under the weight of four coins above it.
Reaction, R = – F = -4m(kgf) = -(4 mg)N
Minor sign indicates that the reaction acts vertically upwards, opposite to the weight.

Question 30.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop ?
Answer:
Here θ = 15°
v = 720 km/h =\(\frac{720 \times 1000}{60 \times 60}\) = 200 ms^-1
g = 9.8 ms^-2
From tan θ = \(\frac{\mathrm{v}^2}{\mathrm{rg}}\)
v² = rg tan θ
r = \(\frac{v^2}{g \tan \theta}=\frac{(200)^2}{9.8 \times \tan 15^{\circ}}\)
= 15232 m = 15.232 km

Question 31.
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What provides the centripetal force required for this purpose – The engine or the rails ? What is the angle of banking required to prevent wearing out of the rail ?
Answer:
The centripetal force is provided by the lateral thrust exerted by the rails on the wheels. By Newton’s 3^rd law, the train exerts an equal and opposite thrust on the rails causing its wear and tear.
Obviously, the outer rail will wear out faster due to the larger force exerted by the train on it.
Here v = 54 km/h =\(\frac{54 \times 1000}{60 \times 60}\) = 15 m/s
g = 9.8 ms^-2
As tan θ = \(\frac{v^2}{r g}=\frac{15 \times 15}{30 \times 9.8}\) = 0.76
∴ θ = tan^-1 0.76 = 37.4°

Question 32.
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ?

Answer:
Here, mass of block m = 25 kg
Mass of man M = 50 kg
Force applied to lift the block
F = mg = 25 × 9.8 = 245 N
Weight of Man W = mg = 50 × 9.8 = 490 N
a) When block is raised by man as shown in Fig. force is applied by the man in the upward direction. This increases the apparent weight of the man. Hence action on the floor.
W¹ = W + F
= 490 + 245
= 735 N

b) When block is raised by man as shown in Fig. force is app ed by the man in the downward direction. This decreases the apparent weight of the man. Hence action on the floor in this case would be
W¹ = W – F = 490 – 245 = 245 N
As the floor yeilds to a normal force of 700 N, the mode (b) has to be adopted by the man to lift the block

Question 33.
A monkey of mass 40 kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break : the money
a) climbs up with an acceleration of 6 m s^-2
b) climbs down with an acceleration of 4 ms^-2
c) climbs up with a uniform speed of 5 ms^-1
d) falls down the rope nearly freely under gravity ?
(ignore the mass of the rope)

Answer:
Here, mass of monkey m = 40 kg
Maximum tension the rope can stand T = 600 N
In each case, actual tension in the rope will be equal to apparent weight of money (R). The rope will break when R exceeds T.
a) When monkey climbs up with a = 6 ms^-2
R = m(g + a)
= 40(10 + 6)
= 640 N (Which is greater than T)
Hence the rope will break.

b) When monkey climbs down with a = 4 ms^-2,
R = m(g – a) = 40(10 – 4) = 240 N
Which is less than T.
∴ The rope will not break.

c) When monkey climbs up with a uniform speed v = 5 ms^-1.
Its acceleration, a = 0
∴ R = mg = 40 × 10 = 400 N,
which is less than T.
∴ The role will not break

d) When monkey falls down the rope nearly freely under gravity
a = g
∴ R = m(g – a) = m(g – g) = zero
Hence the rope will not break.

Question 34.
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig.). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the wall is removed ? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μ_s and μ_k.

Answer:
Here, mass of body A, m₁ = 5 kg
Mass of body B, m₂ = 10 kg

Coefficient of friction between the bodies and the table, μ = 0.15
Horizontal force applied on A,
F = 200 N

a) Force of limiting friction acting to the left
f = μ (m₁ + m₂)g
= 0.15(5 + 10) × 9.8 = 22.05 N
∴ Net force to the right exerted on the partition
F’ = 200 – 22.05 = 177.95 N
Reaction of partition = 177.95 N to the left.

b) Force of limiting friction acting on body A
f₁ = μm₁g = 0.15 × 5 × 9.8
= 7.35 N
∴ Net force exerted by body A on body B.
F” = F -f₁ = 200 – 7.35
= 19265 N
This is to the right
Reaction of body B on body A = 192.65
N to the left when the portion is removed, the system of two bodies will move under the action of net force.
F₁ = 177.95 IM
Acceralation produced in the system
a = \(\frac{F^1}{m_1+m_2}=\frac{177.95}{5+10}\)
= 11.86 ms^-2
Force producing motion in body A
F₁ = m₁ a = 5 × 11.86
= 59.3 N
∴ Net force exerted by A on body B, when partition is removed
= F” – F₁ = 192.65 – 59.3
= 133.35 N.
Hence the reaction of body B on A, when partition is removed = 133.35 N to the left.
Thus answers to (b) do change.

Question 35.
A block of mass 15 kg is placed on a long trolley. The cofficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Answer:
a) Flere, m = 15 kg;
μ = 0.18,
a = 0.5 ms^-2
t = 20 s

Force on the block due to motion of the trolly F’ = ma = 15 × 0.5 = 7.5 N
Force of limiting friction on the block
= F = μR = μmg
= 0.18 × 15 × 9.8 = 26.46 N
This opposes the motion of the block. The block shall not move. The force of static friction F will adjust itself equal and opposite to F’, the applied force.

Flence to a stationary observer on the ground, the block will appear to be at rest relative to the trolly. When trolly moves with uniform velocity, the block will continue to be stationary. Because in that case, forward force is zero. Force of friction alone is acting on the block.

b) An observer moving with the trolly has accelerated motion. The observer is therefore non-inertial.. The law of inertia is no longer valid.

Question 36.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms^-2. At what distance from the starting point does the box fall of the truck?

Answer:
Here mass of the box m = 40 kg
Acceleration of truck a = 2 ms^-2
Distance of box from open end S = 5m
Coeff. of friction μ = 0.15
Force on the box due to accelarated motion of the truck, F = ma = 40 × 2 = 80 N
This force F is in the forward direction Reaction F’ on the box is equal to F = 80 N in the backward direction. This is opposed by force of limiting friction
f = μ R = μ mg
= 0.15 × 40 × 9.8
= 58.8 N in the forward direction
∴ Net force on the box in the backward direction is p = F’ – F = 80 – 58.8 = 21.2 N
Backward acceleration produced in the box
a = \(\frac{p}{m}=\frac{21.2}{40}\) = 0.53 ms^-2
It t is time taken by the box to travel S = 5 metre and fall off the truck, then from
S = ut + \(\frac{1}{2}\) at²
5 = 0 × t + \(\frac{1}{2}\) × 0.53 t²
t = \(\frac{\sqrt{5 \times 2}}{0.53}\) = 4.345
If the truck travels a distance x during this time, then again from
S = ut + \(\frac{1}{2}\) at²
x = 0 × 4.34 + \(\frac{1}{2}\) × 2(4.34)² = 18.84 m

Question 37.
A disc revolves with a speed of 33\(\frac{1}{3}\) rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ?
Answer:
The coin revolves with the record in the case when the force of friction is enough to provide the necessary centripetal force. If this force is not sufficient to provide centripetal force, the coin slips on the record.
Now the frictional force is µR where R is the
normal reaction and R = mg
Hence force of friction = µ mg and centripetal force required is \(\frac{m v^2}{r}\) or mrw²
µ₁w are same for both the coins and we have different values of r for the two coins.
So to prevent slipping i.e., causing coins to rotate µ mg > mrω² or µg > rω² …………….. (1)
For 1^st coin
r = 4 cm = \(\frac{4}{100}\) m
n = 33 \(\frac{1}{2}\) rev/min = \(\frac{100}{3 \times 60}\) rev/sec
w = 2πn = 2π × \(\frac{100}{180}\) = 3.49 S^-1
∴ rw² = \(\frac{4}{100}\) × (3.49)² = 0.49 ms^-2 and
µg = 0.15 × 9.8 = 1.47 ms^-2
As µg > rw², there fore this coin Will revolve with the record.
Note: We have nothing to do with the radius of the record = 15 cm

Question 38.
You may have seen in a circus a motor-cyclist driving in verticle loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a verticle loop if the radius of the chamber is 25 m ?
Answer:
At the uppermost point of the death well, with no support from below, the motorcyclist does not drop down. This is because his weight is being balance by the centrifugal force. Infact, the weight of the motorcyclist is spent up in providing the necessary centripetal force to the motorcyclist and hence he does not drop drown.

At the uppermost point, R + mg = \(\frac{m v^2}{r}\),
where R is the normal reaction (downwards) on the motor cyclist by the ceiling of the chamber.
Speed will be minimum, when N = 0
∴ mg = \(\frac{m v^2}{r}\) or
v = \(\sqrt{\mathrm{rg}}=\sqrt{25 \times 10}\)
= 15.8 m/s

Question 39.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed ?
Answer:
Here m = 70 kg, r = 3
n = 200 rpm = \(\frac{200}{60}\) rps
µ = 0.15
w = ?
The horizontal force N by the wall on the man provides the necessary centripetal force = mrω². The frictional force (f) in this case is vertically upwards opposing the weight (mg) of the man.
After the floor is removed, the man will remain stuck to the wall, when
mg = f < µ N i.e., mg < µ mr ω² or g < µ r ω²
∴ Minimum angular speed of rotation of the cylinder is ω = \(\sqrt{\frac{g}{\mu \mathrm{r}}}=\sqrt{\frac{10}{0.15 \times 3}}\)
= 4.7 rad/s

Question 40.
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency no. Show that a small bead on the wire loop remains at its
lowermost point for ω ≤ \(\sqrt{g / R}\) . What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = 2g/R ?
Answer:
In fig. we have show that radius vector joining the bead to the centre of the wire makes as angle θ with the vertical downward direction. It N is normal reaction, then as it is clear from the figure,

mg = N cos θ ………… (1)
m rω² = N sin θ …………….. (2)
Or m(R sin θ) ω² = N sin θ
Or mRω² = N
From (i) mg = mRω² cos θ
Or cos θ = \(\frac{\mathrm{g}}{\mathrm{R} \omega^2}\) ……………….. (3)
As |cos θ| ≤ 1, therefore, bead will remain at its lower most point for
\(\frac{\mathrm{g}}{\mathrm{R} \omega^2}\) ≤ 1 or w ≤ \(\sqrt{\frac{g}{R}}\)
When ω = \(\sqrt{\frac{2 g}{R}}\), from (iii),
cos θ = \(\frac{g}{R}\left(\frac{R}{2 g}\right)=\frac{1}{2}\)
∴ θ = 60°

Textual Examples

Question 1.
An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a constant rate of 100 ms^-2. What is the acceleration of the astronaut the instant after he is outside the spaceship ? (Assume that there are no nearby stars to exert gravitational force on him).
Answer:
Since there are no nearby starts to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero. By the first law of motion the acceleration of the astronaut is zero.

Question 2.
A bullet of mass 0.04 kg moving with a speed of 90 m s^-1 enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet ?
Answer:
The retardation ‘a’ of the bullet (assumed constant) is given by
a = \(\frac{-u^2}{2 s}=\frac{-90 \times 90}{2 \times 0.6}\) m s^-2 = -6750 m s^-2
The retarding force, by the second law of motion, is = 0.04 kg × 6750 ms^-2 = 270 N
The actual resistive force, and therefore, retardation of the bullet may not be uniform. The answer therefore, only indicates the average resistive force.

Question 3.
The motion of a particle of mass m is described by y = ut + \(\frac{1}{2}\)at². Find the force acting on the particle.
Answer:
We know, y = ut + \(\frac{1}{2}\) gt²
Now, υ = \(\frac{\mathrm{dy}}{\mathrm{dt}}\)u + gt
acceleration, a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = g
Then the force is given by F = \(\frac{\mathrm{dp}}{\mathrm{dt}}\) = ma
F = ma = mg

Question 4.
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s^-1. If the mass of the ball is 0.15 kg determine the impulse imparted to the ball. (Assume linerar motion of the ball) [A.P. Mar. 17]
Answer:
Change in momentum
= 0.15 × 12-(-0.15 × 12) = 3.6 N s
Impulse = 3.6 N s, in the direction from the batsman to the bowler.
This is an example where the force on the ball by the batsman and the time of contact of the ball and the bat are difficult to know, but the impulse is readily calculated.

Question 5.
Two identical billiard balls strike a rigid wall with the same speed but at different angles and get reflected without any change in speed, as shown in Fig. What is (i) the direction of the force on the wall due to each ball ? (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall ?

Answer:
An instinctive answer to (i) might be that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at 30° to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases. How to find the force on the wall ? The trick . is to consider the force on the wall ? The trick is to consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer (i). Let u be the speed of each ball before and after collision with the wall and m the mass of each bah. Choose the x and y axes as shown in the figure, and consider the change in momentum of the ball in each case :
Case (a) :
(P_x)_initial = mu
(P_x)_initial = 0
(P_x)_final = – mu
(P_y)_final = 0
Impulse is the change in momentum vector.
Therefore,
x-component of impulse = -2mu
y-component of impulse – 0
Impulse and force are in the same direction. Clearly, from above, the force on the ball due to the wall is normal to the wall, along the negative x-direction. Using Newton’s third law of motion, the force on the wall due to the ball is normal to the wall along the positive x-direction. The magnitude of force cannot be ascertained since the small time taken for the collision has not been specified in the problem. .
Case (b) :
(P_x)_initial = mu cos 30°
(P_y)_initial = – m u sin 30°
(P_x)_final = – mu cos 30°
(P_y)_final = -m u sin 30°
Note, while p_x changes sign after collision, p_y does not. Therefore,
x-component of impulse = -2 m u cos 30°
y-component of impulse 0
The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative x direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive x direction.
The ratio of the magnitudes of the impulses imparted to the balls in(a) and (b) is
2 mu / (2 m u cos 30°) = \(\frac{2}{\sqrt{3}}\) ≈ 1.2

Question 6.
See Fig. A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the mid point P of the rope, as show. What is the angle the rope makes with the vertical in equilibrium ? (Take g = 10 ms^-2). Neglect the mass of the rope.

Answer:
Figures (b) and (c) are known as free-body diagrams. Figure (b) is the free-body diagram of W and Fig. (c) is the free-body diagram of point P.
Consider the equilibrium of the weight W.
Clearly, T₂ = 6 × 10 = 60 N.
Consider the equilibrium of the point P under the action of three forces – the tensions T₁ and T₂, and the horizontal force 50 N. The horizontal and vertical components of the resultant force must vanish separately :
T₁ cos θ = T₂ = 60N
T₂ sin θ = T₂ = 50N
which gives that tan θ = \(\frac{5}{6}\) or
θ = tan^-1 (\(\frac{5}{6}\)) = 40°
Note the answer does not depend on the length of the rope (assumed massless) nor on the point at which the horizontal force is applied.

Question 7.
Determine the maximum acceleration of the train in which a box lying on the floor will remain stationary, given that the coefficient of static friction between the box and the train’s floor is 0.15.
Answer:
Since the acceleration of the box is due to the static friction,
ma = f_s ≤ μ_s N = μ_s m g
i.e. a ≤ μ_s g
∴ a_maximum = μ_s g = 0.15 × 10m s^-2
= 1.5 ms^-2

Question 8.
See Fig. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined unti at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?

Answer:
The forces acting on a block of mass …. at rest on an inclined plane are (i) the weight mg acting vertically downwards (ii) the normal force N of the plane on the block, and (iii) the static frictional force fs opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight mg along the two directions shown, we have
m g sin θ = f_s, m g cos θ = N
As θ increases, the self-adjusting frictional force f_s increases until at θ = θ_max, f_s achieves its maximum value, (f_s)max = μ_s N.
Therefore,
tan θ_max = μ_s, or θ_max = tan^-1 μ_s
When θ becomes just a little more than there is a small net force on the block and θ_nm begins to slide. Note that θ_max depends only on μ_s and is independent of the mass of the block
θ_max = 15°, μ_s = tan 15° = t = 0.27

Question 9.
What is the acceleration of the book and trolley system shown in the Fig., if the coefficient of kinetic friction between the trolley and the suriace is 0.04 what is the tension in the string. Take g 10 ms^-2). Neglect the mass of the string.

Answer:
As the string is inextensible, and the pully is smooth, the 3 kg block and the 20 kg trolley both have same magnitude of acceleration. Applying second law to motion of the block (Fig. (b).
30 – T = 3a
Apply the second law to motion of the trolley (Fig. (c))
T – f_k = 20a
Now f_k = μ_k N,
Here μ_k = 0.04
N = 20 × 10 = 200N
Thus the equation for the motion of the trolley is
T – 0.04 × 200 = 20 a or T – 8 = 20a
These equations give a = \(\frac{22}{23}\) m s^-2 = 0.96 m s^-2 and T = 27.1 N.

Question 10.
A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The co-efficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn ?
Answer:
On an unbanked road, frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular turn without slipping. If the speed is too large, or if the turn is too sharp (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by
v_max = \(\sqrt{\mu_s R_g}\) v₂ ≤ μ_s Rg
Now, R = 3m, g = 9.8 m s^-2, μ_s = 0.1.
That is, μ_s R g = 2.94 m² s^-2. v = 18 km/h = 5 m s^-1; i.e., = 25 v². The condition is not obeyed. The cyclist will slip while taking the circular turn.

Question 11.
A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the(a) optimum speed of the race-car to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping?
Answer:
On a banked road, the horizontal component of the normal force and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping. At the optimum speed, the normal reaction’s component is enough to provide the needed centripetal force, and the frictional force is not needed. The optimum speed v₀ is given by
v₀ = (R g tan θ)^1/2-
Here R = 300 m, θ = 15°, g = 9.8 m s^-2; we have
v₀ = 28.1 ms^-1
The maximum permissible speed v_max is given by
v_max = R g\(\left(\frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)^{1 / 2}\) = 38.1 m s^-1

Question 12.
See (Fig.) A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of 0.1 ms^-2. What is the action of the block on the floor (a) before and (b) after the floor yields ? Take g = 10 m s^-2. Identify the action- reaction pairs in the problem.

Answer:
(a) The block is at rest on the floor. Its free- body diagram shows two forces on the block, the force of gravitational attraction by the earth equal to 2 × 10 = 20 N; and the normal force R of the floor on the block. By the First Law, the net force on the block must be zero i.e., R = 20 N. Using third law the action of the block (i.e. the force exerted on the floor by the block) is equal to 20 N and directed vertically downwards.

(b)The system (block cylinder) accelerates downwards with 0.1 ms^-2. The free – body diagram of the system shows two forces on the system : the force of gravity due to the earth (270 N); and the normal force R’ by the floor. Note, the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system,
270 – R’ = 27 × 0.1 N
i.e, R’ = 267.3 N
By the third law, the action of the system on the floor is equal to 267.3 N vertically downward.
Action-reaction pairs
For (a) :
(i) the force of gravity (20N) on the block by the earth (say, action); the force of gravity on the earth by the block (reaction) equal to 20 N directed upwards (not shown in the figure).
(ii) the force on the floor by the block (action); the force on the block by the floor (reaction).

For (b):
(i) the force of gravity (270 N) on the system by the earth (say, action), the force of gravity on the earth by the system (reaction), equal to 270 N, directed upwards (not shown in the figure).

(ii) the force on the floor by the system (action); the force on the system by the floor (reaction). In addition, for (b), the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair.

The important thing to remember is that an action-reaction pair consists of mutual fortes which are always equal and opposite between two bodies. Two forces on the same body which happen to be. equal and opposite can never constitute an action-reaction pair. The force of gravity on the mass in (a) or (b) and the normal force on the mass by the floor are not action-reaction pairs. These forces happen to be equal and opposite for (a) since the mass is at rest. They are not so for case (b), as seen already. The weight of the system is 270 N, while the normal force R’ is 267.3 N.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 4th Lesson Motion in a Plane Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 4th Lesson Motion in a Plane

Very Short Answer Questions

Question 1.
The vertical component of a vector is equal to its horizontal component. What is the angle made by the vector with the x-axis?
Answer:
The horizontal component is equal to the vertical component of a vector.

F cos θ = F sin θ.
Tan θ = 1 .
θ = tan^-1(1) = 45°.

Question 2.
A vector V makes an angle 0 with the horizontal. The vector is rotated through an angle 0. Does this rotation change the vector V ?
Answer:
Yes, it changes the vector.

Question 3.
Two forces of magnitudes 3 units and 5 units act at 60° with each other. What is the magnitude of their resultant ? [A.P. Mar. 15]
Answer:
Let P = 3 units, Q = 5 units, θ = 60°
Resultant (R) = \(\sqrt{p^2+Q^2+2 P Q \cos \theta}\)
= \(\sqrt{3^2+5^2+2 \times 3 \times 5 \times \cos 60^{\circ}}\)
= \(\sqrt{9+25+30 \times \frac{1}{2}}=\sqrt{49}\) = 7 units

Question 4.
A = \(\vec{i}+\vec{j}\). What is the angle between the vector and X-axis ? [T.S., A.P. Mar. 17; Mar. 14, 13]
Answer:
A = \(\vec{i}+\vec{j}\)
cos α = \(\frac{A x}{|A|}\) (∵ A_x = 1)
= \(\frac{1}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt{2}}\)
α = cos^-1 \(\left(\frac{1}{\sqrt{2}}\right)\) = 45°

Question 5.
When two right angled vectors of magnitude 7 units and 24 units combine, what is the magnitude of their resultant ? [A.P. Mar. 16]
Answer:
θ = 90°, P = 7 units, Q = 24 units
R = \(\sqrt{P^2+Q^2+2 P Q \cos \theta}\)
R = \(\sqrt{7^2+24^2+2 \times 7 \times 24 \times \cos 90^{\circ}}=\sqrt{49+576}=\sqrt{625}\) = 25 units.

Question 6.
If p = 2i + 4j + 14k and Q = 4i + 4j + 10k find the magnitude of P + Q. [T.S. – Mar. ‘16, ‘15]
Answer:
P = 2i + 4j + 14k, Q = 4i + 4j + 10k,
\(\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}\) = 2i + 4j + 14k + 4i + 4j + 10k
= 6i + 8j + 24k
|\(\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}\)| = \(\sqrt{6^2+8^2+24^2}=\sqrt{676}\)

Question 7.
Can a vector of magnitude zero have non-zero components?
Answer:
No, the components of a vector of magnitude zero have non-zero components..

Question 8.
What is the acceleration projectile at the top of its trajectory?
Answer:
The acceleration of a projectile at the top of its trajectory is vertically downwards.

Question 9.
Can two vectors of unequal magnitude add up to give the zero vector? Can three unequal vectors add up to give the zero vector?
Answer:
No, two vectors of unequal magnitude cannot be equal to zero. According to triangle law, three unequal vectors in equilibrium can be zero.

Short Answer Questions

Question 1.
State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector. [T.S. – Mar. 16; Mar. 14, 15]
Answer:
Statement: If two vectors acting d a point are represented by the adjacent sides of a parallelogram in magnitude and direction, then their resultant is represented by the diagonal of the parallelogram in magnitude and direction dawn from the same vector.

Explanation L Let two forces \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) point O. Let θ be the angle between two nrces. Let the side O = \(\overrightarrow{\mathrm{P}}\) and OB = \(\overrightarrow{\mathrm{Q}}\). The parallelogram OACB is completed. The points O and C are joined. Now OC = \(\overrightarrow{\mathrm{R}}\)

Resultant magnitude :
In fig \(\overrightarrow{O A}=\vec{p}, \overrightarrow{O B}=\vec{Q}, \overrightarrow{O C}=\vec{R}\)
In the triangle COD. OC² = OD² + CD²
0C² = (OA + AD)² + CD² (: OD = OA + AÐ)
OC² = OA² + AD² + 2OA. AD + CD62²
OC² = OA² + AC² + 2OA. AD …………… (1)
From ∆^le CAD, AD² + CD² = AC²
From ∆^le CAD, cos θ = \(\frac{A D}{A C}\)
AD = AC cos θ ……………. (2)
∴ R² = P² + Q² + 2 PQ cosθ
R = \(\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta}\) ………….. (3)

Resultant direction:
Let (L be the angle made by the resultant vector \(\overrightarrow{\mathrm{R}}\) with \(\overrightarrow{\mathrm{P}}\)
Then tan α = \(\frac{C D}{O D}\)
tan α = \(\frac{C D}{O A+A D}\) ……………… (4)
In the triangle CAD, sinO = \(\frac{C D}{A C}\)
CD = AC sin θ
CD = Q sin θ ……………… (5)
∴ tan α = \(\frac{\mathrm{Q} \sin \theta}{\mathrm{P}+\mathrm{Q} \cos \theta}\) (∵ AD = Q COS θ)
α = tan^-1 \(\left(\frac{Q \sin \theta}{P+Q \cos \theta}\right)\) ……………….. (6)

Question 2.
What is relative motion ? Explain It.
Answer:
Relative velocity is defined as the velocity of one body with respect to another body.

Let us consider two observers A and B are making measurements of an event P in space from two frames of reference as shown in figure. At the beginning let the two origins of the two reference frames coincide and are on the same line.

Let the observer B is moving with a constant velocity V_BA with respect to A. Now we can connect the positions of the event P as measured by A with the position of P as measured by B.

As B is moving with constant velocity at the time of observation of event P, the frame B has moved a distance X_BA with respect to A.
X_PA = X_PB + X_BA …………….. (1)
“The position of P as measured by observer A is equal to the position of P as measured by B plus the position of B as measured by A”.
eq (1) can also be written as V_PA = V_PB + V_BA …………….. (2)

Question 3.
Show that a boat must move at an angle with respect to river water in order to cross the river in minimum time.
Answer:
If the boat has to move along the line AB, the resultant velocity of V_BE should be directed along AB and the boat reaches the point B directly. For this to happen, the boat velocity with respect to water V_BW should be directed such that it makes an angle a with the line AB upstream as shown in the figure.

\(\frac{V_{W E}}{V_{B W}}\) = sin α ⇒ α = sin^-1 \(\left(\frac{V_{\mathrm{WE}}}{\mathrm{V}_{\mathrm{BW}}}\right)\)
and V_BE = \(\sqrt{V_{B W}^2-V_{W E}^2}\)
The minimum time taken by the boat to cross the river
t = \(\frac{A B}{V_{B E}}\)
t = \(\frac{A B}{\sqrt{V_{B W}^2-V_{W E}^2}}\)

Question 4.
Define unit vector, null vector and position vector.
Answer:
Unit Vector : A vector having unit magnitude is called unit vector.
\(\hat{A}=\frac{A}{|A|}\) where \(\hat{A}\) is unit vector.

Null vector : A vector having zero magnitude is called null vector.
Position vector : The position of a particle is described by a position vector which is drawn from the origin of a reference frame. The position vector helps to locate the particle in space.
The position of a particle P is represented by
\(\overrightarrow{O P}=\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}\)

Question 5.
If \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\) prove that the angle between \(\vec{a}\) and \(\vec{b}\) is 90°.
Answer:
\(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\)
\(\sqrt{a^2+b^2+2 a b \cos \theta}=\sqrt{a^2+b^2-2 a b \cos \theta}\)
2 ab cos θ = – 2ab cos θ
4 ab cos θ = 0
cos θ = 0 but 4ab ≠ 0
∴ θ = 90°
Hence angle between \(\vec{a}\) and \(\vec{b}\) is 90°.

Question 6.
Show that the trajectory of an object thrown at certain angle with the horizontal is a parabola. [A.P. Mar. 18, 16, 15; T.S. Mar. 18, 15]
Answer:
Consider a body is projected with an initial velocity (u) making an angle 0 with the horizontal. The body does not experience acceleration in horizontal direction. The velocity of the projectile can be resolved in to (i) u cos θ, horizontal component (ii) u sin θ, vertical component. The horizontal component of velocity remains constant through out the motion. Only its vertical component changes due to acceleration due to gravity (g).

The distance travelled along OX in time t is given by
x = u cos θ × t
t = \(\frac{x}{u \cos \theta}\) ………………. (1)
The distance travelled along oy in time t is given by

Y = Ax – Bx² Where A and B are constants.
This is the equation of parabola.
∴ The trajectory of a projectile is parabola.

Question 7.
Explain the terms the average velocity and instantaneous velocity. When are they equal ?
Answer:
Average velocity :
The average velocity of the particle is defined as the ratio of displacement (∆x) to the time interval ∆t
\(\bar{v}=\frac{\Delta x}{\Delta t}=\frac{x_2-x_1}{t_2-t_1}\)
Average velocity is independent of the path followed by the particle between the initial and final positions. It gives the result of the motion.

Instantaneous velocity:
The velocity of a particle at a particular instant of time is known as instantaneous velocity.

The instantaneous velocity may be positive (or) negative in straight line motion.
In uniform motion the instantaneous velocity of a body is equal to the average velocity.

Question 8.
Show that the maximum height and range of a projectile are \(\frac{U^2 \sin ^2 \theta}{2 g}\) and \(\frac{U^2 \sin 2 \theta}{g}\) respectively where the terms have their regular meanings. [Mar. 14]
Answer:
Maximum height:
When the projectile is at the maximum height, its vertical component of velocity v_y = 0
Initial velocity (u) = u sin θ
Distance (s) = H = maximum height
Acceleration (a) = – g
using v² – u² = 2as,
0 – u² sin2 θ = – 2gH
∴ H = \(\frac{u^2 \sin ^2 \theta}{2 g}\)

Horizontal range (R) .
The horizontal distance travelled by the projectile from the point of projection during the time of flight is called range. . .
Range (R) = Horizontal velocity × Time of flight
R = u cos θ × T = u cos θ × \(\frac{2 u \sin \theta}{g}\)
R = \(\frac{u^2 \times 2 \sin \theta \cos \theta}{\mathrm{g}}\)
R = \(\frac{u^2 \sin 2 \theta}{g}\)
If θ = 45°, R_Max = \(\frac{u^2}{g}\)

Question 9.
If the trajectory of a body is parabolic in one reference frame, can it be parabolic in another reference frame that moves at constant velocity with respect to the first reference frame ? If the trajectory can be other than parabolic, what else can it be ?
Answer:
No, when a stone is thrown from a moving bus, the trajectory of the stone is parabolic in one reference frame. That is when a man observes out side foot path.
In another frame of reference, the trajectory is a vertical straight line.

Question 10.
A force 2i + j – k newton d’ts on a body which is initially at rest. At the end of 20 seconds the velocity of the body is 4i + 2j – 2k ms^-1. What is the mass of the body ?
Answer:
F = (2i + j – k) N
t = 20 sec, u = 0
v = (4i + 2j – 2k) m/s
a = \(\frac{v-u}{t}=\frac{(4 i+2 j-2 k)-0}{20}\)
a = \(\frac{2 i+j-k}{10}\) m/s²
F = ma
mass (m) = \(\frac{\mathrm{F}}{\mathrm{a}}\)
= \(\frac{2 i+j-k}{\left(\frac{2 i+k}{10}\right)}\)
m = 10 kg

Problems

Question 1.
Ship A is 10 km due west of ship B. Ship A is heading directly north at a speed of 30 km/h. while ship B is heading in a direction 60° west of north at a speed of 20 km/h.
i) Determine the magnitude of the velocity of ship B relative to ship A.
ii) What will be their distance of closest approach ?
Answer:
i) V_A = 30 kmph, V_B = 20 kmp, θ = 60°
Relative velocity of ship B w.r.to ship A is

Question 2.
If θ is the angle of projection, R the range, h the maximum height of the floor. Then show that (a) tan θ = 4h/R and (b) h = gT^2/8
Answer:

Question 3.
A projectile is fired at an angle of 60° to the horizontal with an initial velocity of 800 m/s:
i) Find the time of flight of the projectile before it hits the ground.
ii) Find the distance it travels before it hits the ground (range)
iii) Find the time of flight for the projectile to reach its maximum height.
Answer:
θ = 60°, u = 800 m/s

Question 4.
For a particle projected slantwise from the ground, the magnitude of its position vector with respect to the point of projection, when it is at the highest point of the path is found to be \(\sqrt{2}\) times the maximum height reached by it. Show that the angle of projection is tan^-1 (2)
Answer:
Range (R) = \(\frac{u^2 \times 2 \sin \theta \cos \theta}{\mathrm{g}}\)
= \(\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}\) ………… (1)
Maximum height (h) = \(\frac{u^2 \sin ^2 \theta}{2 g}\) ……………….. (2)
Given R = \(\sqrt{2}\) h
\(\frac{u^2 \times 2 \sin \theta \cos \theta}{g}=\sqrt{2} \times \frac{u^2 \sin ^2 \theta}{2 g}\)
tan θ = 2 \(\sqrt{2}\)
θ = tan^-1 (2\(\sqrt{2}\))

Question 5.
An object is launched from a cliff 20m above the ground at an angle of 30° above the horizontal with an initial speed of 30 m/s. How far horizontally does the object travel before landing on the ground ? (g = 10 m/s²)
Answer:
h = 20m, θ = 30°, u = 30m/s,
g = 10m/s²
h = -(u sin θ) t + \(\frac{1}{2}\) gt²
20 = – 30 sin 30° × t + \(\frac{1}{2}\) × 10 × t²
20 = – 30 × \(\frac{1}{2}\) × t + \(\frac{1}{2}\) × 10 × t²
4 = – 3t + t²
t² – 3t – 4 = 0
(t – 4) (t + 1) = 0
t = 4 sec (or) t = -1 sec

∴ Range (R) = u cos θ × t
= 30 cos 30° × 4
= 30 × \(\frac{\sqrt{3}}{2}\) × 4
R = 60\(\sqrt{3}\) m

Question 6.
O is a point on the ground chosen as origin. A body first suffers a displacement of 10 \(\sqrt{2}\) m North-East, next 10 m North and finally North-West. How far it is from the origin ?
Answer:
Given OB = 10 \(\sqrt{2}\) m, BC = 10m.

∴ Total displacement (OD) = | OF | + |FE| + || ED ||
OD = 10 + 10 + 10
OD = 30m

Question 7.
From a point on the ground a particle is projected with initial velocity u, such that its horizontal range is maximum. Find the magnitude of average velocity during its ascent.
Answer:
S = \(\sqrt{\left(\frac{R}{2}\right)^2+\left(\frac{R}{4}\right)^2}=\sqrt{\frac{R^2}{4}+\frac{R^2}{16}}\)
= \(\sqrt{\frac{4 R^2+R^2}{16}}=\frac{\sqrt{5} R}{4}\)

Question 8.
A particle is projected from the ground with some initial velocity making an angle of 45° with the horizontal. It reaches a height of 7.5 m above the ground while it travels a horizontal distance of 10m from the point of projection. Find the initial speed of projection, (g = 10m/s²)
Answer:
θ = 45°, g = 10 m/s²
Horizontal distance (x) = 10m.
Vertical distance (y) = 7.5 m.

Additional Problems

Question 1.
State, for each of the following vector quantities, if it is a scalar are vector : volume, mass, speed, acceleration, density, number of mass, velocity, angular frequency. Displacement, angular velocity.
Answer:
Scalars, volume, mass, speed, density, number of moles, angular frequency vectors, acceleration, velocity, displacement, angular velocity.

Question 2.
Pick out the two scalar quantities in the following list  angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Answer:
Work and current are scalar quantities in the given list.

Question 3.
Pick out the only vector quantity in the following list : Temperature, pressure, impulse time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
Answer:
Since, Impulse = change in momentum = force × time. A momentum and force are vector quantities hence impulse is a vector quantity.

Question 4.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful :
a) adding any two scalars,
b) adding a scalar to a vector of the same dimensions,
c) multiplying any vector by any scalar,
d) multiplying any two scalars,
e) adding any two vectors,
f) adding a component of a vector to the same vector.
Answer:
a) No, because only the scalars of same dimensions can be added.
b) No, because a scalar cannot be added to a vector.
c) Yes, when acceleration \(\vec{A}\) is multiplied by mass m, we get a force \(\vec{F}=m \vec{A}\), which is a meaningful operation.
d) Yes, when power P is multiplied by time t, we get work done = Pt, which is a useful operation.
e) No, because the two vectors of same dimensions can be added.
f) Yes, because both are vectors of same dimensions.

Question 5.
Read each statement below carefully and state with reasons, if it is true of false :
a) The magnitude of a vector is always a scalar,
b) each component of a vector is always a scalar,
c) the total path length is always equal to the magnitude of the displacement vector of a particle,
d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time,
e) Three vectors not lying in a plane can never add up to give a null vector.
Answer:
a) True, because magnitude is a pure number.
b) False, each component of a vector is also a vector.
c) True only if the particle moves along a straight line in the same direction, otherwise false.
d) True; because the total path length is either greater than or equal to the magnitude of the displacement vector.
e) True, as they cannot be represented by three sides of a triangle taken in the same order.

Question 6.
Establish the following vector inequalities geometrically or otherwise :
(a) |a + b| ≤ |a| + |b|
(b) |a + b| ≥ ||a| – |b||
(c) |a  – b| ≤ |a| + |b|
(d) |a – b| ≥ ||a| – |b||
When does the equality sign above apply ?
Answer:
Consider two vectors \(\vec{A}\) and \(\vec{B}\) be represented by the sides \(\vec{OP}\) and \(\vec{OQ}\) of a parallelogram OPSQ. According to parallelogram law of vector addition; (\(\vec{A}+\vec{B}\)) will be represented by \(\vec{OS}\) as shown in figure. Thus OP = \(|\vec{A}|\), OQ = PS = \(|\vec{B}|\) and OS = \(|\vec{A}+\vec{B}|\)

a) To prove |\(\vec{A}+\vec{B}\) | ≤ \(|\vec{A}|\) + \(|\vec{B}|\)
we know that the length of one side of a triangle is always less than the sum of the lengths of the other two sides. Hence from ∆OPS, we have OS < OP + PS or OS < OP + OQ or|\(\vec{A}+\vec{B}\)|< |\(\vec{A}+\vec{B}\)| ………………. (i) If the two vectors \(\vec{A}\) and \(\vec{B}\) are acting along a same straight line aind in same directions, then \(|\vec{A}+\vec{B}|=|\vec{A}|+|\vec{B}|\) …………….. (ii) combining the conditions mentioned in (i) and (ii) we get \(|\vec{A}+\vec{B}|=|\vec{A}|+|\vec{B}|\) b) To prove \(|\vec{A}+\vec{B}| \geq\|\vec{A}|+| \vec{B}\|\) From ∆OPS, we have OS + PS> OP or OS > |OP – PS| or OS > (OP – OQ) ………………. (iii)
(∵ PS = OQ)
The modulus of (OP – PS) has been taken because the LH.S is always positive but the R.H.S may be negative if OP < PS. Thus from (iii) we have \(|\vec{A}+\vec{B}|>\|\vec{A}|-| \vec{B}\|\) ………………. (iv)
If the two vectors \(\vec{A}\) and \(\vec{B}\) are acting along a straight line in opposite directions, then
\(|\vec{A}+\vec{B}|=\vec{A}-\vec{B}\) ……………. (v)
combining the conditions mentioned in (iv) and (v) we get
\(|\vec{A}+\vec{B}| \geq \vec{A}-\vec{B}\)

c)

combining the conditions mentioned in (vi) and (vii) we get
\(|\vec{A}-\vec{B}| \leq|\vec{A}|+|\vec{B}|\)

d) To prove \(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}} \geq|| \overrightarrow{\mathrm{A}}|-| \overrightarrow{\mathrm{B}} \mid\)
From ∆OPR, we note that OR + PR > OP or OR > |OP – PR| or OR > |OP – OT| ………………. (viii)
(∵ OT = PR)
The modulus of (OP – OT) has been taken because L.H.S. is positive and R.H.S. may be negative of OP < OT From (viii), \(|\vec{A}-\vec{B}|>|| \vec{A}|-| \vec{B} \|\) …………….. (ix)
If the two vectors \(\vec{A}\) and \(\vec{B}\) are acting along the straight line in the same direction, then \(|\vec{A}-\vec{B}|\)
= \(|\vec{A}|-|\vec{B}|\) ………………. (x)
Combining the conditions mentioned in (ix) and (x) we get
\(\vec{A}-\vec{B} \geq \vec{A}-\vec{B}\)

Question 7.
Given a + b + c + d = 0. Which of the following statements are correct:
a) a, b, c and d must each be a null vector,
b) The magnitude of (a + c) equals the magnitude of (b + d),
c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d,

d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear ?
Answer:

Question 8.
Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate ?
Answer:
Displacement for each girl = \(\vec{PQ}\)
∴ Magnitude of the displacement of each girl = PQ
= diameter of circular ice ground

= 2 × 200 = 400 m
For girl B, the magnitude of displacement is equal to the actual length of path skated.

Question 9.
A cyclist starts from the centre O of a circular park of radius km, reaches the edge p of the pa’rk, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the
(a) net displacement,
(b) average velocity and
(c) average speed of the cyclist ?
Answer:
a) Here, net displacement = zero

Question 10.
On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Answer:
In this problem, the path is regular hexagon ABCDEF of side length 500m let the motorist start from A.
Third turn : The motor cyclist will take the third turn at D. Displacement vector at D = \(\overrightarrow{A D}\) Magnitude of this displacement
= 500 + 500 = 1000 m
Total path length from A to D
= AB + BC + CD = 500 + 500 + 500 = 1500 m.
Sixth turn : The motor cyclist takes the sixth turn at A. So displacement vector is null vector – Total path length
= AB + BC + CD + DE + EF + FA = 6 × 500 = 3000 m.

Eighth turn : The motor cyclist takes the eighth turn at C. The displacement vector = \(\overrightarrow{\mathrm{AC}}\), which is represented by the diagonal of the parallelogram ABCG.
So, |\(\overrightarrow{\mathrm{AC}}\)|
= \(\sqrt{(500)^2+(500)^2+2 \times 500 \times 500 \times \cos 60^{\circ}}\)
= \(\sqrt{(500)^2+(500)^2+250000}\)
= 866.03 m
tan β = \(\frac{500 \sin 60^{\circ}}{500+500 \sin 60^{\circ}}=\frac{500 \times \sqrt{3} / 2}{500(1+1 / 2)}\)
= \(\frac{1}{\sqrt{3}}\) = tan 30° or β = 30°
It means \(\overrightarrow{\mathrm{AC}}\) makes an angle 30° with the initial direction.
Total path length = 8 × 500 = 4000 m.

Question 11.
A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ?
Answer:
Here, actual path length travelled, S = 23 km, displacement = 10 km, time taken, t = 28min = 28/60h
a) Average speed of a taxi
= \(\frac{\text { actual path length }}{\text { time taken }}=\frac{23}{28 / 60}\)
= 49.3 km/hr
b) Magnitude of average velocity
= \(\frac{\text { displacement }}{\text { time taken }}=\frac{10}{(28 / 60)}\)
= 21.4 km/hr
The average speed is not equal to the magnitude of the average velocity. The two are equal for the motion of taxi along a straight path in one direction.

Question 12.
Rain is falling vertically with a speed of 30 ms^-1. A woman rides a bicycle with a- speed of 10 m s^-1 in the north to south direction. What is the direction in which she should hold her umbrella ?
Answer:
The rain is falling along OA with speed 30 ms^-1 and woman rider is moving along OS with speed 10 ms^-1 i.e OA = 30 ms^-1 & OB = 10ms^-1. The woman rider can protect herself from the rain if she holds her umbrella in the direction of relative velocity of rainfall of woman. To draw apply equal and opposite velocity of woman on the rain i.e impress the velocity 10 ms^-1 due to North on which is represented by OC. Now the relative velocity of rain w.r.t woman will be represented by diagonal OD of parallelogram OADL. if ∠AOD = θ, then

= tan 18°26′ or β = 18°26′ with vertical in forward direction.

Question 13.
A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current ? How far down the river does he go when he reaches the other bank ?
Answer:
Time to cross the river, t = \(\frac{\text { width of river }}{\text { speed of man }}\)
= \(\frac{1 \mathrm{~km}}{4 \mathrm{~km} / \mathrm{h}}=\frac{1}{4}\)h = 15min
Distance moved along the river in time
t = v_r × t = 3 km/h × \(\frac{1}{4}\) h = 750 m.

Question 14.
The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s^-1 can go without hitting the ceiling of the hall ?
Answer:
Here, u = 40 ms^-1, H = 25 m; R = ?
Let θ be the angle of projection with the horizontal direction to have the maximum range, with maximum height 25m.

Question 15.
A cricketer can throw a ball to a maximum horizontal distance of 100m. How much high above the ground can the cricketer throw the same ball ?
Answer:
Let u be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projection with horizontal, θ = 45°. Then, R_max = u²/g Here u²/g = 100m ………………………….. (i)
In order to study the motion of the ball along vertical direction, consider a point on the surface of earth as the origin and vertical upward direction as the positive direction of Y – axis. Taking motion of the ball along vertical upward direction we have.
u_y = u, a_y = -g, v_y = 0, t = ? y₀ = 0, y = ?
As u_y = u_y + a_yt
∴ 0 = u + (-g)t or t = u/g
Also y = y₀ + u_yt + \(\frac{1}{2}\) a₀t²
∴ y = 0 + u(u/g) + \(\frac{1}{2}\) (-g) u² /g²
= \(\frac{\mathrm{u}^2}{\mathrm{~g}}-\frac{1}{2} \frac{\mathrm{u}^2}{\mathrm{~g}}\)
= \(\frac{1}{2} \frac{\mathrm{u}^2}{\mathrm{~g}}=\frac{100}{2}\)= 50m [from (i)]

Question 16.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25s. What is the magnitude and direction of acceleration of the stone ?
Answer:
Here, r = 80 cm = 0.8 m ; v = 14/25s^-1
∴ ω = 2πV
= 2 × \(\frac{22}{7} \times \frac{14}{25}=\frac{88}{25}\) rad s^-1
The centripetal accerlation, a = ω²r
= \(\left(\frac{88}{25}\right)^2\) × 0.80
= 9.90 m/s²
The direction of centripetal acceleration is along the string directed towards the centre of circular path.

Question 17.
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Answer:
Here, r = 1 km = 1000 m; v = 900 kmh^-1
= 900 × (1000) m × (60 × 60 s)^-1
= 250 ms^-1
Centripetal acceleration, a = \(\frac{v^2}{r}\)
= \(\frac{(250)^2}{1000}\)
Now, a/g = \(\frac{(250)^2}{1000} \times \frac{1}{9.8}\) = 6.38.

Question 18.
Read each statement below carefully and state, with reasons, if it is true of false :
a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector
Answer:
a) False, the net acceleration of a particle is towards the centre only in the case of a uniform circular motion.
b) True, because while leaving the circular path, the particle moves tangentially to the circular path.
c) True, the direction of acceleration vector in a uniform circular motion is directed towards the centre of circular path. It is constantly changing with time, the resultant of all these vectors will be a zero vector.

Question 19.
The position of a particle is given by r = 3.0 t \(\overline{\mathrm{i}}\) – 2.0t² \(\overline{\mathrm{j}}\) + 4.0 \(\overline{\mathrm{k}}\) m where t is in seconds and the co-efficients have the proper units for r to be in metres, (a) Find the v and a of the particle ? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?
Answer:

∴ θ = 69.5° below the x – axis

Question 20.
A particle starts from the origin at t = 0 s with a velocity of 10.0 \(\overline{\mathrm{j}}\) m/s and moves in the x – y plane with a constant acceleration of (8.0i + 2.0\(\overline{\mathrm{j}}\)) m s^-2. (a) At what time is the x – coordinate of the particle 16 m? What is the y – coordinate of the particle at that time ? (b) What is the speed of the particle at the time ?
Answer:
Here, \(\vec{u}\) = 10.0 \(\overline{\mathrm{j}}\) ms^-1 at t = 0
\(\vec{a}=\frac{\overrightarrow{d v}}{d t}=(8.0 \hat{i}+2.0 \hat{j}) \mathrm{ms}^{-2}\)
So \(d \vec{v}=(8.0 \hat{i}+2.0 \hat{j}) d t\)
Integrating it with in the limits of motion i.e. as time changes from 0 to t, velocity changes is from u to v, we have

Integrating it within the conditions of motion i.e as time changes from 0 to t, displacement is from 0 to r, we have

Question 21.
\(\overline{\mathrm{i}}\) and \(\overline{\mathrm{j}}\) are unit vectors along x – and y – axis respectively. What is the magnitude and direction of the vectors \(\overline{\mathrm{i}}\) + \(\overline{\mathrm{j}}\) and \(\overline{\mathrm{i}}\) – \(\overline{\mathrm{j}}\) ? What are the components of a vector A = 2 \(\overline{\mathrm{i}}\) + 3 \(\overline{\mathrm{j}}\) along the directions of \(\overline{\mathrm{i}}\) + \(\overline{\mathrm{j}}\) and \(\overline{\mathrm{i}}\) – \(\overline{\mathrm{j}}\)? [You may use graphical method]
Answer:
a) Magnitude of \((\hat{i}+\hat{j})=|\hat{i}+\hat{j}|\)
= \(\sqrt{(1)^2+(1)^2}=\sqrt{2}\)

b) Here, \(\vec{B}=2 \hat{i}+3 \hat{j}\)
To find the component vectors of \(\overrightarrow{\mathrm{A}}\) along the vectors \((\hat{i}+\hat{j})\) we first find the unit vector along the vector \((\hat{i}+\hat{j})\). Let be the unit vector along the direction of vector \((\hat{i}+\hat{j})\).

Question 22.
For any arbitrary motion in space, which of the following relations are true :
a) v_average = (1/2) (v (t₁) + v(t₂))
b) v_average = [r (t₂) – r(t₁)] / (t₂ – t₁)
c) v(t)^average = v(0) + a t
d) r (t) = r (0) + v(0) t + (1/2) a t²
e) a_average = [v (t₂) – v(t₁)] / (t₂ – t₁)
(The ‘average’ stands for average of the quantity over the time interval t₁ to t₂)
Answer:
The relations (b) and (e) are true; others are false because relations (a), (c) and (d) hold only for uniform acceleration.

Question 23.
Read each statement below carefully arid state, with reasons and examples, if it is true or fa¹ e :
A scalar quantity is one that
a) is conserved in a process
b) can never take negative values
c) must be dimensionless
d) does not vary from one point to another in space
e) has the same value for observers with different orientations of axes.
Answer:
a) False, because energy is not conserved during inelastic collisions.
b) False, because the temperature can be negative.
c) False, because the density has dimensions.
d) False, because gravitational potential vary from point to point in space.
e) True, because the value of scalar does not change with orientation of axes.

Question 24.
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft ?
Answer:
In figure O is the observation point at the ground. A and B are the positions of aircraft for which ∠AOB = 30°. Draw a perpendicular OC on AB. Here OC = 3400m and ∠AOC = ∠COB = 15°

Time taken by aircraft from A to B is 10s
In ∆AOC, AC = OC tan 15° = 3400 × 0.2679
= 910.86m
AB = AC + CB = AC + AC = 2AC
= 2 × 910.86 m
Speed of the aircraft, v = \(\frac{\text { distance } A B}{\text { time }}\)
= \(\frac{2 \times 910.86}{10}\)
= 182.17 ms^-1 = 182.2 ms^-1

Question 25.
A vector has magnitude and direction. Does K have a location in space ? Can it vary with time? Will two equal vectors a and b at different locations in space necessarily have identical physical effects ? Give examples in support of your answer. .
Answer:
i) A vector in general has no difinite location in space because a vector remains uneffected whenever it is displaced anywhere in space provided its magnitude and direction do not change. However a position vector has a definite location in space.
ii) A vector can vary with time eg the velocity vector of an accelerated particle varies with time.
iii) Two equal vectors at different locations in space do not necessarily have some physical effects. For example two equal forces acting at two different points on a body with can cause the rotation of a body about an axis will not produce equal turning effect.

Question 26.
A vector has both magnitude and direction. Does it mean that any thing that has magnitude and direction is necessarily a vector ? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector ?
Answer:
No. There are certain physical quantities which have both magnitude and direction, but they are not vectors as they do not follow the laws of vectors addition, which is essential for vectors. The finite rotation of a body about an axis is not a vector because the finite rotations do not abey the laws of vectors addition. However, the small rotation of a body is a vector quantity as it obey the law of vector addition.

Question 27.
Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere ? Explain.
Answer:
a) We cannot associate a vector with the length of a wire bent into a loop.
b) We can associate a vector with a plane area. Such a vector is called area vector and its direction is represented by outward drawn normal to the area.
c) we can not associate a vector with volume of sphere however a vector can be associated with the area of sphere.

Question 28.
A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assume the muzzle speed to be fixed, and neglect air resistance.
Answer:
Horizontal range R = \(\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}\)
3 = \(\frac{u^2 \sin 60^{\circ}}{g}=\frac{u^2}{g} \sqrt{3} / 2\) or \(\frac{\mathrm{u}^2}{\mathrm{~g}}=2 \sqrt{3}\)
Since the muzzle velocity is fixed, therefore, Max, horizontal range.
R_max = \(\frac{u^2}{g}=2 \sqrt{3}\) = 3.464m.
So, the bullet cannot hit the target.

Question 29.
A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s^-1 to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g = 10m s^-2).
Answer:
In Fig. 0 be the position of gun and A be the position of plane. The speed of the plane,
v = \(\frac{720 \times 1000}{60 \times 60}\) = 200 ms^-1
The speed of the shell, u = 600 m/s
Let the shell will hit the plane at B after time t if fired at an angle 0 with the vertical from O then the horizontal distance travelled by shel in time t is the same as the distance covered by plane.
i.e. u_x × t = vt or u sin θ t = vt .
or sin θ = \(\frac{v}{u}=\frac{200}{600}\) = 0.3333 = sin 19.5°. or θ = 19.5°.

The plane will not be hit by the bullet from the gun if it is flying at a minimum height which is maximum height (H) attained by bullet after firing from gun.
Here H = \(\frac{u^2 \sin ^2(90-\theta)}{2 g}=\frac{u^2 \cos ^2 \theta}{g}\)
= \(\frac{(600)^2 \times(\cos 19.5)^2}{2 \times 10}\)
= 16000 m
= 16 km

Question 30.
A cyclist is riding with a speed of 27 . km/h. As he approaches a circular turn on the road of radius 80m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?
Answer:
Here v = 27 km/h^-1 = 27 × (1000 m) × (60 × 60s)^-1 = 7.5 ms^-1, r = 80m centripetal acceleration

Let the cyclist applies the brakes at the point P of the Circular turn, then tangential acceleration a_T will act opposite to velocity. Acceleration along the tangent, a_T = 0.5 ms^-2 Angle between both the acceleration is 90° Therefore, the magnitude of the resultant acceleration.
a = \(\sqrt{a c^2+a_T^2}\)
= \(\sqrt{(0.7)^2+(0.5)^2}\)
= 0.86 ms^-2
Let the resultant acceleration make an angle β with the tangent i.e. the direction of velocity of the cyclist, then,
tan β = \(\frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{T}}}=\frac{0.7}{0.5}\) = 1.4
or β = 54° 28′

Question 31.
a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by
b) Shows that the projection angle θ for a projectile launched from the origin is given by θ(t) = tan^-1 \(\left[\begin{array}{cc}
v_{\theta y} & -g t \\
v_{0 x}
\end{array}\right]\)
θ₀ = tan^-1 \(\left(\frac{4 h_m}{R}\right)\)
Where the symbols have their usual meaning

Answer:
a) Let v_ox and v_oy be the initial component velocity of the projectile at O along OX direction and OY direction respectveIy, where OX is horizantal and oy is vertical

Let the projectile go from o to p in time t and v_x, v_y be the component velocity of projectile at P along horizantal and vertical directions respectively. Then v_y = v_oy – gt. and v_x = v_ox
If θ is the angle which the resultant velocity \(\overrightarrow{\mathrm{v}}\) makes with horizontal direction, then
tan θ = \(\frac{v y}{v x}=\frac{v_{o y}-g t}{v_{o x}}\) or
θ = tan^-1 \(\left[\frac{v_{\mathrm{oy}}-g t}{v_{o x}}\right]\)

b) In angular projection, maximum vertical height,

Textual Examples

Question 1.
Rain is falling vertically with a speed of 35 m s^-1. Winds starts blowing after sometime with a speed of 12 m s^-1 in east to west direction. In which direction should a boy waiting at bus stop hold his umbrella?

Answer:
The velocity of the rain and the wind are represented by the vectors v_r and v_w in Fig. a e in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of v_r and v_w is R as shown in the figure. The magnitude of R is
R = \(\sqrt{v_{\mathrm{r}}^2+v_{\mathrm{w}}^2}=\sqrt{35^2+12^2}\) m s^-1 = 37 m s^-1
The direction θ that R makes with the vertical is given by
tan θ = \(\frac{v_w}{v_r}=\frac{12}{35}\) = 0343
Or, θ = tan^-1 (0.343) = 19°
Therefore, the boy should hold his umbrella in the vertical plane at an a glass about 19° with vertical towards the east.

Question 2.
Fine, the magnitude and direction of the resultant of two vectors A and B in terms of their magnitudes and angle θ between them.

Answer:
Let OP and OQ represent the two vectors A and B making an angle θ (fig.). Then using the parallelogram method of vector addition, OS represents the vector R :
R = A + B
SN is normal to OP and PM is normal to OS.
From the geometry of the figure,
OS² = ON² + SN²
but ON = OP + PN = A + B cos θ
SN = B sin θ
OS² = (A + B cos θ)² + (B sin θ)²
or, R² = A² + B² + 2AB cos θ
R = \(\sqrt{A^2+B^2+2 A B \cos \theta}\) …………….. (1)
In ∆ OSN, SN = OS sin α = R sin α, and in ∆ PSN, SN = PS sin θ = B sin θ
Therefore, R sin θ = B sin θ

Equation (1) gives the magnitude of the resultant and Eqs. 5 & 6 its direction. Equation ((1) a) is known as the Law of ‘ cosines and Eq. (4) as the Law of sines.

Question 3.
A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat.
Answer:
The vector vb representing the velocity of the motorboat and the vector v_c representing the water current are shown in Fig. in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure.

We can obtain the magnitude of R using the Law of cosine :

Question 4.
The position of a particle is given by r =\(3.0 t \overline{\mathrm{i}}+2.0 \overline{\mathrm{j}}^2 \mathrm{j}+5.0 \overline{\mathrm{K}}\) where t is in seconds and the coefficients have the proper units for r to he in metres. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at t = 1.0 s.
Answer:
v(t) = \(\frac{d r}{d t}=\frac{d}{d t}\left(3.0 t \hat{i}+2.0 t^2 \hat{j}+5.0 \hat{k}\right)\)
= \(3.0 \hat{\mathrm{i}}+4.0 t \hat{\mathrm{j}}\)
a(t) = \(\frac{d v}{d t}=+4.0 \hat{\mathrm{j}}\)
a = 4.0 ms^-2 along y-direction
At t = 1.0 s, v = \(3.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}\)
It’s magnitude is v = \(\sqrt{3^2+4^2}\) = 5.0 m s^-1 and direction is
θ = tan^-1\(\left(\frac{v_{\mathrm{y}}}{v_{\mathrm{x}}}\right)\) = tan^-1 \(\left(\frac{4}{3}\right)\) ≅ 53° with x-axis.

Question 5.
A particle starts from origin at t = 0 with a velocity 5.0 \(\hat{\mathrm{i}}\) m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0 \(\hat{\mathrm{i}}\) + 2.0\(\hat{\mathrm{j}}\)) m/s².
(a) What is the y- coordinate of the particle at the instant its x-co-ordinate is 84 m ?
(b) What is the speed of the particle at this time?
Answer:
The position of the particle is given by
r(t) = v₀t + \(\frac{1}{2}\) at²
= \(5.0 \hat{i} t+(1 / 2)(3.0 \hat{i}+2.0 \hat{j}) t^2\)
= \(\left(5.0 t+1.5 t^2\right) \hat{i}+1.0 t^2 \hat{j}\)
Therefore, x(t) = 5.0t + 1.5 t²
y(t) = + 1.0t²
Givn x(t) = 84m, t =?
5.0 t + 1.5 t² = 84 ⇒ t = 6s
At t = 6 s, y = 1.0 (6)² = 36.0 m
Now the velocity v = \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
= (5.0 + 3.0t)\(\hat{\mathrm{i}}+2.0 \mathrm{t} \hat{\mathrm{j}}\)
At t = 6s, v = \(23.0 \hat{i}+12.0 \hat{j}\)
speed = |v| = \(\sqrt{23^2+12^2}\) ≅ 26 m s^-1

Question 6.
Rain is falling vertically with a speed of 35 m s^-1. A woman rides a bicycle with a speed of 12 m s^-1 in east to west direction. What is the direction in which she should hold her umbrella?
Answer:
In Fig. v_r represents the velocity of rain and v_b, the velocity of the bicycle, the woman is riding. Both these velocities are with respect to the ground. Since the woman is riding a bicycle, the velocity of rain as experienced by

her is the velocity of rain relative to the velocity of the bicycle she is riding. That is
v_rb = v_r – v_b
This relative velocity vector as shown in Fig. makes an angle θ with the vertical. It is given by
tan θ = \(\frac{v_{\mathrm{b}}}{v_{\mathrm{r}}}=\frac{12}{35}\) = 0.343 or, θ ≅ 19°
Therefore, the woman should hold her umbrella at an angle of about 19° with the vertical towards the west.

Note carefully the difference between this example and the Example 1. In Example 1, the boy experiences the resultant (vector sum) of two velocities while in this example, the woman experiences the velocity of rain relative to the bicycle (the vector difference of the two velocities).

Question 7.
Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement.
Answer:
For a projectile launched with velocity v₀ at an angle θ₀, the range is given by
R = \(\frac{v_0^2 \sin 2 \theta_0}{g}\)
Now, for angles, (45° + α) and (45° – α), 2θ₀ is (90° + 2α) and (90° – 2α), respectively. The values of sin (90° + 2α) and sin (90° – 2α), are the same, equal to that of cos 2a. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.

Question 8.
A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms^-1. Neglecting air resistance, find the time taken by the stone to reach the ground and the speed with which it hits the ground. (Take g = 9.8m s^-2).
Answer:
We choose the origin of the x , and y – axis at the edge of the cliff and t = 0 s at the instant the stone is thrown. Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the vertically upward direction. The x, and y- components of the motion can be treated independently. The equations of motion are :
x (t) = x₀ + υ_0xt
y (t) = y₀ + υ_0y t + (1/2) a_y t²
Here, x₀ = y₀ = 0, υ_0y = 0, a_y = – g = -9.8m s^-2, υ_0x = 15 m s^-1.
The stone hits the ground when y(t) = – 490 m.
– 490 m = – (1/2) (9.8) t².
This gives t = 10 s.
The velocity components are υ_x = υ_0x and υ_y = υ_0y – g t
so that when the stone hits the ground :
u_0x = 15 m s^-1
u_0y = 0 – 9.8 × 10 = -98 m s^-1
Therefore, the speed of the stone is
\(\sqrt{v_x^2+v_y^2}=\sqrt{15^2+98^2}\) = 99 m s^-1.

Question 9.
A cricket ball is thrown at a speed of 28 m s^-1 in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level and (c) the distance from the thrower to the point where the ball returns to the same level.
Answer:
a) The maximum height is given by
h_m = \(\frac{\left(v_0 \sin \theta_0\right)^2}{2 g}=\frac{\left(28 {in} 30^{\circ}\right)^2}{2(9.8)}\) m
= \(\frac{14 \times 14}{2 \times 9.8}\) = 10.0 m
(u0 sin0o)2 (28 in 30°)2 . hm ~ 2g “ 2(9.8) m

b) The time taken to return to the same level is
T_f = (2 υ₀ sinθ₀)/g = (2 × 28 × sin30°)/9.8
= 28 / 9.8 s = 2.9 s

c) The distance from the thrower to the point where the ball returns to the same level is
R = \(\frac{\left(v_0^2 \sin 2 \theta_0\right)}{g}=\frac{28 \times 28 \times \sin 60^{\circ}}{9.8}\)
= 69 m

Question 10.
An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s, (a) What is the angular speed and the linear speed of the motion ? (b) Is the acceleration vector a constant vector ? What is its magnitude ?
Answer:
This is an example nf uniform circular motion. Here R = 12 cm. The angular speed ω is given by
ω = 2π/T = 2π × 7/100 = 0.44 rad/s ,
The linear speed υ is :
υ = ω R = 0.44 s^-1 × 12 cm = 5.3 cm s^-1
The direction of velocity v is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector. However, the magnitude of acceleration is constant :
a – ω² R = (0.44 s^-1)² (12 cm)
= 2.3 cm s^-2

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 3rd Lesson Motion in a Straight Line Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 3rd Lesson Motion in a Straight Line

Very Short Answer Questions

Question 1.
The states of motion and rest are relative. Explain.
Answer:
Rest and motion are relative. They are not absolute. A body can be in the rest or in motion w.r.t. reference frame. A man in a moving train is a rest w.r. to a co-passenger, but he is in motion w.r.t a man on the ground.

Question 2.
How is average velocity different from instantaneous velocity? [Mar. 13]
Answer:
The average velocity does not give any details of the motion of the particle. It gives only the result of the motion. The instantaneous velocity defines how fast the particle moves at a particular instant of time.
In uniform motion, the instantaneous velocity is equal to the average velocity.

Question 3.
Give an example where the velocity of an object is zero but its acceleration is not zero.
Answer:
When the body is projected vertically upwards, at the highest point, its velocity is zero. But its acceleration (a = g) is not equal to zero.

Question 4.
A vehicle travels half the distance L with speed v₁ and the other half with speed v₂. What is the average speed ?
Answer:
Average speed = \(\frac{\text { Total length of the path }}{\text { Total time taken }}=\frac{\mathrm{L}}{\frac{\left(\frac{\mathrm{L}}{2}\right)}{\mathrm{V}_1}+\frac{\left(\frac{\mathrm{L}}{2}\right)}{\mathrm{V}_2}}=\frac{2 \mathrm{~V}_1 \mathrm{~V}_2}{\mathrm{~V}_1+\mathrm{V}_2}\)

Question 5.
A lift coming down is just about to reach the ground floor. Taking the ground floor as origin and positive direction upwards for all quantities, which one of the following is correct
a) x < 0, v < 0, a > 0, b) x > 0, v < 0, a < 0, c) x > 0, v < 0, a > 0, d)x > 0, V > 0, a > 0
Answer:
While lift is moving towards ground floor (origin), position x decreases, velocity decreases. Hence x < 0, v < 0. But a > 0. So(a) is correct option.

Question 6.
A uniformly moving cricket ball is hit with a bat for a very short time and is turned back. Show the variation of its acceleration with time taking the acceleration in the backward direction as positive.
Answer:

Question 7.
Give an example of one-dimensional motion where a particle moving along the positive x-direction comes to rest periodically and moves forward.
Answer:
When simple harmonic oscillator starts from left extreme position and comes to rest at that point periodically and moves forward.

Question 8.
An object falling through a fluid is observed to have an acceleration given by a = g – bv where g is the gravitational accelaration and b is a constant. After a long time it is observed to fall with a constant velocity. What would be the value of this constant velocity ?
Answer:
Given acceleration a = g – bv
\(\frac{\mathrm{dv}}{\mathrm{dt}}\) = g – bv [∵ a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\)]
If an object is moving through fluid with constant velocity, dv = 0
0 = g – bv ∴ v = \(\frac{\mathrm{g}}{\mathrm{b}}\)

Question 9.
If the trajectory of a body is parabolic in one frame, can it be parabolic in another frame that moves with a constant velocity with respect to the first frame ? If not, what can it be ?
Answer:
No. the trajectory is a vertical straight line.

Question 10.
A spring with one end attached to a mass and the other to a rigid support is stretched and released. When is the magnitude of acceleration a maximum ?
Answer:
The magnitude of acceleration is maximum at Extreme Positions.

Short Answer Questions

Question 1.
Can the equations of kinematics be used when the acceleration varies with time ? If not, what form would these equations take ?
Answer:
No, the equations of kinematics be used when the acceleration varies with time.
If an object moves along a straight line with uniform acceleration (a), equations of kinematics are

1. v = v₀ + at;
2. x = v₀t + \(\frac{1}{2}\) at²;
3. v² = v₀² + 2ax

Where x is displacement, v₀ is velocity at t = 0, v is velocity at time t, a is acceleration.
These are kinematic equations of rectilinear motion for constant acceleration.
If an object moves with non-uniform acceleration, the equations of motion are,

1. v = v₀ + at
2. x = x₀ + v₀t + \(\frac{1}{2}\)at²
3. v² = v₀² + 2a (x – x₀)

Question 2.
A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is v₁ and at time t = t is v₂. The average velocity of the particle in this time interval is (v₁ + v₂)/2. Is this correct ? Substantiate your answer.
Answer:
Consider a particle moving with uniform acceleration a.
At t = 0, the (initial) velocity = v₁
At t = t, the (final) velocity = v₂
time = t

∴ The given statement is true.

Question 3.
Can the velocity of an object be in a direction other than the direction of acceleration of the object ? If so, give an example.
Answer:
Yes, the velocity of an object be in a direction other than the direction of acceleration of the object.

Ex. : In the case of the upward motion of a projectile, the angle between velocity and acceleration is 180°. During its journey, the direction of velocity is in upwards and the direction of acceleration is in downwards.

Question 4.
A parachutist flying in an aeroplane jumps when it is at a height of 3 km above ground. He opens his parachute when he is about 1 km above ground. Describe his motion.
Answer:

1. A parachutist is jumping from an air-plane at a height of 3 km from the ground. Upto 1 km above the ground, his motion is like a freely falling body. He falls with constant acceleration 9.8 ms^-2.
2. At a height of 1 km above the ground, when he opened the parachute, the air drag opposes the force of gravity resultant on it. The acceleration of parachutist gradually decreases since velocity increases (a = g – bv) and becomes zero.
3. Further, the parachutist attains terminal speed (constant speed), where air drag (in upward direction) is equal to force of gravity (in downward).
4. Hopefully this terminal speed is slow enough, so he can touch the ground without much difficulty.

Question 5.
A bird holds a fruit in its beak and flies parallel to the ground. It lets go of the fruit at some height. Describe the trajectory of the fruit as it falls to the ground as seen by
(a) the bird
(b) a person on the ground.
Answer:
A bird holds a fruit in its beak and flies parallel to the ground.

It lets go of the fruit at some height, the trajectory of the fruit as it falls to the ground as seen by (a) the bird is a straight line (b) a person on the ground is a parabola.

Question 6.
A man runs across the roof of a tall building and jumps horizontally on to the (lower) roof of an adjacent building. If his speed is 9 ms^-1 and the horizontal distance between the buildings is 10 m and the height difference between the roofs is 9 m, will he be able to land on the next building ? (take g = 10 ms^-2).
Answer:
Difference of heights between two buildings h = 9 m; g = 10 ms^-2

Time of flight of a man t = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 9}{10}}\)
= 1.341 sec
Horizontal speed of a man u = 9 ms^-1
Horizontal distance travelled by a man d_m
= Horizontal speed × Time of flight
= u × t = 9 × 1.341 = 12.07 m
Given, the horizontal distance between the buildings d_b = 10 m
Since d_m > d_b, the man can able to land on the next building.

Question 7.
A ball is dropped from the roof of a tall building and simultaneously another ball is thrown horizontally with some velocity from the same roof. Which ball lands first? Explain your answer.
Answer:
Let height of the building = Displacement of ball = h

For first ball u = 0; S = h. a = g; t = t₁
Substituting these values in S = ut + \(\frac{1}{2}\) at²
⇒ h = 0 + \(\frac{1}{2}\) gt₁²
∴ t₁ = \(\sqrt{\frac{2 h}{g}}\) …………… (1)
For second ball, u_x = u (say) u_y = 0, a_Y = g, S_Y = h; t = t₂
Substituting these values in
S_Y = u_Yt + \(\frac{1}{2}\) a_Yt²
h = 0 + \(\frac{1}{2}\) gt₂²
∴ t₂ = \(\sqrt{\frac{2 h}{g}}\)
From equation (1) and equation (2), t₁ = t₂
∴ Two balls will reach the ground in same time.

Question 8.
A ball is dropped from a building and simultaneously another ball is projected upward with some velocity. Describe the change in relative velocities of the balls as a function of time.
Answer:
For first ball u = u₁; v = v₁; a = g; t = t

Substituting these values in equation
v = u + at, we get
v₁ = u₁ + gt₁
For second ball u = u<sub2; v = v₂; a = – g; t = t₂
Substituting these values in equation, v = u + at,
We get, v₂ = u₂ + gt₂ ………………. (2)
(1) – (2) ⇒ (v₁ – v₂) = (u₁ – u₂) + g(t₁ + t₂)
∴ (v₁ – v₂) – (u₁ – u₂) = g(t₁ + t₂)
[∵ u₁ = 0]
∴ (v₁ – v₂) – (0 – u₂) = g(t₁ + t₂)
∴ The change in final relative velocity and initial relative velocity of two balls = Function of time.

Question 9.
A typical raindrop is about 4 mm in diameter. If a raindrop falls from a cloud which is km above the ground, estimate its momentum when it hits the ground.
Answer:
Diameter of rain drop, D = 4 mm
Radius of ram drop r = 2 mm = 2 × 10^-3 m
Volume of the rain atop, V = \(\frac{4}{3}\) πr³ = \(\frac{4}{3}\) × \(\frac{22}{7}\) × (2 × 10^-3)³
Density of water drop = 10³ kg/m³
Mass of water drop M = Vd = \(\frac{4}{3}\) × \(\frac{22}{7}\) × 8 × 10^-9 × 10³
= 33.5 × 2 × 10^-6 kg
The height of raindrop falls from a cloud, h = 1 km = 1000 m
Velocity of raindrop just before touching the ground V = \(\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 1000}\) = 140 ms^-1
Momentum of raindrop when it hits the ground P = mV = 33.52 × 10^-6 × 140 = 469.28 × 10^-5
= 0.004692 kg ms^-1

Question 10.
Show that the maximum height reached by a projectile launched at an angle of 45° is one quarter of its range.
Answer:

Problems

Question 1.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the (a) magnitude of average velocity and (b) average speed of the man over the time interval 0 to 50 min.
Answer:
From home to market:
X₁ = 2.5 km; v₁ = 5 km h^-1;
t₁ = \(\frac{\mathrm{X_1}}{\mathrm{v_1}}=\frac{2.5}{5}=\frac{1}{2}\) h = 30 min.
From market to home:
X₂ = 2.5 km; v₂ = 7.5 km h^-1
t₂ = \(\frac{\mathrm{X_2}}{\mathrm{v_2}}=\frac{2.5}{7.5}=\frac{1}{3}\) h = 20 min.
a) Magnitude of average velocity
= \(\frac{\Delta \mathrm{X}}{\Delta \mathrm{t}}=\frac{\mathrm{X}_1-\mathrm{X}_2}{\mathrm{t}_1+\mathrm{t}_2}=\frac{2.5-2.5}{50}\) = 0
b) Average speed
= \(\frac{X_1+X_2}{t_1+t_2}=\frac{2.5-2.5}{(30+20) \min }=\frac{5 \mathrm{~km}}{50 \times \frac{h}{60}}\)
= 6 km h^-1

Question 2.
A car travels the first third of a distance with a speed of 10 kmph, the second third at 20 kmph and the last third at 60 kmph. What is its mean speed over the entire distance ? [T.S. Mar 16]
Answer:
v₁ = 10 km ph; v₂ = 20 km ph; v₁ = 60 km ph; v = ?
\(\frac{3}{v}=\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}\)
\(\frac{3}{v}=\frac{1}{10}+\frac{1}{20}+\frac{1}{60}, \frac{3}{v}=\frac{6+3+1}{60}=\frac{3}{v}=\frac{10}{60}\)
∴ v = 18 km ph

Question 3.
A bullet moving with a speed of 150 ms-1 strikes a tree and penetrates 3.5 cm before stopping. What is the magnitude of its retardation in the tree and the time taken for it to stop after striking the tree ?
Answer:
u = 150 m/s, s = 3.5 cm = 0.035 m, v = 0
v² – u² = 2as
0 – 150² = 2 × a × 0.035
a = \(\left|\frac{-150 \times 150}{2 \times 0.035}\right|\)
= -3.214 × 10⁵ m/s² = -3.214 × 10⁵ m/s²
Time = \(\frac{v-u}{a}=\frac{0-150}{3.214 \times 10^5}=\frac{-150}{3.214 \times 10^5}\)
= 4.67 × 10^-4 sec.

Question 4.
A motorist drives north for 30 min at 85 km/h and then stop for 15 min. He continues travelling north and covers 130 km in 2 hours. What is his total displacement and average velocity ?
Answer:
v₁ = 85 kmph, t = 35.0 min, S₂ = 130 km
S₁ = Displacement = \(\frac{85}{60}\) × 30 = 42.5 km
S₂ = 130 km
a) S = S₁ + S₂ = 42.5 + 130 = 172.50 km
b) Avg.velocity = \(\frac{S_1+S_2+S_3}{t_1+t_2+t_3}\)
= \(\frac{42.5+0+130}{\frac{30}{60}+\frac{15}{60}+2}\)
= \(\frac{172.5 \times 60}{165}\)
= 62.7 km/hr

Question 5.
A ball A is dropped from the top of a .building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what fraction of the height of the building did the collison occur ?
Answer:
Let height of the building =H
Let two balls collide at a height = h
For ball A, u = 0; V = V_A s = H – h; t = t; a = g
Substituting these values in s = ut + \(\frac{1}{2}\) at²
H – h = 0 + \(\frac{1}{2}\) gt²
H – h = \(\frac{1}{2}\) gt² …………. (1)
and V_A = gt ……………… (2)
For ball B, u = u; v = v_B; s = h; a = -g
Substituting these values in s = ut + \(\frac{1}{2}\) at²
⇒ h = ut – \(\frac{1}{2}\) gt² ………. (3)
and V_B = u – gt ……….. (4)
given V_A = 2V_B
gt = 2(u – gt)
u = \(\frac{3}{2}\) gt …………… (5)

Question 6.
Drops of water fall at regular intervals from the roof of abuilding of height 16 m. The first drop strikes the ground at the same moment as the fifth drop leaves the roof. Find the distances between successive drops.
Answer:
H = 16 m
Time taken by the first drop to touch the ground t = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 16}{9.8}=\sqrt{3.26}}\)
sec = 1.8 sec
Time interval between two successive drops
= \(\frac{t}{n-1}\) Where n = no. of drops
= \(\frac{1.8}{5-1}=\frac{1.8}{4}\) = 0.45 sec

For second drop h₂ = \(\frac{1}{2}\) gt²
= \(\frac{1}{2}\) × 9.8 × 1.35 × 1.35 = 8.93 m
d₁₂ = 16 – 8.93 = 7.06 = 7 m
For 3^rd drop h₃ = \(\frac{1}{2}\) × 9.8 × 0.90 × 0.90
= 3.97
d₂₃ = 8.93 – 3.97 = 4.961 = 5 m
For 4^th drop h₄ = \(\frac{1}{2}\) × 9.8 × 0.45 × 0.45
= 0.9922
d₃₄ = 3.97 – 0.9922
d₃₄ = 2.9778 = 3 m
Similarly for d₄₅ = 0.9922 – 0 = 0.9922 = 1 m

Question 7.
A hunter aims a gun at a monkey hanging from a tree some distance away. The monkey drops from the branch at the moment he fires the gun hoping to avoid the bullet. Explain why the monkey made a wrong move.
Answer:
Suppose a hunter aims the gun at a monkey hanging from a high tree branch some distance away d. The instant the monkey observes the flash of the gun it drops from the tree.
Here the time taken by the monkey from the tree to the ground is t₁ = \(\sqrt{\frac{2 h}{g}}\) ……………. (1)
The path of bullet from the gun is like a horizontal Projectile, vertical velocity u_y = 0, Let S = h. Let t₂ be the time taken by the bullet to reach the ground.
∴ S = ut + \(\frac{1}{2}\)a t₂²
S = 0 × t + \(\frac{1}{2}\)a t₂²
∴ h = \(\frac{1}{2}\)g t₂²
t₂ = \(\sqrt{\frac{2 h}{g}}\) …………….. (2)
It is observed from equation (1) and (2),
t₁ = t₂
Both bullet and monkey reach the ground simultaneously.
Hence the bullet hits the monkey. Therefore Monkey made a wrong move.

Question 8.
A food packet is dropped from an aeroplane, moving with a speed of 360 kmph in a horizontal direction, from a height of 500 m. Find (i) its time of descent (ii) the horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped.
Answer:
Velocity of aeroplane v = 360 kmph

v = 360 × \(\frac{5}{18}\) = 100 m/s
h = 500 m
i) Time of descent =
t = \(\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 500}{10}}\) = 10 sec
ii) Horizontal range R = u × \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\)
= 100 × 10 sec = 1000 sec

Question 9.
A ball is tossed from the window of a buliding with an initial velocity of 8 ms-1 at an angle of 20° below the horizontal. It strikes the ground 3 s later. From what height was the ball thrown ? How far from the base of the building does the ball strike the ground ?
Answer:
u = 8 m/s, θ = 20°, t = 35
a) Horizontal distance = (u cos θ) t = 8 cos 20° × 3 = 8 × 0.9397 × 3 = 22.6 m

b) Height h = (u sin θ)t + \(\frac{1}{2}\) gt²
= 8 sin 20° × 3 + \(\frac{1}{2}\) × 9.8 × 9
= 8.208 + 44.1 = 52.31 m

c) The ball is thrown from a height of 44.1 m 1
h₁ = (u sin θ)t₁ + \(\frac{1}{2}\) gt₁²
10 = (8 sin 20°)t₁ + \(\frac{1}{2}\) 9.8% t₁²
= 2.736 t₁ + 4.9 t₁²
⇒ 4.9 t₁² + 2.736 t₁ – 10 = 0
t₁ = \(\frac{-2.736 \pm \sqrt{(2.736)^2+4 \times 4.9 \times 10}}{2 \times 4.9}\)
⇒ t₁ = \(\frac{-2.736 \pm \sqrt{203.48}}{9.8}\)
t₁ = \(\frac{-2.736+14.265}{9.8}=\frac{11.5288}{9.8}\)
= 1.176 = 1.18 sec

Question 10.
Two balls are projected from the same point in directions 30° and 60° with respect to the horizontal. What is the ratio of their initial velocities if they
(a) attain the same height ?
(b) have the same range 1
Answer:
θ₁ = 30°, θ₂ = 60°
a) First ball maximum height H₁

Additional Problems

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:
a) a railway carriage moving without jerks between two stations.
b) a monkey sitting on top of a man cycling smoothly on a circular track.
c) a spinning cricket ball that turns sharply on hitting the ground.
d) a tumbling beaker that has slipped off the edge of a table.
Solution:
a) The carriage can be considered a point object because the distance between two stations is very large as compared to the size of the railway carriage.

b) The monkey can be considered as a point object if the cyclist describes a circular track of very large radius because in that case the distance covered by the cyclist is quite large as compared to the size of monkey. The monkey can not be considered as a point object. If the cyclist describes a circular track of small radius, because in that case the distance covered by the cyclist is not very large as compared to the size of the monkey.

c) The cricket ball can not be considered as a point object because the size of the spinning cricket ball is quite appreciable as compared to the distance through which the ball may turn on hitting the ground.

d) A beaker sleeping off the edge of a table cannot be considered as a point, object because the size of the beaker is not negligable as compared to the height of the table.

Question 2.
The position-time (x – t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in fig. Choose the correct entries in the brackets below :

a) (A/B) lives.closer to the school than (B/A).
b) (A/B) starts from the school earlier than (B/A).
c) (A/B) walks faster than (B/A).
d) A and B reach home at the (same/ different) time.
e) (A/B) overtakes (B/A) on the road (once/twice).
Solution:
a) As OP < OQ, A lives closer to the school than B.

b) For x = 0, t = 0 for A; while t has some finite value for B.
Therefore, A starts from the school earlier than B.

c) Since the velocity is equal to slope of x -1 graph in case of uniform motion and slope of x -1 graph for B is greater than for A, hence B walks faster than A.

d) Corresponding to points P and Q, the value of t from x – t graph for A and B is same, which can be checked by drawing lines through P and Q parallel to x axis. Thus both A and B reach home at the same time.

e) The x – t graph for A and B intersect each other only once. Since B starts from the school after- wards, therefore B overtakes A on the road once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5 km h^-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm and returns home by an auto with a speed of 25 km h^-1. Choose suitable scales and plot the x-t graph of her motion.
Answer:
Time taken in reaching office
= \(\frac{\text { Distance }}{\text { Speed }}=\frac{2.5}{5}\) = 0.5 hr.
Time taken in returning from office = \(\frac{2.5}{25}\)
= 0.1 hr = 6 min

It means the woman reaches the office at 9.30 am and returns home at 5.06 p.m. The x – t graph of this motion will be as shown in fig.

Question 4.
A man walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:
The effective distance travelled by drunkard in 8 steps = 5 – 3 = 2 m.

Therefore he take 32 steps to walk 8 meters. Now he will have to cover 5 mt more to reach the pit, for which he has to take only 5 forward steps.

Therefore he will have to take = 32 + 5 = 37 steps to move 13 mts. Thus he will fall into pit after taking 37 steps i.e., after 37 seconds from the start.

Question 5.
A jet airplane travelling at the speed of 500 km h^-1 ejects its products of combustion at the speed of 1500 km h^-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?
Answer:
Lt v_p be the velocity of the products w.r.to ground
Lt us consider the direction of motion of airplane to be positive direction of x-axis.
Here, speed of jet plane v_A = 500 km h^-1
Relative speed of products of combustion w.r.to jet plane v_ρA = -1500 km h^-1
Relative velocity of the products w.r.to air plane is
v_PA – v_P – v_A = -1500
(Or)
v_P = v_A – 1500 = 500 – 1500
= – 1000 km h^-1
Here -ve sign shows the direction of products of combustion is opposite to that of airplane. Thus the magnitude of relative velocity is 1000 km h^-1.

Question 6.
A car moving along a straight highway with speed of 126 km h^-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop ?
Answer:
Here, u = 126 km h^-1 = \(\frac{126 \times 1000}{60 \times 60}\) ms^-1
= 35 ms^-1; v = 0, S = 200 m, a = ?
and t = ?
We know v² = u² + 2as
0 = (35)² + 2 × a × 200 (Or)
a = \(\frac{-(35)^2}{2 \times 200}=\frac{-49}{16}\)
= -3.06 ms^-2
As v = u + at
0 = 35 + \(\left(\frac{-49}{16}\right) \mathrm{t}\) (Or)
t = \(\frac{35 \times 16}{49}=\frac{80}{7}\) = 11.43s

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s^-2. If after 50 s, the guard of B just moves past the driver of A, what was the original distance between them ?
Solution:
For train A; u = 72 km h^-1 = \(\frac{72 \times 1000}{60 \times 60}\)
= 20 m s^-2; t = 50s; a = 0, S = S_A;
As S = ut + \(\frac{1}{2}\) at²
∴ S_A = 20 × 50 + \(\frac{1}{2}\) × 0 × 50²
= 1000 m
For train B; u = 72 kms^-1 = 20 ms^-2;
a = 1 ms^-2, t = 50/S, S = s_-B
As, S = ut + \(\frac{1}{2}\) at²
∴ S_B = 20 × 50 + \(\frac{1}{2}\) × 1 × 50² = 2250 m
Taking the guard of the train B in the last compartment of the train B, it follows that original distance between two trains + length of train A + length of train B = S_B – S_A.
(Or) Original distance between the two trains
400 + 400 = 2250 – 1000 = 1250
(Or) Original distance between the two trains
= 1250 – 800 = 450 m

Question 8.
On a two-lane road, car A is travelling with a speed of 36 km h^-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?
Solution:
Velocity of car A = 36 km h^-1 = 10 ms^-1
Velocity of car B or C = 54 km h^-1 = 15 ms^-1
Relative velocity of B w.r.to
A = 15 – 10 = 5 ms^-1
Relative velocity of C w.r.to
A = 15 + 10 = 25 ms^-1
As, AB = AC = 1 km = 1000 m
Time available to B (Or) C for crossing
A = \(\frac{1000}{25}\) = 40s
If car B accelerates with acceleration a, to cross A before car C does, then
u = 5 ms^-1, t = 40s, s = 1000 m, a = ?
Using s = ut + \(\frac{1}{2}\) at²
1000 = 5 × 40 + \(\frac{1}{2}\) × a × 40² (Or)
1000-200 = 800 a (Or)
a = 1 m/s²

Question 9.
Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h^-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road ?
Solution:
Let v km h^-1 be the constant speed with which the buses ply between the towns A and B.
The relative velocity of the bus (for the motion A to B) w.r.to the cyclist C i.e., in the direction in which the cyclist is going = (v – 20) kmh^-1.
The relative velocity of the bus from B to A W r to the cyclist = (v + 20) kmh^-1.
The distance travelled by the bus in time T(min) = vT
As per question \(\frac{v T}{v-20}\) = 18 (Or) vT
= 18v – 18 × 20 ………………… (i)
and \(\frac{v T}{v+20}\) = 6 (Or) vT = 6v + 20 × 6 …………….. (ii)
Equations (i) and (ii) we get
18v- 18 × 20 = 6v + 20 × 6 (Or)
12v = 20 × 6 + 18 × 20 = 480
(Or) v = 40 kmh^-1
Putting this value of v in (i) we get 40
T = 18 × 40- 18 × 20 = 18 × 20
(Or) T = 18 × 20/40 = 9 min.

Question 10.
A player throws a ball upwards with an initial speed of 29.4 ms^-1.
a) What is the direction of acceleration during the upward motion of the ball ?
b) What are the velocity and accele-ration of the ball at the highest point of its motion ?
c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
d) To what height does the ball rise and after how long does the ball return to the player’s hands ?
(Take g = 9.8 ms^-2 and neglect air resistance).
Solution:
a) Since the ball is moving under the effect of gravity, the direction of acceration due the gravity is always vertically downwards.

b) At the highest point, the velocity of the ball becomes zero and acceleration is equal to the acceleration due to gravity = 9.8 ms^-2 in vertically downward direction.

c) When the highest point is chosen as the location for x = 0 and t = 0 and vertically downward direction to be positive direction of x- axis and upward direction as negative direction of x-axis.
During upward motion, sign of position is negative, sign of velocity is negative and sign of acceleration is positive.
During downward motion, sign of position is positive, sign of acceleration is also positive.

d) Let the t be time taken by the ball to reach the highest point where height from ground to be S.
Taking vertical upward motion of the ball,
we have
u = -29.4 m/s^-1, a = 9.8 m/s^-2,
v = 0, S = 5, t = 2
As v² – u² = 2as
0 – (29.4)² = 2 × 9.8 × s (Or)
S = \(\frac{-(29.4)^2}{2 \times 9.8}\) = -44.1 m
Here -ve sign shows that distance is covered in upward direction.
As v = u + at
∴ 0 = -29.4 + 9.8 × t (or) t = \(\frac{-29.4}{9.8}\) = 3s
It means time of ascent = 3s
When an objective move under the effect of gravity alone, the time of ascent is always equal to time of descent.
Therefore total time after which the ball returns the player’s hand = 3 + 3 = 6s

Question 11.
Read each statement below carefully and state with reasons and examples, if it is true or false :
A particle in one-dimensional motion
a) with zero speed at an instant may have non-zero acceleration at that instant
b) with zero speed may have non-zero velocity.
c) with constant speed must have zero acceleration.
d) with positive value of acceleration must be speeding up.
Solution:
a) True, when a body is thrown vertically upwards in the space. Then at the highest point, the body has zero speed but has downward acceleration equal to acceleration due to gravity.

b) False, because velocity is the speed of body in a given direction. When speed is zero, the magnitude of velocity of body is zero, hence velocity is zero.

c) True, when a particle is moving along a strait line with a constant speed, its velocity remains constant with time. Therefore, acceleration ( = Change in velocity/time) is zero.

d) The statement depends upon the choice of instant of time as origin, when the body is moving along a strait line with positive acceleration. The velocity of the body at on instant of time is v = u + at.

The given statement is not correct. If a is positive and μ is negative, at the instant of time taken as origin. Then for all times before the time for which v vanishes, there is slowing down of the particle i.e., The speed of the particle keeps on decreasing with time. It happens when body is projected vertically upwards. However the given statement is true if u is positive, at the instant of time as origin. When the body is moving along a straight line with positive acceleration. The velocity of body at any instant of time t is v = u + at.

The given statement is not correct if a is positive and u is negative, at the instant of time taken as origin. Then for all times before the time for which v vanishes, there is slowing down of the particle i.e., the speed of particle keeps on decreasing with time. It happens when body is projected vertically upwards. However, the given statement is true if u is positive and a is positive, at the instant of time taken as origin. It is so when the body is falling vertically upwards.

Question 12.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Solution:
Taking vertical downward motion of ball from a height 90 m We have
u = 0, a = 10 m/s², S = 90 m, t = ?, v = ?
t = \(\sqrt{\frac{25}{a}}=\sqrt{\frac{2 \times 90}{10}}=3 \sqrt{25}\) = 4.245
V = \(\sqrt{2 a \mathrm{~s}}=\sqrt{2 \times 10 \times 30}=30 \sqrt{2} \mathrm{~m} / \mathrm{s}\)

Rebound velocity of ball,
u¹ =\(\frac{9}{10} v=\frac{9}{10} \times 30 \sqrt{2}=27 \sqrt{2}\) m/g
Time to reach the highest point is
t¹ = \(\frac{u^1}{a}=\frac{27 \sqrt{2}}{10}=2.7 \sqrt{2}\) = 3.81 S
Total time = t + t¹ = 4.24 + 3.81 = 8.05 S
The ball will take further 3.1 S to fall back to floor, where its velocity before striking the floor = 2.7\(\sqrt{2}\) m/s.
Time to reach the highest point is,
t¹ = \(\frac{u^1}{a}=\frac{27 \sqrt{2}}{10}=2.7 \sqrt{2}\) = 3.81 S
Total time = t + t¹ = 4.24 + 3.81 = 8.05 S
The ball will take further 3.81 S fall back to floor, where its velocity before striking the
floor = 2.7\(\sqrt{2}\) m/s.
Velocity of ball after striking the floor
= \(\frac{9}{10}\) \(\sqrt{2}\) = 24.3 \(\sqrt{2}\) m/s.
Total time elapsed before upward motion of ball.
= 8.05 + 3.81 = 11.86 S
Thus the speed – time graph of this motion will be as shown in fig.

Question 13.
Explain clearly, with examples, the distinction between :
a) magnitude of displacement (sometimes called distance) over an interval of time and the total length of path covered by a particle over the same interval:
b) magnitude of average velocity over an interval of time and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
Answer:
a) Magnitude of displacement of a particle in motion for a given time is shortest distance between the initial and final position of the particle in that time, where as the total length of the path covered by particle is actual path traversed by the particle in the given time. If particle goes from A to B and B to C in time t as shown in fig. Then magnitude of displacement = Distance AC.
Total path length = Distance AB + Distance AC
From the above we note that total path length (AB + AC) is greater than magnitude of displacement (AC).

If there is motion of the particle in one dimension i.e., along a straight line, then magnitude of displacement becomes equal to totalpath length transversed by the particle in the given time.

b) Magnitude of average velocity
= \(\frac{\text { Magnitude of displacement }}{\text { Time interval }}\)
= \(\frac{A C}{t}\) and average speed
= \(\frac{\text { Total path length }}{\text { Time interval }}=\frac{(A B+B C)}{t}\)
As, (AB + BC) AC, So average speed is greates than the magnitude of average velocity. If the particle is moving along a straight line, then in a given time the magnitude of displacement is equal to total path length transvered by particle in that time, so average speed is equal to magnitude of average velocity.

Question 14.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the
a) magnitude of average velocity and
b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 m in, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Solution:
Time taken by man to go from his home to market t₁ = \(\frac{\text { Distance }}{\text { Speed }}=\frac{2.5}{5}=\frac{1}{2}\) h
Time taken by man to go from market to his home t₂ = \(\frac{2.5}{7.5}=\frac{1}{3}\) h
Total time taken = t₁ + t₂ = \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
h = 50 min.
i) 0 to 30 min
a) Average velocity
= \(\frac{\text { Displacement }}{\text { Time }}=\frac{2.5}{1 / 2}\) = 5 km/h
b) Average speed
= \(\frac{\text { Displacement }}{\text { Time }}=\frac{2.5}{1 / 2}\) = 5 km/h

ii) 0 to 50 min
Total distance travelled
= 2.5 + 2.5 = 5 km
Total displacement = zero
a) Average velocity = \(\frac{\text { Displacement }}{\text { Time }}\) = 0
b) Average speed = \(\frac{\text { Distance }}{\text { Time }}=\frac{5}{5 / 6}\) = 6 km/h

iii) 0 to 40 min
Distance moved in 30 min (from home to market) = 2.5 km
Distance moved in 10 min (from market to home) with speed 7.5 km/h = 7.5 × \(\frac{10}{60}\)
= 12.5 km
So displacement = 2.5 – 1.25 = 1.25 km
Distance travelled = 2.5 + 1.25 = 3.75 km
a) Avg velocity = \(\frac{1.25}{(40 / 60)}\) = 1.875 km/h
b) Avg speed = \(\frac{3.75}{(40 / 60)}=\) = 5.625 km/h

Question 15.
In Exercises 3.13and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider in ‘antaneous speed and magnitude of velocity. The instantaneous speed is alway equal to the magnitude of instantaneous velocity. Why ?
Solution:
Instaneous speed (v_ins) of the particle at an instant is the first derivative of the distance w.r.to time at that instant of time
i.e., v_ins = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)
Since in instantaneous speed we take only a small interval of time (dt) during which direction of motion of a body is not supposed to change, here there is no difference between total path length and magnitude of displacement for small interval of time dt. Hence instantaneous speed is always equal to magnitude of instantaneous velocity.

Question 16.
Look at the graphs (a) to (d) (fig.) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Solution:
a) This graph does not represent one dimensional motion because, at the given instant of time, the particle will have two positions, which is not possible in one dimensional motion.

b) This graph does not represent one dimensional motion because, at the given instant of time, particle will have velocity in positive as well as in negative direction which is not possible in one dimensional motion.

c) It also does not represent one dimensional motion, because this graph tells that the particle can have the negative speed but the speed of the particle can never be negative.

d) It also does not represent one dimensional motion, because this graph tells that the total path length decreases after certain time but total path length of a moving particle can never decrease with time.

Question 17.
Figure shows the x – t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves In a straight line for t < 0 and on a parabolic path for t > 0 ? If not, suggest a suitable physical context for this graph.

Solution:
No, because the x – t graph does not represent the trajectory of the path followed by a particle.

From the graph, it is noted that at t = 0, x = 0 context. The above graph can represent the motion of a body falling freely from a tower uncles gravity.

Question 18.
A police van moving on a highway with a speed of 30 km h^-1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 ms^-1 with what speed does the bullet hit the thief’s car ? (Note : Obtain that speed which is relevant for damaging the thief’s car),
Solution:
Muzzle speed of bullet, υ_B = 150 m/s = 540 km h^-1
Speed of police van, υ_p = 30 km/h
Speed of theif car, υ_T = 192 km/h
Since the bullet is sharing the velocity of the police van, its effective velocity is
υ_B = υ_B + υ_P = 540 + 30 = 570 km/h
The speed of bullet w.r.to the theifs car moving in the same direction.
v_BT = v_B – v_T = 570 – 192 = 378 378 km/h
= \(\frac{378 \times 1000}{60 \times 60}\) = 105 m/s

Question 19.
Suggest a suitable physical situation for each of the following graphs (Fig.):

Solution:
In fig(a) : The x – t graph shows that intially x is zero i.e, at rest, then it increases with time, attains a constant value and again reduces to zero with time, then it increases in opposite direction till it again attains a constant value i.e., comes to rest. The similar In fig(b) : The velocity changes sign again and again with passage of time and every time some similar speed is lost. The similar physical situation arises when a ball is thrown up with some velocity, returns back and falls freely on striking the floor, it rebounds with reduced speed each it strikes against the floor.

Question 20.
Figure gives the x – t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, -1.2 s.

Solution:
In the S.H.M. acceleration a = ω²x, where co (i.e., angular frequency) is constant.

i) At time t = 0.35, x is negative, the slope of x -1 graph plot is negative, hence position and velocity are negative, Since a =ω²x, hence, acceleration is positive.
ii) At time t = 1.25, x is positive, the slope of x – t plot is also positive hence position and velocity are positive. Since a = -ω²x, hence acceleration is negative.
iii) At t = 1.2S, x is negative, the slope of x – t plot is also negative. But since both x and t are negative here, hence velocity is positive. Finally acceleration ‘a’ is also positive.

Question 21.
Figure gives the x – t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest and in which is it the least ? Give the sign of average velocity for each interval.

Solution:
We know that average speed in a small interval of time is equal to slope of x -1 graph in that interval of time. The avg speed is the greatest in the interval 3 because slope is greatest and the average speed is least in interval 2 because slope is last three.

Question 22.
Figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?

Solution:
We know that average acceleration in a small interval of time is equal to slope of velocity-time graph in that interval. As the slope of v – t graph is maximum in interval 2 as compared to other intervals 1 and 3, hence the magnitude of average acceleration is greatest in interval 2.
The average speed is greatest in interval 3 for obvious reasons.
In interval 1, the speed of v – t graph is positive. Hence acceleration a is positive. The speed u is positive in this interval due to obvious reasons.

In interval 2, the slope of v – t graph is negative, hence acceleration a is negative. The speed u is positive in this interval due to obvious reasons.

In interval 3, the v – t graph is parallel to time axis, therefore acceleration a is zero in this interval but u is positive due to obvious reasons. At points A, B, C and D the v – t graph is parallel to time axis. Therefore acceleration is zero at ail the four points.

Question 23.
A three-wheeler starts from rest, accelerates uniformly with 1 ms^-2 on a straight road for 10 s and then moves with uniform velocity. Plot the distance covered by the vehicle during the n^th second (n = 1, 2, 3 ) versus n. What do you expect this plot to be during accelerated motion : a straight line or a parabola ?
Solution:
Here u = 0, a = 1 m/s²
Distance covered in n th second is
D_n = u + \(\frac{a}{2}\)(2n-1) = 0 + \(\frac{1}{2}\) (2n – 1) = 0.5 (2n – 1)
Putting n = 1, 2, 3, …………… we can find the value of D_n. The various values of n and corresponding values of D_n are shown below :

On plotting a graph between D_n and n, we get a st. line AB as shown in fig. From (1) D_n∝_n so the graph is a straight line. After 10S the graph is straight line BC parallel to time axis.

Question 24.
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 ms^-1. How much time does the ball take to return to his hands ? If the lift starts moving up with a uniform speed of 5 ms^-1 and the boy again throws the ball up with the maximum speed he can how long does the ball take to return to his hands ?
Solution:
Taking vertical upward direction as the positive direction of x-axis.
When lift is stationary, consider the motion of the ball going vertically upwards and coming down to the hands of the body. We have
u = 49 m/s, a = 9.8 m/s², t = ? x – x₀ = S = 0
As S = ut + \(\frac{1}{2}\) at²
0 = 49 t + \(\frac{1}{2}\) (-9.8)t² (Or) 49t = 4.9 t² (Or)
t = 49/4.9 = 10 sec
When lift starts moving with uniform speed. As the lift starts moving upwards with uniform speed of 5 m/s, there is no change in the relative velocity of the ball w.r.to the boy which remains 49 m/s. Hence even in the case, the ball will return to the boys hand after 10 sec.

Question 25.
On a long horizontally moving belt a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the
a) speed of the child running in the direction of motion of the belt ?
b) speed of the child running opposite to the direction of motion of the belt ?
c) time taken by the child in (a) and (b) ?
Which of the answer alter if motion is viewed by one of the parents ?

Solution:
Let us consider left to right to be the positive direction of x-axis.
a) Here, the velocity of belt υ_B = + 4 km/h; Speed of child w.r. to belt υ_C
= + 9 Km/h = \(\frac{5}{2}\) m/s
Speed of the child w.r.to stationary observer,
υ_C¹ = υ_C + υ_B = 9 + 4 = 13 km/h

b) Here, υ_B = + 4 km/h, υ_C = -9 km/h
Speed of the child w.r.to stationary observer
υ_C¹ = υ_C + υ_B = -9 + 4 = -5 km/h
Here negative sign shows that the child will appears to run in a opposite to the direction f motion of the belt.

c) Distance between the parents S = 540 m Since parents and child are located on the same belt, the speed of the child as observed by stationary observer in either direction (Either from mother to father (or) from father to mother) will be 9 km/h.
Time taken by child in case (a) and (b) is
t = \(\frac{50}{(5 / 2)}\) = 20 S
If motion is observed by one of the parents, answer to case (a) (Or) case (b) will be altered.
It is so because speed of child w.r. to either of mother (or) father is 9 km/h. But answer (c) remains unaltered due to the fact that parents and child are on the same belt and such all are equal affected by the motion of the belt.

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 ms^-1 and 30 ms^-1. Verify that the graph shown in Fig. correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 ms^-2. Give the equations for the linear and curved parts of the plot.

Solution:
Time vertical upward motion of the first tone for time t, we have
x₀ = 200 m, u = 15 m/s, a = -10 m/s², t = t₁, x = x₁
As x = x₀ + ut + \(\frac{1}{2}\) at²
x₁ = 200 + 15 t + \(\frac{1}{2}\) (-10)t² (Or)
x₁ = 200 + 15 t – 5 t² ……………….. (i)
Taking vertical upward motion of the second stone for time t,
We have
x₀ = 200 m, u = 30 m/s^-1, a = -10 m/s^-2, t = t₁, x = x₂
Then x₂ = 200 + 30 t – \(\frac{1}{2}\) × 10 t²
= 200 + 30 t – 5t²
When the first stone hits the ground x₁ = 0,
So t² – 3t – 40 = 0
(Or) (t – 8) (t + 5) = 0 ……………….. (ii)
∴ Either t = 8 S (Or) – 5S
Since t = 0 corresponds to the instant, when the stone was projected. Hence negative time has no meaning in this case. 50 t = 8S. When the second stone hits the ground, x₂ = 0.
0 = 200 + 30 t – 5t² (Or) t² – 6t – 40 = 0 (Or) (t – 10) (t + 4) = 0
Therefore, either t = 10 s (Or) t = -4s
Since t = -4s is meaningless, So t = 10s
Relative position of second stone w.r.to first is
= x₂ – x₁ = 15 t …………………….. (ii)
From (i) and (ii)
Since (x₂ – x₁) and t are linearly related, therefore the graph is a straight line till t = 8s For maximum separation t = 8 S, so maximum separation = 15 × 8 = 120 m After 8 seconds only second tone would be in motion for 2 seconds, so the graph is in accordance with the aquadratic equation x₂ = 200 + 30t – 5t² for the interval of time 8 seconds to 10 seconds.

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown in Fig. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2s to 6s.

What is the average speed of the particle over the intervals in (a) and (b) ?
Solution:
a) Distance travelled by the particle between 0 to 10s will be
= Area of ∆ OAB, whose base is 10s and height is 12 m/s
= \(\frac{1}{2}\) × 10 × 12 = 60 m 60
Average speed = \(\frac{60}{10}\) = 6mS^-1

b) Let s₁ and s₂ be the distances covered by the particle in the time interval t₁ = 2s to 5s and t₂ = 5s to 6s, then total distance covered in time interval t = 2s to 6s will be s = s₁ + s₂ ……………….. (i)

To find s₁: let us consider u₁ is the velocity of particle after 2 seconds and a1 is the acceleration of the particle during the time interval zero to 5 seconds.
Then u₁ = 0, v = 12 m/s,
a = a , a₁ and t = 5s
We have a₁ = \(\frac{v-u}{t}=\frac{12-0}{5}=\frac{12}{5}\)
= 2.4 m/s²
∴ u₁ = υ + a₁t = 0 + 2.4 × 2 = 4.8 m/s^-1 Thus for the distance travelled by particle in 3 seconds (i.e, time interval 2s to 5s), we have
u₁ = 4.8 m/s, t₁ = 3s, a₁ = 2.4 m/s², s₁ = ?
As s₁ = u₁t₁ + \(\frac{1}{2}\) a₁t₁²
S₁ = 4.8 + 3 × \(\frac{1}{2}\) × 2.4 × 32 = 25.2 m

To find s₂ : let a₂ be the acceleration of the particle during the motion t = 5s to t = 10s
We have a₂ = \(\frac{0-12}{10-5}\) = -2.4 m/s²
Taking motion of the particle in time interval t = 5s to t = 6s We have
u₁ = 12 m/s^-1, a₂ = -2.4 m/s², t₂ = 1s, s₂ = ?
As s₂ = u₂t + \(\frac{1}{2}\) a₂t₂²
s₂ = 12 × 1 + \(\frac{1}{2}\) (-2.4) 1² = 10.8 m
∴ Total distance travelled s = 25.2 + 10.8
= 36m
Average velocity = \(\frac{36}{6-2}=\frac{36}{4}\) = 9 m/s

Question 28.
The velocity-time graph of a particle in onedimensional motion is shown in Fig.

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂ :
a) x(t₂) = x(t₁) + v(t₁) (t₂ – t₁) + (\(\frac{1}{2}\)) a(t₂ – t₁)²
b) v(t₂) = v(t₁) + a(t₂ – t₁)
c) v_average = (x(t₂) – x(t₁)) / (t₂ – t₁)
d) v_average = (v(t₂) – v(t₁)) / (t₂ – t₁)
e) x(t₂) = x(t₂) + V_average (t₂ – t₂) + (\(\frac{1}{2}\)) a_average (t₂ – t₁)²
f) x(t₂) = x(t₁) = area under the y – t curve bounded by the t-axis and the dotted line shown.
Solution:
From the graph we note that the slope is not constant and is not uniform, hence the relations (iii), (iv), (v) are correct.

Textual Examples

Question 1.
A car is moving along a straight line. Say OP Fig. 3.1. It moves from O to P in 1 8s and returns from P to Q in 6.0s. What are the average velocity and average speed of the car in going (a) from O to P? arid (b) from O to P and back to Q?
Solution:

Thus, in this case the average speed is equal to the magnitude of the average velocity.

b) In this case,

Question 2.
The position of an object moving along x-axis is given by x = a + bt² where a = 8.5 m, b = 2.5 ms^-2 and t is measured in seconds. What is its velocity at t = 0s and t = 2-.0s. What is the average velocity between t = 2.0s and t = 4.0s ?
Solution:
In notation of differential calculus, the velocity is
υ = \(\frac{d x}{d t}=\frac{d}{d t}\)(a + bt²) = 2b t = 5.0 t ms^-1
At t = 0s, υ = 0 ms^-1 and at t = 2.0s, υ = 10 ms^-1
Average velocity = \(\frac{x(4.0)-x(2.0)}{4.0-2.0}\)
= \(\frac{a+16 b-a-4 b}{2.0}\) = 6.0 × b
= 6.0 × 2.5 = 15 ms^-1

Question 3.
Obtain equations of motion for cons-tant acceleration using method of calculus.
Answer:
By definition a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\)
dυ = a dt
Integrating both sides
\(\int_{v_0}^v d v=\int_0^t a d t\)
= \(a \int_0^t d t\) (a is constant)
υ – υ₀ = at
υ = υ₀ + at
Further υ = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)
dx = υ dt
Integrating both sides
\(\int_{x_0}^x d x=\int_0^t v d t=\int_0^t\left(v_0+a t\right) d t\)
x – x₀ = υ₀ t + \(\frac{1}{2}\) a t²
x = x₀ + υ₀ t + \(\frac{1}{2}\) a t²
We can write
a = \(\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}=v \frac{d v}{d x}\)
or, υ dυ = a dx
Integrating both sides,
\(\int_{v_0}^v v d v=\int_{x_0}^x a d x\)
\(\frac{v^2-v_0^2}{2}\) = a(x – x₀)
υ² = υ₀² + 2a(x – x₀)
The advantage of this method is that it can be used for motion with non-uniform acceleration also.
Now, we shall use these equations to some important cases.

Question 4.
A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. [T.S. A.P. Mar. 15]
(a) How high will the ball rise ? and
(b) how long will it be before the ball hits the ground ? Take g = 10 ms^-2. (Actual valqp is 9.8 ms^-2)
Solution:
a) Let us take the y—axis in the vertically upward direction with zero at the vertically upward direction with zero at the ground, as shown in fig. 3.13.
Now υ₀ = + 20 ms^-1
a = -g = -10 ms^-2,
υ = 0 ms^-1
If the ball rises to height y from the point of launch, then using the equation
υ² + υ₀² + 2a(y – y₀)
we get
0 = (20)² + 2(-10) (y – y₀)
Solving, we get, (y – y₀) = 20 m

b) We can solve this part of the problem in two ways. Note carefully the methods used.

First Method : In the first method, we split the path in two parts : the upward motion (A to B) and the downward motion (B to C) and calculate the corresponding time taken t₁ and t₂. Since the velocity at B is zero, we have :
υ = υ₀ + at
0 = 20 – 10 t₁
t₁ = 2s
Or,
This is the time in going from A to B. From B, or the point of the maximum height, the ball falls freely under the acceleration due to gravity. The ball is moving in negative y direction. We use equation
y = y₀ + υ₀t + \(\frac{1}{2}\) at²
We have, y₀ = 45 m, y = 0, υ₀ = 0,
a = -g = – 10 ms^-2
0 = 45 + (\(\frac{1}{2}\)) (-10) t₂²
Solving, we get t₂ = 3s
Therefore, the total time taken by the ball
before it hits the ground = t₁ + t₂ = 2s + 3s = 5s.

Second method : The total time taken can also be calculated by nothing the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation.
y = y₀ + υ₀t + \(\frac{1}{2}\) at²
Now y₀ = 25 m, y = 0 m
υ₀ = 20 ms^-1, a = -10 ms^-2, t = ?
0 = 25 + 20t + (\(\frac{1}{2}\))(-10)t²
Or, 5t² -20t – 25 = 0
Solving this quadratic equation for t, we get t = 5s

Question 5.
Free-fall : Discuss the motion of an object under fall. Neglect air resistance.
Solution:
An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the earth’s radius, g can be taken to be constant, equal to 9.8 ms^-2. Free fall is thus a case of motion with uniform acceleration.

We assume that the motion is in y-direction, more correctly in -y-direction because we choose upward direction as positive Since that acceleration due to gravity is always downward, it is in the negative direction and we have
a = – g = -9.8 ms^-2
The object is released from rest at y = 0. Therefore, υ₀ = 0 and the equations of motion become :


Motion of an object under free fall.
a) Variation of acceleration with time.
b) Variation of velocity with time.
c) Variation of distance with time.

Question 6.
Galileo’s law of odd numbers : The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity (namely 1:3:5:7 )ⁿ Prove it.
Solution:
We have y = –\(\frac{1}{2}\) gt² Using this equation, we can calculate the position of the object after different time intervals, 0, τ, τ2, τ3 which are given in second column of table 3.2. If we take (-1/2) gτ² as y₀ – the position coordinate after first time interval τ, then third column gives the positions in the unit of y₀. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3 : 5 : 7 : 9 :11 as shown in the last column.

Question 7.
Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (-υ₀) and the braking capacity or deceleration, – a that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of υ₀ and a.
Solution:
Let the distance travelled by the vehicle before it stops be ds. Then, using equation of motion υ² = υ₀² + 2 ax and noting that υ = 0, we have the stopping distance
d_s = \(\frac{-v_0^2}{2 a}\)
Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).

For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15,20 and 25 m/s which are nearly consistent with the above formula.

Question 8.
Reaction time : When a situation demands our immediate action. It takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then, the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual.

You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Fig. 3.15). After you catch it, find the distance d travelled by the ruler. In a particular case, d was found to be 21.0 cm. Estimate reaction time.

Solution:
The ruler drops under free fall. Therefore υ₀ = 0 and a = -g = -9.8 ms^-2. The distance travelled d and the reaction time t, are related by d = –\(\frac{1}{2}\) gt_r² Or t_r = \(\sqrt{\frac{2 \mathrm{~d}}{\mathrm{~g}}}\)s.
Given d = 21.0 cm and g = 0.8 ms^-2 the reaction time is
t_r = \(\sqrt{\frac{2 \times 0.21}{9.8}}\) s ≅ 0.2 s

Question 9.
Two parallel rail tracks run north- south. Train A moves north with a speed of 54 km h^-1 and train B moves south with a speed of 90 kmh^-1. What is the
a) velocity of B with respect to A ?
b) velocity of ground with respect to B ?
c) velocity of a monkey running on the roof of the train A against. Its motion (with a velocity of 18 kmh^-1 with respect to the train A) as observed by a man standing on the ground ?
Solution:
Choose the positive direction of x-axis to be from south to north. Then,
υ_A = + 54 km h^-1 = 15 ms^-1
υ_B = – 90 km h^-1 = -25 ms^-1
Relative velocity of B with respect to A = υ_A – υ_B = – 40 ms^-1, i.e., the train B appears to A to move with a speed of 40 ms^-1 from north to souch.
Relative velocity of ground with respect to B = 0 – υ_B = 25 ms^-1
In.(c), let the velocity of the monkey with respect to ground be υ_M. Relative velocity of the monkey with respect to A,
υ_MA = υ_M – υ_A = – 18 km h^– = – 5 mh^-1
Therefore, υ_M = (1 5 – 5)mh^-1 = 10 mh^-1.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements

Very Short Answer Questions

Question 1.
Distinguish between accuracy and precision. [A.P. Mar. 16, 15; T.S. Mar. 15, 13]
Answer:
Accuracy:

1. The accuracy of measurement of any physical quantity made by any measuring instrument is a measure of how close the measured value is to the true value of the quantity.
2. The accuracy depends on errors.

Precision:

1. The precision of the measuring instrument denotes upto what limit (or) resolution the quantity can be measured with the given instrument.
2. The precision does not depend on errors.

Question 2.
What are the different types of errors that can occur in a measurement?
Answer:
Mainly there are three types of errors.

1. Systematic errors
2. Random errors
3. Gross errors.

Question 3.
How can systematic errors be minimized or eliminated? [Mar. 14]
Answer:
Systematic errors can be minimised by improving experimental techniques, selecting better instruments and removing personal bias as far as possible. For a given set up, these errors may be estimated to a certain extent and the necessary corrections may be applied to the readings.

Question 4.
Illustrate how the result of a measurement is to be reported indicating the error involved. [T.S., A.P. – Mar. 17]
Answer:
To measure any physical quantity, we compare it with a standard (unit) of that quantity. No measurement is perfect as the errors involved in the process cannot be removed completely. Hence inspite of our best efforts, the measured value is always some what different from its actual value (or) true value.

Question 5.
What are significant figures and what do they represent when reporting the result of a measurement ? [Mar. 13]
Answer:
The digits of a number that are definitely known plus one more digit that is estimated are called significant digits (or) significant figures.
Example : Time period of a simple pendulum is 1.62, the digits 1 and 6 are reliable while the digit 2 is uncertain. The measured value has three significant figures.

Question 6.
Distinguish between fundamental units and derived units. [T.S. – Mar. 16]
Answer:

1. Units of fundamental quantities are called fundamental units. Fundamental units can neither derived from one another, nor can they be resolved into other units.
2. Units of derived quantities are called derived units.

Question 7.
Why do we have different units for the same physical quantity ?
Answer:
We have different systems like C.G.S system, M.K.S system, F.P.S system and S.l system. Hence we have different physical units for the same physical quantity.

Question 8.
What is dimensional analysis ?
Answer:
Dimensional analysis is the representation of derived physical quantities in terms of units of fundamental quantities.

With the help of dimensional analysis to check the correctness of the equation, convert one system of units into other system and derive certain equations relating physical quantities.

Question 9.
How many orders of magnitude greater is the radius of the atom as compared to that of the nucleus ?
Answer:
Size of atomic nucleus = 10^-14 m
Size of atom = 10^-10 m
Hence size of atom is 1CT4 m greater the size of the nucleus.

Question 10.
Express unified atomic mass unit in kg.
Answer:
1 unified atomic mass unit = \(\frac{1}{12}\) of the mass of carbon – 12 atom.
1 a.m.u = 1.66 × 10^-27 kg

Short Answer Questions

Question 1.
The vernier scale of an instrument has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, then using this instrument what would be the minimum inaccuracy in the measurement of distance ?
Answer:
Value of each main scale division = 0.5 mm
No. of vernier scale divisions = 50
In vernier callipers, L.C = \(\frac{S}{N}=\frac{\text { Value of one main scale division }}{\text { No. of vernier scale divisions }}\)
L.C. = \(\frac{0.5}{50}\) = 0.01 m.m
The minimum inaccuracy in the measurement of distance is 0.01 m.m. (Or)
Value of 1 MSD = 0.5 m.m
NV.S.D = (N – 1) M.S.D
50 V.S.D – 49 M.S.D
1 V.S.D = \(\) M.S.D. = \(\frac{49}{50}\) × 0.5
L.C = 1 M.S.D – 1 V.S.D
(0.5 – \(\frac{49}{50}\) × 0.5) = (1 – \(\frac{49}{50}\))0.5 = \(\frac{1}{50}\) × 0.5
L.C. = 0.01 m.
∴ Minimum inaccuracy in the measurement of distance is 0.01 m.m.

Question 2.
In a system of units, the unit of force is 100 IM. unit of length is 10 m and the unit of time is 100s. What is the unit of mass in this system ?
Answer:
Force (F) = 100 N; Length (L) = 10 m; Time (T) = 100 s F = ma
mass (m) = \(\frac{F}{a}=\frac{F}{L T^{-2}}\); m = \(\frac{\mathrm{FT}^2}{\mathrm{~L}}=\frac{100 \times(100)^2}{10}\) = 10⁵ kg

Question 3.
The distance of a galaxy from Earth is of the order of 10²⁵ m. Calculate the order of magnitude of the time taken by light to reach us from the galaxy.
Answer:
Distance of galaxy from earth = 10²⁵ m
Velocity of light (C) = 3 × 10⁸ m/s
Time taken by the light (t) = \(\frac{\text { Distance of galaxy from earth }}{\text { velocity of light }}\)
t = \(\frac{\mathrm{d}}{\mathrm{c}}\) ⇒ t = \(\frac{10^{25}}{3 \times 10^8}\) = 0.33 × 10¹⁷ s ⇒ t = 3.3 × 10¹⁶ sec.

Question 4.
The Earth-Moon distance is about 60 Earth radius. What will be the approximate diameter of the Earth as seen from the Moon?
Answer:
Distance between Earth-Moon = 60 R
Radius of the earth = R
r = 60R = 60 × 6400 × 10³ (R = 6400km)
θ = 1 sec = \(\frac{1}{60}\) Min = \(\frac{1}{60 \times 60}\) degree ⇒ θ = \(\frac{1}{60 \times 60} \times \frac{\pi}{180}\) radian
Then r = \(\frac{l}{\theta}\) ⇒ l = rθ ⇒ l = 60 × 6400 × 10³ × \(\frac{1}{60 \times 60} \times \frac{\pi}{180}\)
l = 11.16 × 10³ km ⇒ Diamter (l) = 11.16 × 10³ km

Question 5.
Three measurements of the time for 20 oscillations of a pendulum give t₁ = 39.6 s, t₂ = 39.9 s and t₃ = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurements?
Answer:
No. of oscillations = 20
t₁ = 39.6 sec, t₂ = 39.9 sec, t₃ = 39.5 sec
Mean value = \(\frac{t_1+t_2+t_3}{3}=\frac{39.6+39.9+39.5}{3}=\frac{119}{3}\) = 39.66
Mean value = 39.7 sec;
Precision = 0.1 sec.
Accuracy is the closeness of measured value with true value.
Hence 39.6 s is accuracy.

Question 6.
1 calorie = 4.2J where 1J = 1 kg m²s². Suppose we employ a system of units in which the unit of mass is \(\hat{a}\) kg, the unit of length is \(\hat{a}\) m and the unit of time is \(\tilde{a}\) s, show that a calorie has a magnitude 4.2 \(\hat{\mathrm{a}}^{-1} \hat{a}^{-2} \tilde{\mathrm{a}}^{-2}\) in the new system.
Answer:
1 Calorie = 4. 2J ⇒ 1 J = 1 kg m² s^-2
1 calorie = 4.2 kg m² s^-2
In new system, 1 calorie = 4.2 \(\hat{a} \hat{a}^2 \hat{a}^{-2}\)

Question 7.
A new unit of length is chosen so that the speed of light in vacuum is 1 ms-2. If light takes 8 min and 20 $ to cover this distance, what is the distance between the Sun and Earth in terms of the new unit ?
Answer:
V = Speed of light in vacuum = 1 m/s
Time taken (t) = 8 min 20 sec = 500 sec
Distance between the sun and earth (d) = \(\frac{\mathrm{V}}{\mathrm{t}}\) ⇒ d = \(\frac{1}{500}\) = 0.002 m.

Question 8.
A student measures the thickness of a human hair using a microscope of magnification 100. He makes 20 observations and finds that the average thickness (as viewed in the microscope) is 3.5 mm. What is the estimate of the thickness of hair ?
Answer:
Magnification of microscope = M = 100
Observed thickness = 3.5 m.m
Magnification (M) = \(\frac{\text { Observed thickness }}{\text { Real thickness }}\) ⇒ 100 = \(\frac{3.5}{\text { Real thickness }(t)}\) ⇒ t = \(\frac{3.5}{100}\) = 0.035 m.m.

Question 9.
A physical quantity X is related to four measureable quantities a, b, c and d as follows. X = a² b³ c^5/2d^-2
The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. What is the percentage error in X ?
Answer:
X = a² b³ c^5/2d^-2
\(\frac{\Delta \mathrm{a}}{\mathrm{a}}\) × 100 = 1%, \(\frac{\Delta \mathrm{b}}{\mathrm{b}}\) × 100 = 2%, \(\frac{\Delta \mathrm{c}}{\mathrm{c}}\) × 100 = 3%, \(\frac{\Delta \mathrm{d}}{\mathrm{d}}\) × 100 = 4%
Percentage error in X is
(\(\frac{\Delta \mathrm{X}}{\mathrm{X}}\)) × 100 = 2(\(\frac{\Delta \mathrm{a}}{\mathrm{a}}\) × 100) + 3(\(\frac{\Delta \mathrm{b}}{\mathrm{b}}\) × 100) + \(\frac{5}{2}\) (\(\frac{\Delta \mathrm{b}}{\mathrm{b}}\) × 100) + 2 \(\frac{\Delta \mathrm{d}}{\mathrm{d}}\) × 100
= 2 × 1 + 3 × 2 + \(\frac{5}{2}\) × 3 + 2 × 4 = 2 + 6 + \(\frac{15}{2}\) + 8
⇒ (\(\frac{\Delta \mathrm{X}}{\mathrm{X}}\) 100 = 23.5 %

Question 10.
The velocity of a body is given by v = At² + Bt + C. If v and t are expressed in SI what are the units of A, B and C ?
Answer:
Given V = At² + Bt + C
According to principle of homogeneity,

1. V = At² ⇒ A =\(\frac{\mathrm{V}}{\mathrm{t}^2}=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^2}\) = [LT^-3] ⇒ A = ms^-3
2. V = Bt ⇒ B = \(\frac{\mathrm{V}}{\mathrm{t}}=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}}\) = [LT^-1] ⇒ B = ms^-2
3. V = C ⇒ C = LT^-1 ⇒ C = ms^-1

Problems

Question 1.
In the expression P = E l² m^-5 G^-2 the quantities E, l, m and G denote energy, angular momentum, mass and gravitational constant respectively. Show that P is a dimension-less quantity.
Solution:
P = E L² m^-5 G^-2
Energy (E) = [ML² T^-2]
Angular momentum (L) = ML² T^-1
Mass = [M]
Universal gravitational constant
(G) = [M^-1L³T^-2]
P = [ML² T^-2] [ML² T^-1]² [M]^-5 [M^-1 L³ T^-2]^-2
P = M^1+2-5+2 L^2+4-6 T^-2-2+4
P = [M⁰ L⁰ T⁰]
Hence P is dimensional less quantity.

Question 2.
If the velocity of light c. Planck’s constant h and the gravitational ‘ constant G are taken as fundamental quantities : then express mass, length and time in terms of dimensions of these quantities.
Solution:
i) M ∝ G^x C^y h^z
[M¹ L⁰ T⁰] = [M^-1 L³ T^-2]^x [LT^-1]^y [ML² T^-1]^z
[M¹L⁰T⁰] = M^-x+z L^3x+y+2z T^-2x-y-z = 0
– x + z = 1, 3x + y + 2z = 0, – 2x – y – z = 0
Solving these equations, we get
x = \(\frac{-1}{2}\), y = \(\frac{1}{2}\), z = \(\frac{1}{2}\)
M = G^{\(\frac{-1}{2}\)} C^{\(\frac{1}{2}\)} h^{\(\frac{1}{2}\)}
M = \(\sqrt{\frac{\mathrm{hc}}{\mathrm{G}}}\)

ii) Length (l) ∝ G^x C^y h^z
[M⁰ L¹ T⁰] = [M^-1 L³ T^-2]^x [LT^-1]^y [ML² T^-1]^z
[M⁰L¹T⁰] = M^-x+z L^3x+y+2z T^-2x-y-z
Applying principle of homogeneity
-x + z = 0, 3x + y + 2z = 0, -2x – y – z = 0
Solving these equations, we get
x = \(\frac{1}{2}\), y = \(\frac{-3}{2}\), z = \(\frac{1}{2}\)
From equation (1), l = G^{\(\frac{1}{2}\)} C^{\(\frac{-3}{2}\)} h^{\(\frac{1}{2}\)}
Length (l) = \(\sqrt{\frac{\mathrm{Gh}}{\mathrm{C}^3}}\)

iii) Time (T) ∝ G^x C^y h^z
[M⁰ L⁰ T¹] = [M^-1 L³ T^-2]^x [LT^-1]^y [ML² T^-1]^z
[M⁰L⁰T¹] = M^-x+z L^3x+y+2z T^-2x-y-z
Applying principle of homogeneity
-x + z = 0, 3x + y + 2z = 0, -2x – y – z = 1
Solving these equations, we get
x = \(\frac{1}{2}\), y = \(\frac{-5}{2}\), z = \(\frac{1}{2}\)
From equation (1), l = G^{\(\frac{1}{2}\)} C^{\(\frac{-5}{2}\)} h^{\(\frac{1}{2}\)}
Length (l) = \(\sqrt{\frac{\mathrm{Gh}}{\mathrm{C}^5}}\)

Question 3.
An artificial satellite is revolving around a planet of mass M and radius R in a circular orbit of radius r. Using dimensional analysis show that the period of the satellite
T = \(\frac{k}{R} \sqrt{\frac{r^3}{g}}\)
where k is a dimensionless constant and g is acceleration due to gravity.
Solution:
T = \(\frac{k}{R} \sqrt{\frac{r^3}{g}}\)
L.H.S. : Time period = [T]
R.H.S. : \(\frac{k}{R} \sqrt{\frac{r^3}{g}}=\frac{1}{L} \sqrt{\frac{L^3}{L T^{-2}}}\) = [T]
∴ L.H.S. = R.H.S.
Above equation is correct.

Question 4.
State the number of significant figures in the following.
a) 6729
b) 0.024
c) 0.08240
d). 6.032
e) 4.57x
Solution:
a) 6729 – 4     Significant figures
b) 0024 – 2            ”
c) 008240 – 4         ”
d) 6.032 –  4           ”
e) 4.57 × 10⁸ – 3     ”

Question 5.
A stick has a length of 12.132 cm and another has a length of 12.4 cm. If the two sticks are placed end to end what is the total length ? If the two sticks are placed side by side, what is the difference in their lengths?
Solution:
a) Let lengths of the rods are
l₁ = 12.132 cm, l₂ = 12.4 cm
Here has one decimal place and l₁ has to be rounded to have only two decimal places.
l = l₁ + l₂ = 12.13 + 12.4 = 24.53
This is to be rounded off to have one decimal place only.
∴ The result is 24.5 cm

b) l₁ = 12.132 cm, l₂ = 12.4 cm
Here l₁ is rounded only two decimal places,
l₂ – l₁ = 12.4 – 12.13 = 0.27
This should be rounded off to have only one decimal place, l₂ – l₁ = 0.3

Question 6.
Each side of a cube is measured to be 7.203 m. What is
(i) the total surface area and
(ii) the volume of the cube, to appropriate significant figures ?
Solution:
Length of a side = 7.203 m
(i) Total surface area = 6a²
= 6 × (7.203)²
= 311.29
This should be rounded off to four significant figure as 7.203 as 4 significant figures.
∴ The result = 311.3m²

(ii) Volume of the cube = a³
= (7.203)³
= 373.71
This should be rounded off to four significant figures has 7.203 has four significant figures.
∴ Volume = 373.7 m³

Question 7.
The measured mass and volume of a body are 2.42 g and 4.7 cm³ respectively with possible errors 0.01 g and 0.1 cm³. Find the maximum error in density.
Solution:
M = 2.42 g, V = 4.7 cm³
∆M = 0.01 g, ∆V = 0.1 cm³
Density (ρ) = \(\frac{M}{V}\)
\(\frac{\Delta \rho}{\rho}=\frac{\Delta \mathrm{M}}{\mathrm{M}}+\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{0.01}{2.42}+\frac{0.1}{4.7}\)
= 0.004 + 0.02 = 0.024
% error is \(\frac{\Delta \rho}{\rho}\) × 100 = 0.024 × 100
= 2.4% ≈ 2%

Question 8.
The error in measurement of radius of a sphere is 1%. What is the error in the measurement of volume ?
Solution:
Radius of the sphere \(\frac{\Delta \mathrm{r}}{\mathrm{r}}\) × 100 = 1%
Volume (V) = \(\frac{4}{3}\) πr³
\(\frac{\Delta \mathrm{V}}{\mathrm{V}}\) × 100 = 3 × \(\frac{\Delta \mathrm{r}}{\mathrm{r}}\) × 100 = 3 × 1% = 3%

Question 9.
The percentage error in the mass and speed are 2% and 3% respectively. What is the maximum error in kinetic energy calculated using these quantities ?
Solution:
\(\frac{\Delta \mathrm{M}}{\mathrm{M}}\) × 100 = 2%, \(\frac{\Delta \mathrm{V}}{\mathrm{V}}\) × 100 = 3%;
Kinetic energy = \(\frac{1}{2}\) mV²
\(\frac{\Delta \mathrm{K}}{\mathrm{K}}\) × 100 = \(\frac{\Delta \mathrm{M}}{\mathrm{M}}\) × 100 + 2 \(\frac{\Delta \mathrm{V}}{\mathrm{V}}\) × 100
= 2 + 2(3) = 2 + 6 = 8%

Question 10.
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). If the size of the hydrogen molecule is about 1A°, what is the ratio of molar volume to the atomic volume of a mole of hydrogen ?
Solution:
Molar volume = 22.4 lit = 22.4 × 1000 c.c.
= 22.4 × 10^-3 m³
Diameter of the hydrogen molecule
= 1 A° = 10^-10 m
Radius (r) = \(\frac{D}{2}=\frac{10^{-10}}{2}\) = 0.5 × 10^-10 m
Volume (V) = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × 3.14 × (0.5 × 10^-10)³
v = \(\frac{1.57}{3}\) × 10^-30 = 0.5233 × 10^-30 m³
Atomic volume = V × Avagadro’s number (N)
= 0.5233 × 10^-30 × 6.023 × 10²³ = 3.151 × 10^-7
∴ \(\frac{\text { Molar volume }}{\text { Atomic volume }}=\frac{22.4 \times 10^{-3}}{3.151 \times 10^{-7}}\)
= 7.108 × 10^-4

Additional Problems

Note : In stating numerical answers, take care of significant figures.

Question 1.
Fill in the blanks
a) The volume of a cube of side 1 cm is equal to …………….. m³
b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ……………. (mm)²
c) A vehicle moving with a speed of 18 km h^-1 covers ……………. m in 1 s
d) The relative density of lead is 11.3. Its density is …………….. g cm^-3 or …………….. kg m^-3.
Solution:
a) Here, length of slide L = 1 cm = 10^-2 m
Volume of cube = L³ = (10^-2 m)³ = 10^-6 m³

b) Here, r = 2.0 cm = 20 mm
h = 10.0 cm = 100 mm
Surface area of solid cylinder = (2rcr) × h
= 2 × \(\frac{22}{7}\) × 20 × 100 mm²
= 1.26 × 10⁴ mm²

c) Here, speed V = 18 km
h^-1 = \(\frac{18 \times 1000 \mathrm{~m}}{60 \times 60 \mathrm{~s}}\) = 5 ms^-1
∴ Distance covered in 1 second = 5 m

d) Here, relative density = 11.3
density = 11.3 g k.c = \(\frac{11.3 \times 10^{-3}}{\left(10^{-2} \mathrm{~m}\right)^3}\)
= 11.3 × 10³ kg m^-3

Question 2.
Fill in the blanks by suitable conversion of units.
a) 1 kg m² s^-2 = ………….. g cm² s^-2
b) 1 m = ………….. 1y
c) 3.0 ms^-2 = ………….. km h^-2
d) G = 6.67 × 10^-11 N m² (kg)^-2 = …………. (cm)³ s^-2 g^-1.
Solution:
a) 1 kg m² s²
= 1 × 10³ g (10² cm)²s^-2 = 10⁷ g cm² s^-2

b) We know, 1 light year = 9.46 × 10¹⁵ m
∴ 1 m = \(\frac{1}{9.46 \times 10^{15}}\) light year
= 1.053 × 10^-16 light year

c) 3ms^-2 = 3 × 10^-3 km \(\left(\frac{1}{60 \times 60} h\right)^{-2}\)
= 3 × 10^-3 × 3600 × 3600 km h^-2

d) G = 6.61 × 10^-11 Nm² kg^-2
= 6.67 × 10^-11 (kg ms^-2) m² kg^-2
= 6.67 × 10^-11 m³ s^-2 kg^-1
= 6.67 × 10^-11 (100 cm)³ s^-2 (1000 g)^-1
= 6.67 × 10^-8 cm³ s^-2 g^-1

Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m² s^-2. Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals b m, the unit of time is g s. Show that a calorie has a magnitude 4.2 a^-1b^-2 g² in terms of the new units.
Solution:
Here 1 calorie = 4.2 J = 4.2 Kg m^-3S^-1 …………… (1)
As new unit of mass = 1kg
∴ 1 kg = \(\frac{1}{\mathrm{a}}\) new unit of mass = a^-1
similarly, 1m = \(\frac{1}{\mathrm{b}}\) = b^-1
1s = \(\frac{1}{\mathrm{g}}\) = g^-1
Putting these values in eq(1), we obtains.
1 calorie = 4.2 (a^-1 new unit of mass)
(b^-1 new unit of length)²
(g^-1 new unit of time)²
1 calorie = 4.2 a^-1 b^-2 g².

Question 4.
Explain this statement clearly :
‘To call a dimensional quantity ‘large1 or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary :
a) atoms are very small objects
b) a jet plane moves with great speed
c) the mass of Jupiter is very large
d) the air inside this room contains a large number of molecules
e) a proton is much more massive than an electron
f) the speed of sound is much smaller than the speed of light.
Solution:
i) The statement is true. This is because a dimensionless quantity can be large or small only in comparison to some standard. For example, angle is dimensionless ∠θ = 60° is larger than ∠θ = 30°, but smaller than ∠θ = 90°

ii) a) The size of an atom is smaller than the sharp tin of a pin.
b) A jet plane moves faster than a superfast train.
c) The mass of Jupiter is very large compared to the mass of earth.
d) The air inside this room contains more number of molecules than in one mole of air.
e) The statement is already correct.
f) The statement is already correct.

Question 5.
A new unit of length is chosen such that the speed of light in vacuum in unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ?
Solution:
We are given that velocity of light in vacuum, c = 1 new unit of length s^-1
Time taken by light of Sun to reach the Earth,
t = 8 min 20 s = 8 × 60 + 20 = 500 s
∴ Distance between the Sun and the Earth,
x = C × t =1 new unit of length s^-1 × 500 s
= 500 new units of length

Question 6.
Which of the following is the most
precise device for measuring length :
a) a vernier cuHipers with 20 divisions on the sliding scale
b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
c) an optical instrument that can measure length to within a wave-length of light ?
Solution:
The most precise device is that whose least count is minimum. Now
a) Least count of this vernier callipers
= 1 SD – 19 SD = 1 SD = \(\frac{19}{20}\) SD = \(\frac{1}{20}\) SD
= \(\frac{1}{20}\) mm = \(\frac{1}{200}\) cm = 0.005 cm

b) Least count of screw gauge
= \(\frac{\text { Pitch }}{\text { No. of divisions on circular scale }}\)
= \(\frac{1}{100}\) mm = \(\frac{1}{1000}\) cm = 0.01 cm

c) Wavelength of light,
λ = 10^-5 cm = 0.00001 cm
Obviously, the most precise measurement is with optical instrument.

Question 7.
A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?
Soution:
Magnification m = \(\frac{\text { observed with }(\mathrm{y})}{\text { real width }(\mathrm{x})}\)
x = \(\frac{\mathrm{y}}{\mathrm{m}}=\frac{3.5 \mathrm{~mm}}{100}\) = 0.035 mm

Question 8.
Answer the following :
a) You are given a thread and a metre scale. How will you estimate the diameter of the thread ?
b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?
c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?
Solution:
a) The diameter of a thread is so small that it cannot be measured using a metre scale. We wind a number of turns of the thread on the metre scale. So that the turns are closely touching one another. Measure the length (l) of the windings on the scale which contains n number of turns.
∴ Diameter of thread = \(\frac{1}{n}\)

b) As least count
= \(\frac{\text { Pitch }}{\text { Number of divisions on circular scale }}\)
∴ Theoretically speaking, least count decreases on increasing the number of divisions on the circular scale. Hence accuracy would increase. Practically, it may not be possible to take the reading precisely due to low resolution of human eye.

c) A large number of observations will give more reliable result than smaller number of observations. This is because of probability of making a positive random error of certain magnitude is equal to that of making a negative random errors are likely to cancel and the result may be more reliable.

Question 9.
The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on to a screen and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement.
Solution:
Here, area of object = 1.75 cm² and area of image = 1.55 m² = 1.55 × 10⁴ cm²
∴ Areal magnification = \(\frac{\text { area of image }}{\text { area of object }}\)
= \(\frac{1.55 \times 10^4}{1.75}\) = 8857
Linear magnification = \(\sqrt{8857}\) = 94.1

Question 10.
State the number of significant figures in the following :
a) 0.007 m²
b) 2.64 × 10²⁴ kg
c) 0.2370 g cm^-3
d) 6.320 J
e) 6.032 N m^-2
f) 0.0006032 m²
Solution:
The number of significant figures is as given below.
a) one
b) three
c) four
d) four
e) four
f) four

Question 11.
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Solution:
Here, length l = 4.234 m;
Breath, b = 1.005 m
Thickness, t = 2.01 cm = 2.01 × 10^-2 m
Area of the sheet = 2(l × b + b × t + t × l)
= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
2(4.3604739) = 8.7209478 m²
As area can contain a maximum of three significant digits, therefore, rounding off, we get
Area = 8.72 m²
Also, volume = l × b × t
V = 4.234 × 1.005 × 0.0201
V = 0.0855289
V = 0.855 m³

Question 12.
The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ?
Solution:
Here, mass of the box, m = 2.3 kg
Mass of one gold piece,
m₁ = 20.15 g = 0.02015 kg
Mass of another gold piece,
m₂ = 20.17g = 0.02017 kg
a) Total mass = m + m₁ + m₁
= 2.3 + 0.02015 + 0.02017
= 2.34032 kg
As the result is correct only upto one place of decimal, therefore, on rounding off total mass = 2.3 kg

b) Difference in masses
= m₂ – m₁ = 20.17 – 20.15
= 0.02g (correct upto two places of decimal)

Question 13.
A physical quantity P is related to four observables a, b, c and d as follows :
P = a³b²/ (\(\sqrt{c}\)d )
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?
Solution:
Here, P = \(\frac{a^3 b^2}{\sqrt{c} d}\)
Maximum fractional error in P is given by \(\frac{\Delta \mathrm{P}}{\mathrm{P}}\)
= \(3 \frac{\Delta \mathrm{a}}{\mathrm{a}}+2 \frac{\Delta \mathrm{b}}{\mathrm{b}}+\frac{1}{2} \frac{\Delta \mathrm{c}}{\mathrm{c}}+\frac{\Delta \mathrm{d}}{\mathrm{d}}\)
= \(3\left(\frac{1}{100}\right)+2\left(\frac{3}{100}\right)+\frac{1}{2}\left(\frac{4}{100}\right)+\frac{2}{100}\)
= \(\frac{13}{100}\) = 0.13
Percentage error in
P = \(\frac{\Delta \mathrm{P}}{\mathrm{P}}\) × 100 = 0.13 × 100 = 13%
As the result (13% error) has two significant figures, therefore if P turns out to be 3.763, the result would be rounded of to 3.8.

Question 14.
A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion :
a) y = a sin 2 π t/T) b) y = a sin vt,
c) y = (a/T) sin t/a,
d) y = (a\(\sqrt{2}\)) (sin 2πt /T + cos 2 πt/T)
(a = maximum displacement of the particle, v = speed of the particle. T = time- period of motion). Rule out the wrong formulas on dimensional grounds.
Solution:
The argument of a trigonometrical function i.e., angle is dimensionless. Now in
a) \(\frac{2 \pi t}{T}=\frac{T}{T}\) = 1 = (M⁰L⁰T⁰)
…………… dimensionless
b) vt = (LF^-1) (T) = L = (M⁰L¹T⁰)
………….. not dimensionless

c) \(\frac{\mathrm{t}}{\mathrm{a}}=\frac{\mathrm{T}}{\mathrm{L}}\) = (L^-1T^-1)
………….. not dimensionless

d) \(\frac{2 \pi t}{\mathrm{~T}}=\frac{\mathrm{T}}{\mathrm{T}}\) = 1 = [M⁰L⁰T⁰]
…………… dimensionless

Question 15.
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einsterin). A boy recalls the relation almost correctly but for-gets where to put the constant c. He writes :
m = \(\frac{m_0}{\left(1-v^2\right)^{1 / 2}}\)
Guess where to put the missing c.
Solution:
According to the principle of homogenity of dimensions power of M, L, T on either side of the formula must be equal. For this, on RHS, the denominator (1 – v²)^1/2 should be dimensionless. Therefore, instead of (1 – v²)^1/2,
we should write \(\left(\frac{1-v^2}{c^2}\right)^{1 / 2}\)
Hence the correct formula would be
m = \(\frac{m_0}{\left(\frac{1-v^2}{c^2}\right)^{1 / 2}}\)

Question 16.
The unit of length convenient on the atomic scale is known as an angstrom and is denoted byA° : 1A° = 1010m. The size of a hydrogen atom is about 0.5 A°. What is the total atomic volume in m³ of a mole of hydrogen atoms ?
Solution:
Here r = 0.5 A° = 0.5 × 10^-10 m
Volume of each atom of hydrogen = \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\) × 3.14(0.5 × 10^-10) = 5.236 × 10^-31 m³
Number of hydrogen atoms in one gram mole of hydrogen = Avagadro’s number
= 6.023 × 10²³
∴ Atomic volume of one gram mole of hydrogen atom
= 5.236 × 10^-31 × 6.023 × 10²³
= 3.154 × 10^-7 m³

Question 17.
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L of (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to about 1A°). Why is this ratio so large ?
Solution:
Atomic volume = \(\frac{4}{3}\)πR³ × N
= \(\frac{4}{3}\) (0.5 × 10^-10)³ × 6.023 × 10²³
= 3.154 × 10^-7 m³
Molar volume = 22.4 lit = 22.4 × 10^-3 m³
\(\frac{\text { Molar volume }}{\text { Atomic volume }}=\frac{22.4 \times 10^{-3}}{3.154 \times 10^{-7}}\) = 7.1 × 10⁴
The ratio is large due to large intermolecular separations.

Question 18.
Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby tr6es, houses etc., seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving; these distant objects seem to move with you).
Solution:
The line joining the object to the eye is called the line of sight. When a train moves rapidly, the line of sight of a near by tree changes its direction of motion rapidly. Therefore, the trees appear to run in opposite direction. On the contrary, the line of sight of far off objects does not change its direction. So much, due to extremely large distance from the eye. Hence distant hill tops, moon, the stars etc., appear stationary.

Question 19.
The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart 21. in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit 3 × 10¹¹ m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1″ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1″ (second) of arc from opposite ends of a baseline equal to the distance from the Earth to Sol. the Sun. How much is a parsec in terms of metres ?
Solution:
Here, length of baseline
= distance from each to the sun
= 1 A.U = 1.5 × 10¹¹ m
Parallax angle, θ = 1
\(\frac{1^1}{60}=\frac{1^{\circ}}{60 \times 60}=\frac{\pi}{180} \times \frac{1}{60 \times 60}\) radian
r = 1 par sec = ?; From l = rθ = \(\frac{l}{\theta}\)
= \(\frac{1.5 \times 10^{\prime \prime}}{\pi / 180 \times 60 \times 60} \mathrm{~m}\) = 3.1 × 10¹⁶ m
Hence 1 parsec = 3.1 × 10¹⁶ m

Question 20.
The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parasecs ? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?
Solution:
x = 4.29 ly = 4.29 × 9.46 × 10¹⁵ m
= \(\frac{4.29 \times 9.46 \times 10^{15}}{3.08 \times 10^{16}}\) par sec
= 1.323 par sec
θ = \(\frac{l}{r}=\frac{2 A u}{x}\)
= \(\frac{2 \times 1.496 \times 10^4}{4.29 \times 9.46 \times 10^{15}}\) radian = 1.512 sec

Question 21.
Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modem science where precise measurements of length, time, mass etc., are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Solution:
Precise measurements of physical quantities like length, mass and time are the primary requirements for development of quantitative laws of physics or any other science. For example, in the measurement of distance of moon from earth by laser beam, very accurate measurement of time taken is required. Similarly, for measuring distance, elevation and velocity of an aeroplane by radar method, time measurement has to be accurate. For measuring distances of near by stars, accurate measurement of parallax angle is required.

In the field of crystallography, precise measurement of length is needed to determine interatomic distances using a mass spectrometer, the precision measurement of masses of atoms are made.

Question 22.
Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
a) the total mass of rain-bearing clouds over India during the Monsoon
b) the mass of an elephant
c) the wind speed during a storm
d) the number of strands of hair on * your head
e) the number of air molecules in your classroom.
Solution:
a) During the Monsoon, meteorologist record about 100 cm of rainfall, i.e.,
h – 100 cm = 1m
Area of our country,
A = 3.3 million square km
= 3.3 × 10⁶ (10³m²)
= 3.3 × 10¹² m⁶
∴ Volume of rain water,
v = A × h = 3.3 × 10¹² 1m³
As density of water,
P = 103 km/m³
∴ Mass of rain water
= vP = 3.3 × 10¹² × 10³ kg
= 3.3 × 10¹⁵ kg
This must be the total mass of rain bearing clouds over India.

b) To estimate the mass of an elephant, we take a boat of known base area A. Measure the depth of boat in water. Let it be x₁. Therefore, volume of water displaced by the boat , V₁ = Ax₁ move the elephant into this boat. The boat gets deeper into water. Measure the depth of boat now into water, Let it be x₁
∴ Volume of water displaced by boat and elephant V₂ = Ax₂
∴ Volume of water displaced by the elephant V = V₂ – V₁ = A(x₂ – x₁)
If ρ is density of water, then mass of elephant
= mass of water displaced by it
= V_ρ = A(X₂ – X₁) ρ

c) The wind speed during a storm can be estimated using a gas filled balloon. In figure OA is normal position of a gas filled balloon, when there is no wind. As the wind blows to the right, the balloon drifts to position B in one second. The angle of drift ∠AOB = θ is measured, if h is the height of the balloon, then AB = d = hθ This is the distance travelled by the balloon in one second it must be the wind speed.

d) For this, we measure the area of the head . that carrier the hair let it be A. Using a screw guage, we measure thickness of hair, let it bad.
∴ Area of cross section of hair = πd²
Assuming that the distribution of hair over the head is uniform.
The number of strands of hair
= \(\frac{\text { total area }}{\text { area of cross section of each hair }}=\frac{\mathrm{A}}{\pi \mathrm{d}^2}\)
Calculations show that number of strands of hair on human head is of the order of one million.

e) Measure the volume of room. We know that one mole of air at NTP occupies a volume of 22.4 lit i.e., 22.4 × 10^-3 m³
∴ Number of air molecules in 22.4 × 10^-3 m³ = 6.023 × 10²³
Number of air molecules in volume v of room = \(\frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}}\)v

Question 23.
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10⁷ K and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid of liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 10³⁰ kg, radius of the Sun = 7.0 × 10⁸ m.
Solution:
Here, M = 2.0 × 10³⁰ kg; R = 7.0 × 10⁸ m
Density ρ = ?
ρ = \(\frac{\text { mass }}{\text { volume }}=\frac{M}{\frac{4}{3} \pi R^3}=\frac{3 M}{4 \pi R^3}\)
= \(\frac{3 \times 2.0 \times 10^{30}}{4 \times 3.14 \times\left(7 \times 10^8\right)^3}\)
= 1.392 × 10³ kg/m³
This is the order of density of solids and liquids; and not gases.
The high density of Sun is due to inward gravitational attraction on outer layers, due to inner layers of the Sun.

Question 24.
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72, of arc. Calculate the diameter of Jupiter.
Solution:
Here, r = 824.7 × 10⁶ km
θ = 35.72″ radian
= \(\frac{35.72}{60 \times 60} \times \frac{\pi}{180}\) radian
Diameter, l = ?; As l = rθ
∴ l = 824.7 × 10⁶ × \(\frac{35.72 \times \pi}{60 \times 60 \times 180}\) km
= 1.429 × 10⁵km

Question 25.
A man walking briskly in rain with speed u must slant his umbrella forward making an angle q with the vertical. A student derives the following relation between q and v : tan θ = v and checks that the relation has a correct limit: as v → 0, θ →0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.
Solution:
The relation tan θ = v has a correct limit, as v → 0, θ → 0.
However, RHS = tan θ = [M⁰L⁰T⁰] and L.H.S = v = [M⁰L¹T^-1]
Therefore, the relation is not correct dimensionally.
We shall find the correct relation is
tan θ = \(\frac{\mathrm{v^2}}{\mathrm{rg}}\)

Question 26.
It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ?
Solution:
Error in 100 years = 0.02s
Error in 1 sec

Hence the accuracy of the standard cesium clock is measuring a time interval of 1s is 10^-12 s.

Question 27.
Estimate the average mass density of a sodium atom assuming its size to be about 2.5 A°. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m^-3. Are the two densities of the same order of magnitude ? If so why ?
Solution:
Atomic volume = \(\frac{4}{3}\)πR³ × N
= \(\frac{4}{3}\) × \(\frac{22}{7}\) (1.25 × 10^-10)³ 6.023 × 10²³ m
= 4.93 × 10^-6 m³
Average mass density
= \(\frac{\text { mass }}{\text { volume }}=\frac{23 \times 10^{-3}}{4.93 \times 10^{-6}}\)
= 4.67 × 10³ kg/m³
The two densities are not of the same order. This is due to interatomic spacing in the crystalline phase.

Question 28.
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10^-15 m. Nuclear sizes obey roughly the following empirical relation :
r = r₀A^1/3
where r is the radius of the nucleus. A its mass number, and r₀ is a constant equal to about 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different density of a sodium atom obtained in Exercise. 27.
Answer:
Let m be the average mass of nucleon (neutron or proton)
As the nucleus contains A nucleons,
∴ mass of nucleus μ = mA
Radius of nucleus r = r₀A^1/3
Nuclear density ρ = \(\frac{\text { mass }}{\text { volume }}=\frac{\mu}{\frac{4}{3} \pi r^3}\)
= \(\frac{3 m A}{4 \pi\left(r_0 A^{1 / 3}\right)^3}=\frac{3 m}{4 \pi r_0^3}\)
As m and r₀ are constant, therefore, nuclear density is constant for all nuclei.
using m = 1.66 × 10^-27 kg and r₀
= 1.2f = 1.2 × 10^-15 m
We get ρ = \(\frac{3 m}{4 \pi r_0^3}=\frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14\left(1.2 \times 10^{-15}\right)^3}\)
= 2.29 × 10¹⁷ kg m^-3
As ρ is constant for all nuclei, this must be the density of sodium nucleus also.
Density of sodium atom
ρ’ = 4.67 × 10³ kg m^-3
= \(\frac{\rho}{\rho^{\prime}}=\frac{2.29 \times 10^{17}}{4.67 \times 10^3}\) = 4.67 × 10^-3

Question 29.
A LASER is a source or very intense, monochromatic and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth ?
Solution:
Here, t = 2.56s
Velocity of laser light in vacuum,
c = 3 × 10⁸ m/s
The radius of lunar orbit is the distance of moon from the earth. Let it be x.
As x = \(\frac{c \times t}{2}\)
∴ x = \(\frac{3 \times 10^8 \times 2.56}{2}\) = 3 84 × 10⁸ m

Question 30.
A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in waiter = 1450 ms^-1).
Solution:
Here, t = 77.0 s, x = ? V = 1450 ms^-1
As x = \(\frac{\mathrm{V} \times \mathrm{t}}{2}\)
= \(\frac{1450 \times 77.0}{2}\)m; x = 55825 m

Question 31.
The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzlingn features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ?
Solution:
Here, x = ?
Time taken t = 30 billion years = 3 × 10⁹ yr
= 3 × 10⁹ × 365 × 24 × 60 × 60 s
Velocity of light in vacuum, c = 3 × 10⁸ m/s
= 3 × 10⁵ km/s
As distance = velocity × time
x = (3 × 10⁵) × 3 × 10⁹ × 365 × 24 × 60 × 60 km.
= 2.84 × 10²² km

Question 32.
It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 3 and 4. determine the approximate diameter of the moon.
Solution:
Distance of moon from earth,
ME = 3.84 × 10⁸ m
Distance of Sun from earth,

SE = 1.496 × 10¹¹ m
Distance of Sun AB = 1.39 × 10⁹ m
The situation during total solar eclipse is shown in figure.
As ∆^s ABE of CDE are similar, therefore,
\(\frac{A B}{C D}=\frac{S E}{M E}\)
CD = AB × \(\frac{M E}{S E}=\frac{1.39 \times 10^9 \times 3.84 \times 10^8}{1.496 \times 10^{11}}\)
= 3.5679 × 10⁶ m = 3567.9 km

Question 33.
A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational costant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~ 15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants ?
Solution:
Trying out with basic constants of atomic physics (speed of light c, charge on electron e, mass of electron m_c, mass of proton m_p) and universal gravitational constant G, we can arrive at a quantity which has the dimensions of time. One such quantity is
t = \(\left(\frac{c^2}{4 \pi \epsilon_0}\right)^2 \times \frac{1}{m_p m_{e^2} c^3 G}\)
Put e = 1.6 × 10^-19 c
\(\frac{1}{4 \pi \epsilon_0}\) = 9 × 10⁹,
c = 3 × 10<sup8 m/s and
G = 6.67 × 10^-11 Nm² kg^-2
t = (1.6 × 10^-19)⁴ × (9 × 10⁹)² × \(\frac{1}{1.67 \times 10^{-27} \times\left(9 \times 10^{-31}\right) \times\left(3 \times 10^8\right)^3 \times 6.67 \times 10^{-11}}\)
t = 2.18 × 10¹⁶ sec; This time is of the order of age of universe.

Textual Examples

Question 1.
Calculate the angle at (a) 1° (degree) (b) V (minute of arc or arcmin) and (c) 1″ (second of arc or are second) in radians. Use 360° = 2π rad, 1° = 60′ and 1′ = 60″.
Answer:
(a) We have 360° = 2π rad
1° = (π/180) rad = 1.745 × 10^-2 rad

(b) 1′ = 60′ = 1.745 × 10^-2 rad
V = 2.908 × 10^-4 rad, 2.91 × 10^-4 rad

(c) 1′ = 60″ = 2.908 × 10^-4 rad
1″ = 4.847 × 10^-6 rad, 4.85 × 10^-6 rad.

Question 2.
A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m and looks at O and C again. Since O is very distant, the direction BO is practically the same as AO; but he finds the line of sight of C shifted from the original line of sight by an angle θ = 40° (θ is known as parallax’) estimate the distance of the tower C from his original position A.

Answer:
Parallax angle θ = 40°; AB = AC tan θ
AC = AB/tan θ = 100 m/tan 40°
= 100 m/0.8391 = 119 m.

Question 3.
The moon is observed from two diametrically opposite points A and B on Earth. The angle 6 subtended at the moon by the two directions of observation is 1°54‘. Given the diameter of the Earth to be about 1.276 × 10⁷ m, compute the distance of the moon from the Earth.
Solution:
We have θ = 1°54′ = 114
= (114 × 60)” × (4.85 × 10^-6) rad
= 3.32 × 10^-2 rad.
Since 1″ = 4.85 × 10^-6,; rad
Also b = AB = 1.276 × 10⁷ m
Hence from D = b/θ, we have the earth-moon distance,
D = b/θ
= \(\frac{1.276 \times 10^7}{3.32 \times 10^{-2}}\) = 3.84 × 10⁸ m.

Question 4.
The Sun’s angular diamater is mea-sured to be 1920″. The distance D of the Sun from the Earth is 1.496 × 10″ m. What is the diameter of the Sun ?
Answer:
Sun’s angular diameter α = 1920”
= 1920 × 4.85 × 10^-6 rad
= 9.31 × 10^-3 rad
Sun’s diameter d = α
D = (9.31 × 10^-3) × (1.496 × 10¹¹) m
= 1.39 × 10⁹ m.

Question 5.
If the size of a nucleus in the range of 10^-15 to 10^-14 m) is scaled up to the tip of a sharp pin, what roughly is the size of an atom ? Assume tip of the pin to be in the range 10^-5 m to 10^-4 m.
Answer:
The size of a nucleus is in the range of 10^-15m and 10^-14 m. The tip of a sharp pin is taken to be in the range of 10^-5 m and 10^-4 m. Thus we are scaling up by a factor of 10¹⁰. An atom roughly of size 10^-10 m will be scaled up to a size of 1 m. Thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre long.

Question 6.
Two clocks are being tested against a standard clock located in a national laboratory. At 12 : 00 : 00 noon by the standard dock, the readings of the two clocks are :

If you are doing an experiment that requires ‘precision time interval’ measurements, which of the two clocks will you prefer ?
Answer:
The range of variation over the seven days of observations is 162 s for clock 1 and 31 s for clock 2. The average reading of clock 1 is much closer to the standard time than the average reading of clock 2. The important point is that a clock’s zero error is not as significant for precision work as its variation, because a ‘zero-error’ can always be easily corrected. Hence clock 2 is to be preferred to clock 1.

Question 7.
We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s. 2.56 s, 2.42 s, 2.71 is and 2.80 s. Calculate the absolute errors relative error or percentage error.
Answer:
The mean period of oscillation of the pendulum
T = \(\frac{(2.63+2.56+2.42+2.71+2.80) s}{5}\)
= \(\frac{13.22}{5}\) s = 2.624 s = 2.62 s
As the periods are measured to a resolution of 0.01 s, all times are to the second decimal; it is proper to put this mean period also to the second decimal.
The errors in the measurements are
2.63 s – 2.62 s = 0.01 s
2.56 s – 2.62 s = -0.06 s
2.42 s – 2.62 s = -0.20 s
2.71 s – 2.62 s = 0.09 s
2.80 s – 2.62 s = 0.18 s
Note that the errors have the same units as the quantity to be measured.
The arithmetic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes) is
∆T_mean = [(0.01 + 0.06 + 0.20 + 0.09 + 0.18)s]/5 = 0.54 s/5 = 0.11 s
That means, the period of oscillation of the simple pendulum is (2.62 ± 0.11) s i.e. it lies between (2.62 + 0.11) s and (2.62 – 0.11) s or between 2.73 s and 2.51 s. As the arithmetic mean of all the absolute errors is 0.11 s, there is already an error in the tenth of a second. Hence there is no point in giving the period to a hundredth. A more correct way will be to write T = 2.6 ± 0.1 s

Note that the last numeral 6 is unreliable, since it may be anything between 5 and 7. We indicate this by saying that the measurement has two significant figures. In this case, the two significant figures are 2, which is reliable and 6, which has an error associated with it. You will learn more about the significant figures in section 2.7.
For this example, the relative error or the percentage error is δa = \(\frac{0.1}{2.6}\) × 100 = 4.

Question 8.
The temperatures of two bodies measured by a thermometer are t₁ = 20 °C ± 0.5 °C and t₂ = 50 °C ± 0.5°C. Calculate the temperature difference and the error therein.
Answer:
t’ = t₂ – t₁
= (50 °C ± 0.5 °C) – (20 °C ± 0.5 °C)
t’ = 30 °C ± 1 °C.

Question 9.
The resistance R = V/I where V = (100 ± 5) V and I = (10 ± 0.2) A. Find the percentage error in R.
Answer:
The percentage error in V is 5 and in I it is 2.
The total error in R would therefore be 5 + 2 = 7%.

Question 10.
Two resistors of resistances R₁ = 100 ± 3 ohm and R₂ = 200 ± 4 ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the relation R
= R₁ + R₂ and for (b) \(\frac{1}{R^{\prime}}+\frac{1}{R_1}+\frac{1}{R_2}\) and \(\frac{\Delta R^{\prime}}{R^{\prime 2}}=\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2}\)
Answer:
a) The equivalent resistance of series, combination
R = R₁ + R₂ = (100 ± 3)
ohm + (200 ± 4) ohm = 300 ± 7 ohm.

b) The equivalent resistance of parallel combination

Then, R’ = 66.7 ± 1.8 ohm
(Here, ∆R is expressed as 1.8 instead of 2 to keep in conformity with the rules of significant figures).

Question 11.
Find the relative error in Z, if Z = A⁴B^1/3/CD^3/2.
Answer:
The relative error in Z is ∆Z/Z = 4(∆A/A) + (1/3) (∆B/B) + (∆C/C) + (3/2) (∆D/D).

Question 12.
The period of oscillation of a simple pendulum is T = \(2 \pi \sqrt{L / g}\). Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. What is the accuracy in the determination of g ?
Answer:
g = 4π²L/T²; Here, T = \(\frac{t}{n}\) and ∆T = \(\frac{\Delta t}{n}\)
Therefore, \(\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{\Delta \mathrm{t}}{\mathrm{t}}\)
The errors in both L and t are the least count errors. Therefore, (∆g/g) = (∆L/L) + 2(∆T/T)
= \(\frac{0.1}{20.0}+2\left(\frac{1}{90}\right)\) = 0.027
Thus, the percentage error in g is 100 (∆g/g)
= 100(∆L/L) + 2 × 100 (∆T/T) = 3.

Question 13.
Each side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the cube to appro¬priate significant figures ?
Answer:
The number of significant figures in the measured length is 4. The calculated area and the volume should therefore be rounded off to 4 significant figures.
Surface area of the cube = 6(7.203)² m²
= 311.299254 m²
= 311.3 m²
Volume of the cube = (7.203)³ m³
= 373.714754 m³
= 373.7 m³

Question 14.
5.74 g of a substance occupies 1.2 cm³. Express its density by keeping the significant figures in view.
Answer:
There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume. Hence the density should be expressed to only 2 significant figures.
Density = \(\frac{5.74}{1.2}\) cm^-3 = 4.8 g cm^-3.

Question 15.
Let us consider an equation \(\frac{1}{2}\) mv² = mgh where m is the mass of the body, v its velocity, g is the acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct.
Answer:
The dimensions of LHS are
[M] [LT^-1]² = [M] [L²T^-2] = [M L²T^-2]
The dimensions of RHS are
[M] [LT^-2] [L] = [M] [L²T^-2] = [M L²T^-2]
The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.

Question 16.
The SI unit of energy is J = kg m²s^-2; that of speed v is ms-1 and of acceleration a is ms^-2. Which of the formulae for kinetic energy (K) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body):
(a) K = m²V³
(b) K = (1/2) mv²
(c) K = ma
(d) K = (3/16) mv²
(e) K = (1/2) mv² + ma.
Answer:
Every correct formula or equation must have the same dimensions on both sides of the equation. Also, only quantities with the same physical dimensions can be added or subtracted. The dimensions of the quantity on the right side are [M² L³ T^-3] for (a) : [ML²T^-2] for (b) and (d); [MLT^-2] for (c). The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added. Since the kinetic energy K has the dimensions of [ML²T^-2], formulas (a), (c) and (e) are ruled out. Note that dimensional arguments cannot tell which of the two, (b) or (d), is the correct formula. For this, one must turn to the actual definition of kinetic energy. The correct formula for kinetic energy is given by (b).

Question 17.
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its period using method of dimensions.
Answer:
The dependence of time period T on the quantities l, g and m as a product may be written as :
T = k lx g^y m^z
Where k is dimensionless constant and x, y and z are the exponents.
By considering dimensions on both sides, we have
[L°M°T¹] = [L¹]x [L¹T^-2]y [M¹]² = L^x+yT^-2y M^z
On equating the dimensions on both sides, we have
x + y = 0; -2y = 1 and z = 0
So that x = \(\frac{1}{2}\), y = \(\frac{1}{2}\), z = 0
Then, T = kl^1/2 g^1/2 or T = \(\mathrm{k} \sqrt{\frac{l}{\mathrm{~g}}}\)
Note that value of constant k can hot be obtained by the method of dimensions. Here it does not matter if some number multiplies the right side of this formula because that does not affect its dimensions.
Actually, k = 2π so that T = \(2 \pi \sqrt{\frac{l}{g}}\)