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Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids

Very Short Answer Questions

Question 1.
State Hooke’s law of elasticity.
Answer:
“With in the elastic limit stress directly proportional to the strain”.
Stress ∝ strain
Stress = k × strain
k = [latex]\frac{\text { Stress }}{\text { Strain }}[/latex]
Where k is modulus of elasticity.

Question 2.
State the units and dimensions of stress.
Answer:

1. Stress = [latex]\frac{\text { Force }}{\text { Area }}=\frac{F}{A}[/latex]
   S.l units → N/m² (or) Pascal
2. Dimensional formula
   Stress = [latex]\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}[/latex] = [ML^-1T^-2].

Question 3.
State the units and dimensions of modulus of elasticity.
Answer:
Modulus of elasticity (k) = [latex]\frac{\text { Stress }}{\text { Strain }}[/latex]
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2]

Question 4.
State the units and dimensions of Young’s modulus.
Answer:
Young’s modules (y) = [latex]\frac{\text { LongitudinalStress }}{\text { Longitudinal Strain }}=\frac{\frac{F}{A}}{\frac{e}{L}}[/latex]
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2]

Question 5.
State the units and dimensions of modulus of rigidity.
Answer:
Modulus of rigidity (G) = [latex]\frac{F}{A \theta}=\frac{\text { Shearing Stress }}{\text { Shearing Strain }}[/latex]
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2].

Question 6.
State the units and dimensions of Bulk modulus.
Answer:
Bulk modulus (B) = [latex]\frac{\text { Bulk Stress }}{\text { Bulk Strain }}=\frac{-P V}{\Delta V}[/latex]
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2].

Question 7.
State the examples of nearly perfect elastic and plastic bodies.
Answer:

• Nearly perfect elastic bodies are quartz fibre.
• Nearly perfect plastic bodies are dough and day.

Short Answer Questions

Question 1.
Define Hooke’s Law of elasticity, proportionality, permanent set and breaking stress.
Answer:
Hooke’s law : “With in the elastic limit stress is directly proportional to the strain”.
Stress ∝ strain
Stress = k × strain
Where k is modulus of elasticity.
Proportionality limit: The maximum stress developed in a body till it obeys Hookes law is called proportionality limit.
Permanent Set : Permanent deformation produced when a body is stretched beyond its elastic limit.
Breaking stress : The maximum stress a body can bear before it breaks.

Question 2.
Define modulus of elasticity, stress, strain and Poisson’s ratio.
Answer:
Modulus of elasticity : It is the ratio stress applied on a body to the strain produced in the body.
k = [latex]\frac{\text { Stress }}{\text { Strain }}[/latex]
S.I unit → N/m² (or) Pascal
Stress : When a body is subjected to an external force, the force per unit area is called stress.
Stress = [latex]\frac{\text { Force }}{\text { Area }}=\frac{F}{A}[/latex]
S.I unit → N/m² (or) Pascal
Strain : When deforming forces act on a body, the fractional deformation produced in the body. It has no units

Poisson’s ratio (σ) : The ratio between lateral strain to longitudinal strain of a body is called poisson’s ratio.
σ = [latex]\frac{\text { Lateral Strain }}{\text { Longitudinal Strain }}=\frac{\frac{-\Delta \mathrm{r}}{\mathrm{r}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}[/latex]

Question 3.
Define Young’s modulus. Bulk modulus and Shear modulus.
Answer:
Young’s modulus (y) : With in the elastic limit, the ratio of longitudinal stress to longitudinal strain is called young’s modulus.
y = [latex]\frac{\text { Longitudinal Stress }}{\text { Longitudinal Strain }}=\frac{\frac{F}{A}}{\frac{e}{L}}[/latex]
y = [latex]\frac{\mathrm{FL}}{\mathrm{A} \cdot \mathrm{e}}[/latex]
S.I unit → N/m² (or) Pascal

Bulk modulus (B) : With in the elastic limit, it is defined as the ratio of Bulk stress to Bulk strain
B = [latex]\frac{\text { Bulk Stress }}{\text { Bulk Strain }}[/latex]
B = [latex]\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{-\Delta \mathrm{V}}{\mathrm{V}}}=\frac{-\mathrm{PV}}{\Delta \mathrm{V}}[/latex] (∵ -ve sign indicates volume decreases)
S.I unit → N/m² (or) Pascal

Rigidity modulus (G) : With in the elastic limit, it is defined as the ratio of shearing stress to shearing strain.
G = [latex]\frac{\text { Shearing Stress }}{\text { Shearing Strain }}[/latex]
G = [latex]\frac{\frac{F}{A}}{\theta}=\frac{F}{A \theta}[/latex]
S.I unit → N/m² (or) Pascal

Question 4.
Define stress and explain the types of stress. [T.S. Mar. 16]
Answer:
Stress : The restoring force per unit area is called stress
∴ Stress = [latex]\frac{\text { Restoring Force }}{\text { Area }}=\frac{F}{A}[/latex]
Stress is classified into three types.

1. Longitudinal stress
2. Volume (or) Bulk stress
3. Tangential (or) shearing stress

1. Longitudinal stress (or) Linear stress : When a normal stress changes the length of a body, then it is called longitudinal stress.
Longitudinal stress = [latex]\frac{F}{A}[/latex]

2. Volume (or) Bulk stress : When a normal stresschanges the volume of a body, then it is called volume stress.
Volume stress = [latex]\frac{\text { Force }}{\text { Area }}[/latex] = pressure.

3. Tangential (or) shearing stress : When the stress is tangential to the surface due to the application of forces parallel to the surface, then the stress is called tangential stress.
Tangential stress = [latex]\frac{F}{A}[/latex].

Question 5.
Define strain and explain the types of strain.
Answer:
Strain : It is the ratio of change in dimension to its original dimension.
Strain = [latex]\frac{\text { Changes in dimension }}{\text { Original dimension }}[/latex]
Strain is of three types.
1. Longitudinal strain : It is the ratio of change in length to its original length.
Longitudinal strain = [latex]\frac{\text { Changes in length }}{\text { Original length }}=\frac{e}{L}[/latex]

2. Shearing strain (or) Tangential strain : When simultaneous compression and extension in mutually perpendicular direction takes place in a body, the change of shape it under goes, is called shearing strain.

Shearing strain (θ) = [latex]\frac{1}{L}[/latex]

3. Bulk (or) volume strain : It is the ratio of change in volume to its original volume is called bulk strain. It is called Bulk (or) volume strain.
Bulk strain = [latex]\frac{\text { Change in Volume }}{\text { Original Volume }}=\frac{\Delta V}{V}[/latex]

Question 6.
Define strain energy and derive the equation for the same. [Mar. 14]
Answer:
The potential energy stored in a body when stretched is called strain energy.
Let us consider a wire of length L and cross – sectional area A. Let x be the change in length of the wire by the application of stretching force F.

Strain energy per unit volume = [latex]\frac{1}{2} \times \frac{F}{A} \cdot \frac{x}{L}[/latex]
= [latex]\frac{1}{2}[/latex] × stress × strain.

Question 7.
Explain why steel is preferred to copper, brass, aluminium in heavy-duty machines and in structural designs.
Answer:
The elastic behavior of materials plays an important role in everyday life. Designing of buildings, the structural design of the columns, beams and supports require knowledge of strength of material used. The elasticity of the material is due to stress developed with in the body, when extenal force acts on it. A material is of more elastic nature if it develops more stress (or) restoring force. Steel develops more stress than copper, brass, aluminium for same strain. So steel is more elastic.
y = [latex]\frac{\text { Stress }}{\text { Strain }}[/latex]

Question 8.
Describe the behaviour of a wire under gradually increasing load. [A.P. – Mar. ’18, ’16, ’15; TS – Mar. ’18, ’15, ’13]
Answer:
When the load is increased in steps, a graph is drawn between stress on y-axis and corresponding strain on x-axis.

1. Proportionality limit : In the linear position OA, stress is proportional to strain, i.e. Hookes law is obeyed by the wire upto point A. The graph is a straight lint. When ever the stretching force at A is removed, the wire regains its original length.
A is called proportionality limit.

2. Elastic limit : In the graph B is the elastic limit.
Through the wire doesnot obey Hooke’s law at B. The wire regains its original length after removing the stretching force at B. upto point B the wire is under elastic behaviour.

3. Permanent set (or) yield point: In the graph c is the yield point. If the stretching force at c is removed, the wire doesnot regain its original length and the length of the wire changes permanently. In this position the wire flows like a viscous liquid. After the point c, the wire is under plastic behavior, c is called permanent set (or) yield point.

4. Breaking point: When the stress increased, the wire becomes thinner and thinner. When the stress increases to a certain limit the wire breaks. The stress at which the wire breaks is called breaking stress and the point D is called breaking point.

5. Elastic fatigue : The state of temperary loss of elastic nature of a body due to continuous strain is called elastic fatigue. When a body is subjected to continuous strain with in the elastic limit, it appears to have lost Hastic property temporarily to some extent and becomes weak.

Question 9.
Two identical solid balls, one of ivory and the other of wet-day are dropped from the same height onto the floor. Which one will rise to greater height after striking the floor and why ?
Answer:
Ivory ball rise to greater height after striking the floor. The ivory ball regain its original shape after striking the floor. The elastic property of ivory ball is more. Where as wet-day ball does not regain its original shape after striking the floor.

So wet-day ball acts like plastic body.

Question 10.
While constructing buildings and bridges a pillar with distributed ends is preferred to a pillar with rounded ends. Why ?
Answer:
Use of pillars (or) columns is also very common in buildings and bridges. A pillar with rounded ends supports less load than that with a distributed shape at the ends. The precise design of a bridge (or) a building has to take into account the conditions under which it will function, the cost and long period, reliability of usable materials.

Question 11.
Explain why the maximum height of a mountain on earth is approximately 10 km ?
Answer:
The maximum height of a mountain on earth is 10 km, can also be provided by considering the elastic properties of rocks. A mountain base is not under uniform compression and this provides some shearing stress to rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow.

At the bottom of a mountain of height h, the force per unit area due to the weight of the mountain is hρg. Where ρ is the density of the mountain. The material at the bottom experiences this force in the vertical direction and the sides of the mountain are free.
There is a shear component, approximately hρg itself.
Elastic limit for 3 typical rock is 30 × 10⁷ N/m²
hρg = 30 × 10⁷ (ρ = 3 × 10³ kg/m³)
h = [latex]\frac{30 \times 10^7}{3 \times 10^3 \times 10}[/latex]
h = 10 km.

Question 12.
Explain the concept of Elastic Potential Energy in a stretched wire and hence obtain the expression for it.
Answer:
“When a wire is put under a tensile stress, work is done against the inter-atomic forces. The work is stored in the wire in the form of elastic potential energy”.

Expression for elastic potential energy : Consider a wire of length L and area of cross section A is subjected to a deforming force F along the length of the wire. Let the length of the wire is elongated by l.
Young’s modulus (y) = [latex]\frac{\mathrm{FL}}{\mathrm{Al}}[/latex]
F = [latex]\frac{\mathrm{yAl}}{\mathrm{L}}[/latex] ……………. (1)
Work done due to further elongation of small length dl
Work done (dw) = F × dl = ([latex]\frac{\mathrm{yAl}}{\mathrm{L}}[/latex])dl ……………… (2)
Total work done in increasing the length of the wire from L to (L + l)
w = [latex]\int_0^1 \frac{\mathrm{yAl} }{\mathrm{L}} \mathrm{dl}=\frac{\mathrm{yA}}{2} \times \frac{l^2}{\mathrm{~L}}[/latex]
w = [latex]\frac{1}{2} \times \mathrm{y} \times\left(\frac{l}{\mathrm{~L}}\right)^2 \times \mathrm{Al}[/latex]
= [latex]\frac{1}{2}[/latex] y × stress² × volume of the wire
w = [latex]\frac{1}{2}[/latex] × stress × strain × volume of the wire.
This work is stored in the wire in elastic potential energy (u).

Long Answer Question

Question 1.
Define Hooke’s law of elasticity and describe an experiment to determine the Young’s modulus of the material of a wire.
Answer:
Hooke’s law : With in the elastic limit, stress is directly proportional to the strain.
Stress ∝ strain
Stress = k × strain
Where k is modulus of elasticity.
Determination of young’s modulus of the material of a wire:

Young’s Modulus of the Material of a wire

1. It consists of two long straight wires of same length and same area of cross-section suspended side by side from a rigid .support.
2. The wire A (reference wire) carries a metre scale M and a pan to place a weight.
3. The wire B (experimental wire) carries a pan in which known weights can be placed.
4. A vernier scale v is attached to a pointer at the bottom of the experimental wire B and the main scale M is fixed to the wire A.
5. The weights placed in the pan, the elongation of the wire is measured by the vernier arrangement.
6. The reference wire is used to compensate for any change in length that may occur due to change in room temperature.
7. Both the reference and experimental wires are given an initial small load to keep the wires straight and the vernier reading is noted.
8. Now the experimental wire is gradually loaded with more weights, the vernier reading is noted again.
9. The difference between two vernier readings gives the elongation produced in the wire.
10. Let r and L be the radius and initial length of the experimental wire. Let M be the mass that produced an elongation ∆L in the wire.
    Young’s modulus of the material of the experimental wire is given by
    y = [latex]\frac{\text { Longitudinal Stress }}{\text { Longitudinal Strain }}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}[/latex]
    y = [latex]\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}[/latex]
    y = [latex]\frac{\mathrm{MgL}}{\pi r^2 \times \Delta \mathrm{L}}[/latex]
    From above equation young’s modulus of the material of the wire is determined.

Problems

Question 1.
A copper wire of 1mm diameter is stretched by applying a force of 10 N. Find the stress in the wire.
Solution:
D = 1 m.m = 10^-3m, r = [latex]\frac{D}{2}[/latex] = 0.5 × 10^-3 m.
F = 10 N
Stress = [latex]\frac{F}{A}=\frac{F}{\pi r^2}[/latex]
= [latex]\frac{10}{3.14 \times\left(0.5 \times 10^{-3}\right)^2}[/latex]
= 1.273 × 10⁷ N/m².

Question 2.
A tungsten wire of length 20 cm is stretched by 0.1 cm. Find the strain on the wire.
Solution:
L = 20 × 10^-2 m, ∆L = 0.1 × 10^-2 m
Strain = [latex]\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{0.1 \times 10^{-2}}{20 \times 10^{-2}}[/latex] = 0.005.

Question 3.
If an iron wire is stretched by 1 %, what is the strain on the Wire ?
Solution:
Strain = [latex]\frac{\Delta \mathrm{L}}{\mathrm{L}}[/latex] = 1 %
Strain = [latex]\frac{1}{100}[/latex] = 0.01

Question 4.
A brass wire of diameter 1mm and length 2 m is streched by applying a force of 20N. If the increase in length is 0.51 mm. find
(i) the stress,
(ii) the strain and
(iii) the Young’s modulus of the wire.
Solution:
D = 1 m.m, r = [latex]\frac{D}{2}[/latex] = 0.5 × 10^-3 m
L = 2 m, F = 20 N, ∆L = 0.51 m.m = 0.51 × 10^-3 m

= 9.984 × 10¹⁰ N/m²

Question 5.
A copper wire and an aluminium wire have lengths in the ratio 3 : 2, diameters in the ratio 2 : 3 and forces applied in the ratio 4: 5. Find the ratio of increase in length of the two wires. (Y_cu = 1.1 × 10¹¹ Nm^-2, Y_Al = 0.7 × 10¹¹ Nm^-2).
Solution:

= [latex]\frac{4}{5} \times \frac{3}{2} \times\left(\frac{0.7 \times 10^{11}}{1.1 \times 10^{11}}\right) \times\left(\frac{3}{2}\right)^2[/latex]
[latex]\frac{\Delta \mathrm{L}_1}{\Delta \mathrm{L}_2}=\frac{189}{110}[/latex]

Question 6.
A brass wire of cross-sectional area 2 mm2 is suspended from a rigid support and a body of volume 100 cm3 is attached to its other end. If the decrease in the length of the wire is 0.11 mm, when the body is completely immersed in water, find the natural length of the wire.
(Y_brass = 0.91 × 10¹¹ Nm^-2, ρ_water = 10³ kg m^-3).
Solution:
A = πr² = 2 × 10^-6 m², V = 100 × 10^-6 = 10^-4 m³
∆L = 0.11 × 10^-3 m, y_Brass = 0.91 × 10¹¹ N/m², ρ = 10³ kg/m³
y = [latex]\frac{M g L}{A \times \Delta L}=\frac{v \rho g L}{A \times \Delta L}[/latex]
L = [latex]\frac{\mathrm{yA} \Delta \mathrm{L}}{\mathrm{v \rho g}}=\frac{0.91 \times 10^{11} \times 2 \times 10^{-6} \times 0.11 \times 10^{-3}}{10^{-4} \times 10^3 \times 9.8}[/latex]
L = 2.04 m.

Question 7.
There are two wires of same material. Their radii and lengths are both in the ratio 1:2. If the extensions produced are equal, what is the ratio of loads ?
Solution:

Question 8.
Two wires of different material have same lengths and areas of cross¬section. What is the ratio of their increase in length when forces applied are the same ?
(Y₁ = 0.9 × 10¹¹ Nm^-2, Y₂ = 3.6 × 10¹¹ Nm^-2)
Solution:
y₁ = 0.9 × 10¹¹ Nm^-2
y₂ = 3.6 × 10¹¹ Nm^-2
y = [latex]\frac{F L}{A \times \Delta L}[/latex]
∆L ∝ [latex]\frac{1}{y}[/latex]
[latex]\frac{(\Delta L)_1}{(\Delta L)_2}=\frac{y_2}{y_1}=\frac{3.6 \times 10^{11}}{0.9 \times 10^{11}}=\frac{4}{1}[/latex]

Question 9.
A metal wire of length 2.5 m and area of cross-section 1.5 × 10^-6 m² is stretched through 2 mm. if its Young’s modulus is 1.25 × 10¹¹ N.m², find the tension in the wire.
Solution:
L = 2.5 m, A = 1.5 × 10^-6 m²
∆L = 2 × 10^-9 m
y = 1.25 × 10¹¹ N.m²
y = [latex]\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}[/latex]
F = [latex]\frac{\mathrm{yA} \Delta \mathrm{L}}{\mathrm{L}}[/latex]
= [latex]\frac{1.25 \times 10^{11} \times 1.5 \times 10^{-6} \times 2 \times 10^{-3}}{2.5}[/latex]
F = 150 N

Question 10.
An aluminium wire and a steel wire of the same length and cross-section are joined end-to-end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 1.35 mm, find the ratio of the
(i) stress in the two wires and
(ii) strain in the two wires.
(Y_Al = 0.7 × 10¹¹ N.m², Y_Steel = 2 × 10¹¹ Nm²).
Solution:
Y_Al = 0.7 × 10¹¹ N.m², Y_Steel = 2 × 10¹¹ Nm²
∆L₁ + ∆L₂ = 1.35 mm ……………… (1)

Question 11.
A 2 cm cube of some substance has its upper face displaced by 0.15 cm due to a tangential force of 0.3 N while keeping the lower face fixed, Calculate the rigidity modulus of the substance.
Solution:
L = 2 × 10^-2 m, A = L² = 4 × 10^-4 m²
∆x = 0.15 × 10^-2 m
F = 0.3 N
G = [latex]\frac{\frac{F}{A}}{\frac{\Delta x}{L}}=\frac{F L}{A \Delta x}[/latex] (∵ θ = [latex]\frac{\Delta x}{L}[/latex])
G = [latex]\frac{0.3 \times 2 \times 10^{-2}}{4 \times 10^{-4} \times 0.15^6 \times 10^{-2}}[/latex]
G = 10⁴ N/m²

Question 12.
A spherical ball of volume 1000 cm³ is subjected to a pressure of 10 Atmosphere. The change in volume is 10^-2 cm³. If the ball is made of iron, find its bulk modulus.
(1 atmosphere = 1 × 10⁵ Nm^-2).
Solution:
v = 1000 cm³ = 1000 × 10^-6 = 10^-3 m³
p = 1 atm = 1 × 10⁵ = 10⁵ N/m²
-∆v = 10^-2 cm³ = 10^-2 × 10^-6 = 10^-8 m³
Bulk modulus (B) = [latex]\frac{-p v}{\Delta v}[/latex]
= [latex]\frac{10^5 \times 10^{-3}}{10^{-8}}[/latex]
B = 10¹⁰ N/m².

Question 13.
A copper cube of side of length 1 cm is subjected to a pressure of 100 atmosphere. Find the change in its volume if the bulk modulus of copper is 1.4 × 10¹¹ Nm^-2. (1 atm = 1 × 10⁵ Nm^-2).
Solution:
l = 1 cm = 10^-2 m
V = Volume of the cube = l³ = 1cm³
= 10^-6 m³
P = 100 atm = 100 × 10⁵ = 10⁷ N/m²
B = 1.4 × 10¹¹ N/m²
B = [latex]\frac{-P V}{\Delta V}[/latex]
-∆V = [latex]\frac{P V}{B}=\frac{10^7 \times 10^{-6}}{1.4 \times 10^{11}}[/latex]
-∆V = 0.7143 × 10^-10 m³.

Question 14.
Determine the pressure required to reduce the given volume of water by 2%. Bulk modulus of water is 2.2 × 10⁹ Nm^-2.
Solution:
[latex]\frac{-\Delta V}{V}[/latex] = 2 % = [latex]\frac{2}{100}[/latex]
B = 2.2 × 10⁹ Nm²
B = [latex]\frac{-P V}{\Delta V}[/latex]
P = -B × [latex]\frac{\Delta V}{V}[/latex]
= 2.2 × 10⁹ × [latex]\frac{2}{100}[/latex]
P = 4.4 × 10⁷ N/m².

Question 15.
A steel wire of length 20 cm is stretched to increase its length by 0.2 cm. Find the lateral strain in the wire if the Poisson’s ratio for steel is 0.19.
Solution:
L = 20 cm = 20 × 10^-2 m
∆L = 0.2 × 10^-2 m
σ = 0.19
σ = [latex]\frac{\text { Lateral strain }}{\text { Longitudinal strain }\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)}[/latex]
Lateral strain = σ × [latex]\frac{\Delta L}{L}[/latex]
= [latex]\frac{0.19 \times 0.2 \times 10^{-2}}{20 \times 10^{-2}}[/latex]
= 0.0019

Additional Problems

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10^-5 m²,
Solution:
Given, for steel wire, a₁ = 3.0 × 10^-5 m², l₁ = 4.7 m, ∆l₁ = ∆l, F₁ = F
For copper wire, a₂ = 4.0 × 10^-5 m², l₂ = 3.5 m, ∆l₂ = ∆l, F₂ = F .
Let y₁, y₂ be the young modulus of steel wire and copper wire respectively.
∴ y₁ = [latex]\frac{F_1}{a_1} \times \frac{l_1}{\Delta l_2}=\frac{F}{3.0 \times 10^{-5}} \times \frac{4.7}{\Delta l}[/latex] ………….. (i)
and y₂ = [latex]\frac{F_2 \times l_2}{a_2 \times \Delta l_2}=\frac{F \times 3.2}{4 \times 10^{-5} \times \Delta l}[/latex]
[latex]\frac{\mathrm{y}_1}{\mathrm{y}_2}=\frac{4.7 \times 4 \times 10^{-5}}{3.5 \times 3.0 \times 10^{-5}}[/latex] = 1.8
Here y₁ : y₁ = 1.8 : 1.

Question 2.
Figure shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material ?

Solution:
a) From graph, for stress = 150 × 10⁶ Nm^-2 the corresponding strain = 0.002
young’s modulus y = [latex]\frac{\text { Stress }}{\text { Strain }}=\frac{150 \times 10^6}{0.002}[/latex]
= 7.5 × 10¹⁰ Nm^-2

b) Approximate yeild strength will be equal to the maximum stress it can substain with out crossing the elastic limit. Therefore, the approximate yeild strength
= 300 × 10⁶ Nm^-2
= 3 × 10⁸ Nm^-2

Question 3.
The stress-strain graphs for materials A and B are shown in Fig.

The graphs are drawn to the same scale.
a) Which of the materials has the greater Young’s modulus ?
b) Which of the two is the stronger material ?
Solution:
a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence young’s modulus (= stress/ strain) is greater for A than that of B.

b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.

Question 4.
Read the following two statements below carefully and state, with reasons, if it is true or false.
a) The Young’s modulus of rubber is greater than that of steel;
b) The stretching of a coil is determined by its shear modulus.
Solution:
a) False, because for a given stress there is more strain in rubber than steel and modulus of elasticity is inversly proportional to strain.

b) True because the strecting of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elasticity is involved.

Question 5.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of steel wire is 1.5 m and that of brass Wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Solution:
For steel wire, total force on steel wire,
F₁ = 4 + 6 = 10 kg, f = 10 × 9.8 N
l₁ = 1.5 m, ∆l₁ = ?, 2r₁ = 0.25cm or r₁ =(0.25/2)cm = 0.125 × 10^-2 m
y₁ = 2.0 × 10¹¹ pa
For brass wire, F₂ = 6.0 kg, f = 6 × 9.8 N
2r₂ = 0.25 cm or r₂ = (0.25/2) cm = 0.125 × 10^-2 m,
y₂ = 0.91 × 10¹¹ pa, l₂ = 1.0 m, ∆l₂ = ?

Question 6.
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a verticle wall. A mass of 100 kdis then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Solution:
A = 0.10 × 0.10 = 10^-2 m², F = mg = 100 × 10 N
Shearing strain = [latex]\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\left(\frac{\mathrm{F}}{\mathrm{A}}\right)}{\eta}[/latex]
or ∆L = [latex]\frac{F L}{A \eta}[/latex]
= [latex]\frac{(100 \times 10) \times(0.10)}{10^{-2} \times\left(25 \times 10^9\right)}[/latex] = 4 × 10^-7 m.

Question 7.
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the corn pressional strain of each column.
Solution:
Load on each column, F = [latex]\frac{50,000}{4}[/latex] kgwt
= [latex]\frac{50,000 \times 9.8}{4}[/latex] N
A = π(r₂² – r₁²) = [latex]\frac{22}{7}[/latex](0.60)² – (0.30)²]
= [latex]\frac{22}{7}[/latex] 0.27 m²
Compression strain = [latex]\frac{\frac{F}{A}}{y}=\frac{F}{A y}[/latex]
= [latex]\frac{50,000 \times 9.8}{4 \times\left(\frac{22}{7} \times 0.27\right) \times 2.0 \times 10^{11}}[/latex]
= 7.21 × 10^-7.

Question 8.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Solution:
Here, A = 15.2 × 19.2 × 10^-6 m², F = 44,
500 N, η = 42 × 10⁹ Nm^-2
Strain = [latex]\frac{\text { Stress }}{\text { Modulus of elasticity }}[/latex]
= [latex]\frac{\frac{F}{A}}{\eta}=\frac{F}{A \eta}=\frac{44500}{\left(15.2 \times 19.2 \times 10^{-6}\right) \times 42 \times 10^9}[/latex]
= 3.65 × 10^-3.

Question 9.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress Is not to exceed 10⁸ N m^-2, what is the maximum load the cable can support ?
Solution:
Maximum load macimum stress × area of cross-section
= 10⁸πr²
= 10⁸ × [latex]\frac{22}{7}[/latex] × (1.5 × 10^-2)²
= 7.07 × 104 N.

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
Solution:
As each wire has same tension F, so each wire has same extansion due to mass of rigid bar. As each wire is of same length, hence each wire has same strain, if D is the diameter of wire, then
y = [latex]\frac{4 \mathrm{~F} / \pi \mathrm{D}^2}{\text { Strain }}[/latex] or D² ∝ 1/y
[latex]\frac{D_{\mathrm{cu}}}{\mathrm{D}_{\mathrm{iron}}}=\sqrt{\frac{\mathrm{y}_{\mathrm{iron}}}{\mathrm{y}_{\mathrm{cu}}}}[/latex]
= [latex]\sqrt{\frac{190 \times 10^9}{110 \times 10^9}}=\sqrt{\frac{19}{11}}[/latex] = 1.31.

Question 11.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
Solution:
Here, m = 14.5 kg, l = r = 1m, v = 2 rps, A = 0.065 × 10^-4 m²
Total pulling force on mass, when it is at the lowest position of the vertical circle is
F = mg + mrω²
= mg + mr4πv²
= 14.5 × 9.8 + 14.5 × 1 × 4 × ([latex]\frac{22}{7}[/latex])² × 2²
= 142.1 +2291.6
= 2433.7 N
y = [latex]\frac{F}{A} \times \frac{l}{\Delta l}[/latex] or ∆l = [latex]\frac{F l}{A y}[/latex]
= [latex]\frac{2433.7 \times 1}{\left(0.065 \times 10^{-4}\right) \times\left(2 \times 10^{11}\right)}[/latex]
= 1.87 × 10^-3 m
= 1.87 mm.

Question 12.
Compute the bulk modulus of water from the following data : Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Solution:
Here, V = 100 lit = 100 × 10^-3 m³, P = 100 atm = 100 × 1.013 × 10⁵ Pa
V + ∆V = 100.5 litre or ∆V= (V + ∆V) – V
= 100.5 – 100
= 0.5 litre = 0.5 × 10^-3 m³
We known that bulk modulus, B = [latex]\frac{\mathrm{PV}}{\Delta \mathrm{V}}[/latex]
= [latex]\frac{100 \times 1.013 \times 10^5 \times 100 \times 10^{-3}}{0.5 \times 10^{-3}}[/latex]
= 2.026 × 10⁹ Pa
Bulk modulus of air = 1.0 × 10⁵ Pa
[latex]\frac{\text { Bulk modulus of water }}{\text { Bulk modulus of air }}=\frac{2.026 \times 10^9}{1.0 \times 10^5}[/latex]
= 2.026 × 10¹⁴.
It is so because gases are much more compressible than those of liquids. The molecules in gases are very poorly coupled to their neighbours as compared to those of gases.

Question 13.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^-3 ?
Solution:
Here, P = 80.0 atm = 80.0 × 1.013 × 10⁵ pa,
compressibility = [latex]\left(\frac{1}{B}\right)[/latex] = 45.8 × 10^-11 pa^-1
Density of water at surface,
ρ = 1.03 × 10³ kg m^-3
Let p be the density of water at the given depth, if v and v’ are volumes of certain mass M of ocean water at surface and at a given

Putting this value in (i) we get
1 – [latex]\frac{1.03 \times 10^3}{\rho^{\prime}}[/latex] = 3.712 × 10^-3 or
ρ’ = [latex]\frac{1.03 \times 10^3}{1-3.712 \times 10^{-3}}[/latex] = 1.034 × 10³ kg m^-3.

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Solution:
Here, P = 10 atm = 10 × 1.013 × 10⁵ pa,
B = 37 × 10⁹ Nm^-2
Volumetric strain = [latex]\frac{\Delta V}{V}=\frac{P}{B}[/latex]
= [latex]\frac{10 \times 1.013 \times 10^5}{37 \times 10^9}[/latex] = 2.74 × 10^-5
∴ Fractional change in volume = [latex]\frac{\Delta V}{V}[/latex]
= 2.74 × 10^-5

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa.
Solution:
Here, L = 10 cm = 0.10m; P = 7 × 10⁶ pa B = 140 Gpa = 140 × 10⁹ pa
As B = [latex]\frac{\mathrm{PV}}{\Delta \mathrm{V}}=\frac{\mathrm{Pl}^3}{\Delta \mathrm{V}}[/latex] or ∆V = [latex]\frac{\mathrm{Pl}^3}{\mathrm{~B}}[/latex]
= [latex]\frac{\left(7 \times 10^6\right) \times(0.10)^3}{140 \times 10^9}[/latex] = 5 × 10^-8 m³
= 5 × 10^-2 mm³

Question 16.
How much should be pressure on a litre of water be changed to compress it by 0.10% ?
Solution:
Here, V = 1 litre = 10^-3m³;
∆V/V = 0,10/100 = 10^-3
B = [latex]\frac{P V}{\Delta V}[/latex] or P = B [latex]\frac{\Delta V}{V}[/latex]
= (2.2 × 10⁹) × 10^-3 = 2.2 × 10⁶pa

Question 17.
Anvils made of single crystals of diamond, with the shape as shown in Fig. are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil ?

Solution:
Here, D = 0.5 mm = 0.5 × 10^-3m = 5 × 10^-4m
F = 50,000 N = 5 × 10⁴N
Pressure at the tip of anvil.
P = [latex]\frac{F}{\pi D^2 / 4}=\frac{4 F}{\pi D^2}[/latex]
P = [latex]\frac{4 \times\left(5 \times 10^4\right)}{(22 / 7) \times\left(5 \times 10^{-4}\right)^2}[/latex] = 2.5 × 10¹¹pa.

Question 18.
A rod of length 1.05 m having negligible mass is supported at its end by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm², respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Solution:
For Steel wire A, l₁ = l, A₁ = 1 mm²
Y₁ = 2 × 10¹¹Nm^-2
For aluminium wire B, l₂ = l;
A₂ = 2mm²; y₂ = 7 × 10¹⁰ Nm^-2
a) Let mass m be suspended from the rod at the distance × from the end where wires A is connected. Let F₁ and F₂ be the tension in two wires and there is equal stress in two wires, then
[latex]\frac{F_1}{A_1}=\frac{F_2}{A_2} \text { or } \frac{F_1}{F_2}=\frac{A_1}{A_2}=\frac{1}{2}[/latex] …………………. (i)
Taking moment of forces about the point of suspension of mass from the rod, we have
F₁x = F₂ (1.05 – x) or [latex]\frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{1}{2}[/latex]
or 2.10 – 2x = x or x = 0.70m = 70 cm

b) Let mass m be supended from the rod at distance × from the end where wire A is connected. Let F₁ and F₂ be the tension in the wires and there is equal strain in the two wires i.e.
[latex]\frac{F_1}{A_1 Y_1}=\frac{F_2}{A_2 Y_2}[/latex] or [latex]\frac{F_1}{F_2}=\frac{A_1}{A_2} \frac{Y_1}{Y_2}[/latex]
= [latex]\frac{1}{2} \times \frac{2 \times 10^{11}}{7 \times 10^{10}}=\frac{10}{7}[/latex]
As the rod is stationary, so F₁x = F₂(1.05 – x)
or [latex]\frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{10}{7}[/latex]
or 10 x = 7.35 – 7x
or x = 0.4324 m
x = 43.2cm.

Question 19.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10^-2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the- mid-point.
Solution:
Refer the figure, let x be the depression at the mid point i.e CD = x
In fig. AC = CB = Z = 0.5m
m = 100g = 0.100 kg
AD = BD = (l² + x²)^1/2
Increase in length, ∆l = AD + DB – AB
= 2 AD – AB
= 2 (l² + x²)^1/2 – 2l
= 2l(1 + [latex]\frac{x^2}{l^2}[/latex])^1/2 – 2l
= 2l (1 + [latex]\frac{x^2}{2 l^2}[/latex]) – 2l = [latex]\frac{x^2}{l}[/latex]
Strain = [latex]\frac{\Delta l}{2 l}=\frac{x^2}{2 l^2}[/latex]
If T is the tension in the wire, then 2T cos θ
= mg or T = [latex]\frac{\mathrm{mg}}{2 \cos \theta}[/latex]

Question 20.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 10⁷ Pa ? Assume that each rivet is to carry one quarter of the load.
Solution:
Here, r = 6/2 = 3mm = 3 × 10^-3 m, max.
stress = 6.9 × 10⁷ Pa
Max . load on a rivet = Max stress × area of cross section
= 6.9 × 10⁷ × (22/7) × (3 × 10^-3)²
∴ Maximum tension
= 4 (6.9 × 10⁷ × [latex]\frac{22}{7}[/latex] × 9 × 10^-6)
= 7.8 × 10³N.

Question 21.
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom ?
Solution:
Here, P = 1.1 × 10⁸ Pa, V = 0.32 m³,
B = 16 × 10¹¹Pa
∆V = [latex]\frac{\mathrm{PV}}{\mathrm{B}}[/latex]
= [latex]\frac{\left(1.1 \times 10^8\right) \times 0.32}{1.6 \times 10^{11}}[/latex]
= 2.2 × 10^-4m³

Textual Examples

Question 1.
A structural steel rod has a radius of 10 mm and a length of 10 m. A 100 KN force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel ¡s 2.0 × 10¹¹ Nm².
Answer:
a) Given Stress = [latex]\frac{F}{A}=\frac{F}{\pi r^2}[/latex]
= [latex]\frac{100 \times 10^3 \mathrm{~N}}{3.14 \times\left(10^{-2} \mathrm{~m}\right)^2}[/latex] = 3.18 × 10⁸ Nm^-2

b) The elongation
∆L = [latex]\frac{(\mathrm{F} / \mathrm{A}) \mathrm{L}}{\mathrm{Y}}[/latex]
= [latex]\frac{\left(3.18 \times 10^8 \mathrm{Nm}^2\right)(1 \mathrm{~m})}{2 \times 10^{11} \mathrm{Nm}^{-2}}[/latex]
= 1.59 × 10^-3 m
= 1.59 mm

c) The strain is given by
Strain = ∆L/L = (1.59 × 10^-3) km
= 1.59 × 10^-3 = 0.16%

Question 2.
A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.
Answer:
From y = [latex]\frac{\sigma}{\varepsilon}[/latex]
we have stress = strain × Young’s modulus.
W/A = Y_c × (∆L_c/L_c) = Y_s × (∆L_s/L_s)
where the subscripts c and s refer to copper and stainless steel respectively,
∆L_c/∆L_s = (Y_s/Y_s) () (L_c/L_s)
Given L_c = 2.2 m, L_s = 1.6 m,
Y_c = 1.1 × 10¹¹ N.m^-2 and Y_s = 2.0 × 10¹¹ N.m^-2.
∆L_c/∆L_s = [latex]\frac{2.0 \times 10^{11}}{1.1 \times 10^{11}}=\frac{2.2}{1.6}[/latex] = 2.5
∆L_c + ∆L_s = 7.0 × 10^-4 m
Solving the above equations,
∆L_c = 5.0 × 10^-4 m and ∆L_s = 2.0 × 10^-4 m.
∴ W = (A × Y_c × ∆L_c)L_c
= π(1.5 × 10^-3)² × [latex]\left[\frac{\left(5.0 \times 10^{-4} \times 1.1 \times 10^{11}\right)}{2.2}\right][/latex] = 1.8 × 10² N.

Question 3.
In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in Fig.) The combined mass of all the persons performing the act and the tables, planks etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load.

Answer:
Total mass of all the performers, tables, plaques = 280 kg
Mass of the performer = 60 kg
Mass supported by the legs of the performer at the bottom of the pyramid = 280 – 60
= 220 kg
Weight of this supported mass = 220 kg wt.
= 220 × 9.8 N = 2156 N
Weight supported by each thighbone of the performer = [latex]\frac{1}{2}[/latex] (2156) N = 1078 N.
The Young’s modulus for bone is Y = 9.4 × 10⁹ Nm^-2 (compressive)
Length of each thighbone L = 0.5 m the radius of thigbone = 2.0 cm
Thus the cross-sectional area of the thighbone
A = π × (2 × 10^-2)² m²
= 1.26 × 10^-3 m²
Using [latex]\frac{(F \times L)}{(A \times \Delta L)}[/latex] the compression in each thigbone (∆L) can be computed as
∆L = [latex]\frac{F \times L}{Y \times A}[/latex]
= [(1078 × 0.5)/(9.4 9 × 10⁹ × 1.26 × 10^-3)]
= 4.55 × 10^-5 m or 4.55 × 10^-3 cm.
This is a very small change ! The fractional decrease in the thighbone is ∆L/L = 0.000091 or 0.0091%.

Question 4.
A square slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × 10⁴ N. The lower edge is riveted to the floor. How much will the upper edge be displaced ?
Answer:
The area (A) = 50 cm × 10 cm
= 0.5 m × 0.1 m
= 0.05 m²
Therefore, the stress appIid is
= (9.4 × 10⁴N/0.05 m²)
= 1.80 × 10⁶ N.m²
We know that shearing strain = (∆x/L)
= Stress/G.
Therefore the displacement
∆x = (Stress × L)/G = [latex]\frac{\left(1.8 \times 10^6 \mathrm{Nm}^{-2} \times 0.5 \mathrm{~m}\right)}{5.6 \times 10^9 \mathrm{Nm}^{-2}}[/latex]
= 1.6 × 10^-4 m = 0.16 mm.

Question 5.
The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression. ∆V/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 Nm^-2. (Take g = 10 ms^-2)
Answer:
The pressure exerted by a 3000 m column of water on the bottom layer
ρ = hρg
= 3000 m × 1000 kg m^-3× 10 ms^-2
= 3 × 10⁷ Nm^-2
Fractional compression ∆V/V, is
∆V/V = stress [latex]\frac{\left(3 \times 10^7 \mathrm{Nm}^{-2}\right)}{2.2 \times 109 \mathrm{Nm}^{-2}}[/latex]
= 1.36 × 10^-2 or 1.36%

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 9th Lesson Gravitation Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 9th Lesson Gravitation

Very Short Answer Questions

Question 1.
State the unit and dimension of the universal gravitational constant (G).
Answer:
F = [latex]\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}^2}[/latex]
units of G = Nm² Kg^-2
dimensional formula of G = [latex]\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{[\mathrm{M}][\mathrm{M}]}[/latex] = [M^-1 L³ T^-2]

Question 2.
State the vector form of Newtons’s law of gravitation.
Answer:
Vector form of Newton’s law of gravitation is
F = [latex]\frac{-G m_1 m_2}{r^3} \hat{r}[/latex] where [latex]\hat{r}[/latex] is unit vector.

Question 3.
If the gravitational force of Earth on the Moon is F, what is the gravitational force of moon on earth ? Do these forces to attraction-reaction pair ?
Answer:
F. Yes, they form action and reaction pair.

Question 4.
What would be the change in acceleration due to gravity (g) at the surface, if the radius of Earth decreases by 2% keeping the mass of Earth constant ?
Answer:
g₁r₁², l = g₂r₂², r₂ = [latex]\frac{98}{100}[/latex] r₁
[latex]\frac{g_2}{g_1}=\frac{r_1^2}{r_2^2}=\frac{r_1^2}{\left(\frac{98}{100}\right) r_1^2}=\frac{100 \times 100}{98 \times 98}[/latex]
[latex]\frac{g_2}{g_1}[/latex] = 1.04
[latex]\frac{g_2}{g_1}[/latex] – 1 = 1.04 – 1
[latex]\frac{g_2-g_1}{g_1}[/latex] = 0.04

Question 5.
As we go from one planet to another, how will
a) the mass and
b) the weight of a body change ?
Answer:
a) The mass remains constant.
b) The weight (w = mg), changes from one planet to another planet.

Question 6.
Keeping the length of a simple pendulum constant, will the time period be the same on all planets ? Support your answer with reason.
Answer:
No, Time period depends on acceleration due to gravity (g).
T = 2π [latex]\sqrt{\frac{l}{g}}[/latex]
g value varies from planet to planet. So time period changes.

Question 7.
Give the equation for the value of g at a depth ‘d’ from the surface of Earth. What is the value of ‘g’ at the centre of Earth ?
Answer:

1. g_d = g(1 – [latex]\frac{\mathrm{d}}{\mathrm{R}}[/latex]) where d = Depth
   R = Radius of the Earth
2. At the centre of the Earth g = 0.

Question 8.
What are the factors that make ‘g’ the least at the equator and maximum at the poles ? > .
Answer:

1. g value is maximum at poles due to
   a) Rotation of the Earth
   b) Earth is flattened at the poles
   c) The equatorial radius is less at the poles.
2. g value minimum at equator due to
   a) Rotation of the earth
   b) Bulging near the equator.

Question 9.
“Hydrogen is in abundance around the sun but not around Earth”. Explain.
Answer:
The escape velocity on the sun is 620 km/s and escape velocity on the Earth is 11.2 km/s. The r.m.s velocities of hydrogen (2 km/s) is less than escape velocity on the Sun. So hydrogen is more abundant around the Sun and less around the Earth.

Question 10.
What is the time period of revolution of a geostationary satellite ? Does it rorate from West to East or from East to West ?
Answer:
Time period of geo-stationary satellite is 24 hours. It can rotate from west to east.

Question 11.
What are polar satellites ?
Answer:
Polar satellites are low altitude satellites (500 to 800 km), but they go around the poles of the earth in a north-south direction. Its time period is around 100 minutes.

Short Answer Questions

Question 1.
State Kepler’s laws of planetary motion.
Answer:
The three laws of Kepler can be stated as follows.

1. Law of orbits : All planets move in elliptical orbits with the sun situated at one of the foci.
   
2. Law of areas : The line that joins any planet to the sun sweeps equal areas in equal intervals of time.
3. Law of periods : The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.
   T² ∝ R³

Question 2.
Derive the relation between acceleration due to gravity (g) at the surface of a planet and Gravitational constant (G).
Answer:
Consider a body of mass m on the surface of the planet. Let R be the radius of the Earth and M be the mass of the Earth.
Force acting on the body due to gravitational pull of the planet is
F = m g → (1)
According to Newton’s gravitational law, Force on the body is F = [latex]\frac{\mathrm{GMm}}{\mathrm{R}^2}[/latex] → (2)

From eq’s (1) and (2), we have
m g = [latex]\frac{\mathrm{GMm}}{\mathrm{R}^2}[/latex]
g = [latex]\frac{\mathrm{GM}}{\mathrm{R}^2}[/latex]
This is the relation between g and G
Mass of the earth (M) = Volume × density of the earth
M = [latex]\frac{4}{3}[/latex] π R² × ρ
g = [latex]\frac{4}{3}[/latex] π G R ρ

Question 3.
How does the acceleration due to gravity (g) change for the same values of height(h) and depth (d).
Answer:
a) g_h = g(1 – [latex]\frac{2 \mathrm{~h}}{\mathrm{R}}[/latex]), g_d = g(1 – [latex]\frac{\mathrm{d}}{\mathrm{R}}[/latex])
Same values of height and depth, h = d
g_h = g (1 – [latex]\frac{2 \mathrm{~d}}{\mathrm{R}}[/latex]) and g_d = g(1 – [latex]\frac{\mathrm{d}}{\mathrm{R}}[/latex])
∴ g_d > g_h

b) For large height and large depth
g_h = [latex]\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}[/latex] and g_d = g(1 – [latex]\frac{\mathrm{d}}{\mathrm{R}}[/latex])
If h = d = R
g_h = [latex]\frac{\mathrm{g}}{\left(1+\frac{\mathrm{R}}{\mathrm{R}}\right)^2}[/latex] and g_d = g(1 – [latex]\frac{\mathrm{R}}{\mathrm{R}}[/latex]) = 0
∴ g_h > g_d

Question 4.
What is orbital velocity ? Obtain an expression for it. [Mar. 14]
Answer:
Orbital velocity (V₀) : The horizontal velocity required for an object to revolve around a planet in a circular orbit is called orbital velocity.

Expression for orbital velocity :
Consider a body (satellite) of mass m, revolves round the earth in a circular orbit. Let h be the height of the satellite from the surface of the earth. Then (R + h) is the radius of the orbit.
The Gravitational force of attraction of the earth on the body is given by F = [latex]\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^2}[/latex] ………….. (1)
Where M = Mass of the earth, R = Radius of the earth, G = universal gravitational constant. If V₀ is the orbital velocity of the body.
The centripetal force on the body is given by F = [latex]\frac{\mathrm{mv}_{\mathrm{o}}^2}{(\mathrm{R}+\mathrm{h})}[/latex] …………… (2)
In order to make the body revolve in the same orbit, its centripetal force must be equal to the gravitational force

Question 5.
What is escape velocity ? Obtain an expression for it.
Answer:
Escape velocity : It is the minimum velocity with which a body should be projected, so that it moves into the space by overcoming the earth’s gravitational field.

Expression for escape velocity :
Consider a body of mass m thrown with a velocity v₂
Then K.E = [latex]\frac{1}{2}[/latex] m v_e² …………. (1)
The gravitational force of attraction of the earth of mass M and Radius R on a body of mass m at its surface is F = [latex]\frac{\mathrm{GMm}}{\mathrm{R}^2}[/latex] ……………… (2)
Gravitational P. E. = work done on the body
∴ P. E. = F × R = [latex]\frac{\mathrm{GMm}}{\mathrm{R}^2}[/latex] × R
P.E. = [latex]\frac{\mathrm{GMm}}{\mathrm{R}}[/latex] …………….. (3)
A body just escapes when its K. E. = P. E
[latex]\frac{1}{2}[/latex] m v_e² = [latex]\frac{\mathrm{GMm}}{\mathrm{R}^2}[/latex]
v_e² = [latex]\frac{2 \mathrm{GM}}{\mathrm{R}}[/latex] (∵ g = [latex]\frac{\mathrm{GM}}{\mathrm{R}^2}[/latex])
v_e = [latex]\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}[/latex]
v_e = [latex]\sqrt{2 g R}[/latex] (gR = [latex]\frac{\mathrm{GM}}{\mathrm{R}}[/latex])
v_e = [latex]\sqrt{2} \times \sqrt{g R}[/latex] (∵ v₀ = [latex]\sqrt{g R}[/latex])
v_e = [latex]\sqrt{2}[/latex] × v₀
∴ Escape velocity is [latex]\sqrt{2}[/latex] times the orbital velocity.

Question 6.
What is a geostationary satellite ? State its uses. [T.S. Mar. 18, 15; A.P. Mar. 16]
Answer:
Geo-stationary satellite : If the period of revolution of an artificial satellite is equal to the period of rotation of earth, then such a satellite is called geo-stationary satellite.
Time period of geo-stationary satellite is 24 hours.
Uses :

1. Study the upper layers of atmosphere
2. Forecast the changes in atmosphere
3. Know the shape and size of the earth.
4. Identify the minerals and natural resources present inside and on the surface of the earth.
5. Transmit the T. V. programmes to distant objects
6. Under take space research i.e. to know about the planets, satellites, comets etc.

Question 7.
If two places are at the same height from the mean sea level; One is a mountain and other is in air at which place will ‘g’ be greater ? State the reason for your answer.
Answer:
The acceleration due to gravity on mountain is greater than that of air.
g = [latex]\frac{\mathrm{GM}}{\mathrm{R}^2}[/latex] ………….. (1)
Mass (M) = volume × density (ρ)
M = [latex]\frac{4}{3}[/latex]π R³ × ρ
g = [latex]\frac{\mathrm{G}}{\mathrm{R}^2}[/latex] × [latex]\frac{4}{3}[/latex] π R³ ρ
g = – [latex]\frac{4}{3}[/latex] π R G ρ …………….. (2)
g ∝ ρ
So density is more at mountains. So g is more on mountain.

Question 8.
The weight of an object is more at the poles than at the equator. At which of these can we get more sugar for the same weight ? State the reason for your answer.
Answer:
Weight of the object at poles = m_p g_p (∵ w = mg)
Weight of the object at equator = m_e g_e
Given weight of the object at poles > weight of the object at equator
m_p g_p > m_eg_e
We know that g_p > g_e
Then m_p < m_e
Hence we can get more sugar at equator.

Question 9.
If a nut becomes loose and gets detached form a satellite revolving around the earth, will it fall down to earth or will it revolve around earth ? Give reasons for your answer.
Answer:
When a nut is detached from a satellite revolving around the earth, the nut is also moving with the speed of the satellite as the orbit of a satellite does not depend upon its mass. Hence nut is moving in the same orbit under centripetal force.

Question 10.
An object projected with a velocity greater than or equal to 11.2 kms it will not return to earth. Explain the reason.
Answer:
The escape velocity on the surface of the earth (v_e) = 11.2 km/s. Any object projected with the velocity greater then (or) equal to 11.2 km/s it will not come back. Because it has overcome the earth’s gravitational pull.
So an object never come back to earth.

Long Answer Questions

Question 1.
Define gravitational potential energy and derive an expression for it associated with two particles of masses m₁ and m₂.
Answer:
Gravitational potential energy : Gravitational potential energy of a body at a point in a gravitational field of another body is defined as the amount of work done in brining the given body from infinity to that point without acceleration.

Expression for gravitational potential energy : Consider a gravitational field due to earth of mass M, radius R. The mass of the earth can be supposed to be concentrated at its centre 0. Let us calculate the gravitational the potential energy of the body of mass m placed at point p in the gravitational field, where OP = r and r > R. Let OA = x and AB = dx.
The gravitational force on the body at A will be
F = [latex]\frac{\mathrm{GMm}}{\mathrm{X}^2}[/latex] ……………… (1)
Small amount of work done in bringing the body without acceleration through a small distance dx is given by
dw = Force × displacement
dw = F × dx
dw = [latex]\frac{\mathrm{GMm}}{\mathrm{X}^2}[/latex] × dx ……………… (2)
Total work done in bringing the body from infinity to point P is given by

This work done is stored in the body as its gravitational potential energy (u)
∴ Gravitational potential energy (u) = [latex]\frac{\mathrm{GMm}}{\mathrm{r}}[/latex] ……………….. (4)
Gravitational potential energy associated with two particles of masses m, and m2 separated by a distance r is given by
u = -[latex]\frac{G m_1 m_2}{r}[/latex] ……………….. (5) (if we choose u = 0 as r → ∞).

Question 2.
Derive an expression for the variation of acceleration due to gravity (a) above and (b) below the surface of the Earth.
Answer:
i) Variation of g with height:
When an object is on the surface of the earth, it will be at a distance r = R radius of the earth, then we have g = [latex]\frac{\mathrm{GM}}{\mathrm{R}^2}[/latex]
Where G = universal gravitational constant, M = Mass of the earth
When the object is at a height h above the surface of the earth, Then r = R + h

g value decreases with altitude.

ii) Variation of g with depth :
Let us assume that the earth to be a homogeneous uniform sphere of radius R, mass M and of uniform density ρ.
We know that g = [latex]\frac{\mathrm{GM}}{\mathrm{R}^2}[/latex] = [latex]\frac{4}{3}[/latex] π ρ G R ………………… (1)
Consider a body of mass m be placed at a depth d.

The value of g decreases with depth.

Question 3.
State Newton’s Universal Law of gravitation. Explain how the value of the Gravitational constant (G) can be determined by Cavendish method.
Answer:
Newton’s law of gravitation :
“Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversly proportional to the square of the distance between them”
Determination of G value by cavendish method :

1. In 1798 Henry Cavendish determined the value of G experimentally.
2. The bar AB has two small lead spheres attached at its ends.
3. The bar is suspended from a rigid support by a fine wire.
4. Two large lead spheres are brought close to the small ones but on opposite sides as shown in figure.
5. The big spheres attract the nearby small ones by equal and opposite force as shown in figurer.
6. There is no net force on the bar but only a torque which is clearly equal to F times the length of the bar. When F is the force of attraction between a big sphere and its neighbouring small sphere.
7. Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque.
   Restoring torque = τ θ ………………… (1)
   Where τ is restoring couple per unit twist 0 is the angle
8. If d is the seperation between big and small balls having masses M and m.
   Gravitational force (F) = [latex]\frac{\mathrm{GMm}}{\mathrm{d}^2}[/latex] ……………… (2)
   ix) If L is the length of the bar A B, then the torque arising out of F is F multiplied by L. At equilibrium, this is equal to the restoring torque.
   [latex]\frac{\mathrm{GMm}}{\mathrm{d}^2}[/latex] = τ θ
   observations of θ thus enables one to calculate G.
   The measurement of G = 6.67 × 10^-11 Nm²/ Kg²

Problems

(Gravitational Constant ‘G’ = 6.67 × 10^-11 Nm²/ Kg^-2; Radius of earth ‘R’ = 6400 km; Mass of earth ‘ME’ = 6 × 10²⁴ kg)

Question 1.
Two spherical balls each of mass 1 kg are placed 1 cm apart. Find the gravitational force of attraction between them.
Solution:
m₁ = m₂ = 1 kg, d = 1 cm = 1 × 10^-2 m
F = [latex]\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}^2}[/latex]
F = [latex]\frac{6.67 \times 10^{-11} \times 1 \times 1}{\left(1 \times 10^{-2}\right)^2}[/latex] = 6.67 × 10^-7N

Question 2.
The mass of a ball is four times the mass of another ball. When these balls are separated by a distance of 10 cm, the force of gravitation between them is 6.67 × 10^-7 N. Find the masses of the two balls.
Solution:
m₁ = m, m₂ = 4m, d = 10 cm = 10 × 10^-2 m,
F = 6.67 × 10^-7 N
G = 6.67 × 10^-11 Nm²/kg ²
F = [latex]\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}^2}[/latex]
6.67 × 10^-7 = [latex]\frac{6.67 \times 10^{-11} \times \mathrm{m} \times 4 \mathrm{~m}}{\left(10 \times 10^{-2}\right)^2}[/latex]
4 m² = 10²
m² = [latex]\frac{100}{4}[/latex] = 25
m = 5 kg
∴ m₁ = m = 5 kg
m₂ = 4m = 4 × 5 = 20 kg

Question 3.
Three spherical balls of masses 1 kg, 2kg and 3 kg are placed at the corners of an equilateral triangle of side 1 m. Find the magnitude of gravitational force exerted by the 2 kg and 3kg masses on the 1 kg mass.
Solution:
The force of attraction at 2 kg on the 1 kg particle
F₂ = [latex]\frac{\mathrm{Gmn}{\mathrm{~d}^2}[/latex] = [latex]\frac{\mathrm{G} \times 1 \times 2}{1^2}[/latex]
F₂ = 2 G

Question 4.
At a certain height above the earth’s surface, the acceleration due to gravity is 4% of its value at the surface of earth. Determine the height.
Solution:
g_h = 4% of g = [latex]\frac{4}{100}[/latex]g, R = 6400 km
g_h = [latex]\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}[/latex]
[latex]\frac{4 \mathrm{~g}}{100}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}[/latex]
[latex]\left(1+\frac{h}{R}\right)^2=\frac{100}{4}[/latex] = 25
1 + [latex]\frac{h}{R}[/latex] = 5
[latex]\frac{h}{R}[/latex] = 4
h = 4 × R = 4 × 6400 = 25,600 km.

Question 5.
A satellite is orbiting the earth at a height of 1000km. Find its orbital speed.
Solution:
h = 1000 km
Oribital velocity (v₀) = [latex]\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}[/latex]
G = 6.67 × 10^-11 Nm²/kg ², M = 6 × 10²⁴ kg
R + h = 6400 + 1000 = 7400 km
= 7400 × 10³m
v₀ = [latex]\sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{7400 \times 10^3}}[/latex]
v₀ = [latex]\sqrt{0.5408 \times 10^{10}}[/latex] = 73.54 × 10³ m/s
v₀ = 7.354km/s

Question 6.
A satellite orbits the earth at a height equal to the radius of earth. Find it’s
(i) orbital speed and
(ii) Period of revolution
Solution:
Height h = R

Question 7.
The gravitational force of attraction between two objects decreases by 36% when the distance between them is increased by 4 m. Find the original distance between them.
Solution:
F₁ = F, F₂ = [latex]\frac{64}{100}[/latex] F
d₁ = d, d₂ = (d + 4) m

5d = 4d + 16
d = 16 m.

Question 8.
Four identical masses of m are kept at the corners of a square of side a. Find the gravitational force exerted on one of the masses by the other masses.
Solution:

Question 9.
Two spherical balls of 1 kg and 4kg are separated by a distance of 12 cm. Find the distance of a point from the 1 kg mass at which the gravitational force on any mass becomes zero.
Solution:
m₁ = 1 kg, m₂ = 4kg, r = 12 cm
∴ x = [latex]\frac{r}{\sqrt{\frac{m_2}{m_1}}+1}[/latex] from m₁
= [latex]\frac{12}{\sqrt{\frac{4}{1}}+1}=\frac{12}{2+1}=\frac{12}{3}[/latex] = 4 cm
At x = 4 cm the gravitational force is zero.

Question 10.
Three uniform spheres each of mass m and radius R are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any one of the spheres due to the other two.
Solution:

Question 11.
Two satellites are revolving round the earth at different heights. The ratio of their orbital speeds ¡s 2 : 1. If one of them is at a height of 100 km, what is the height of the other satellite ?
Solution:

4R + 400 = R + h₂
h₂ = 3R + 400 = 3 × 6400 + 400
= 19200 + 400
h₂ = 19,600 km.

Question 12.
A satellite is revolving round in a circular orbit with a speed of 8 km s-1 at a height where the value of acceleration due to gravity is 8 m s-2. How high is the satellite from the Earth’s surface ? (Radius of planet = 6000 km)
Solution:
v₀ = 8 km/s = 8000 m/s
g_h = 8 m/s², R = 6000 km = 6000 × 10³ m
∴ v₀ = [latex]\sqrt{\frac{G M}{R+h}}=\sqrt{g(R+h)}[/latex]
v₀² = g(R + h)
(8000)² = 8(6000 × 10³ + h)
6000 × 10³ + h = 8 × 10⁶
h = (8 – 6) 10⁶
h = 2 × 10⁶m
h = 2000 × 10³ = 2000 km.

Question 13.
(a) Calculate the escape velocity of a body from the Earth’s surface, (b) If. the Earth were made of wood, its mass would be 10% of its current mass. What would be the escape velocity, if the Earth were made of wood ?
Solution:
R = 6400 × 10³m,

Additional Problems

Question 1.
Answer the following :
a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?
b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ?
c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull, (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ?
Solution:
a) We cannot shield a body from the gravitational influence of nearby matter because the gravitational force on the body due to near by matter is independent of the presence of other matter, whereas it is not so in the case of electrical forces it means the gravitational screens are not possible.

b) Yes, if the size of the spaceship orbiting around the earth is large enough, an astronaut inside the spaceship can detect the variation in g.

c) Tidal effect depends inversly on the cube of the distance, unlike force which depends inversly on the square of the distance. Since the distance of moon from the ocean water is very small as compared to the distance of sun from the ocean water on earth. Therefore, the tidal effect of moon’s pull is greater than the tidal effect of the sun.

Question 2.
Choose the correct alternative :
a) Acceleration due to gravity increase^ decreases with increasing altitude.
b) Acceleration due to gravity increases/decreases with increas¬ing depth (assume the earth to be a sphere of uniform density).
c) Acceleration due to gravity is independent of mass of the earth/ mass of the body.
d) The formula – G Mm (1/r₂ – 1/r₁) is more/less accurate than the formula mg (r₂ – r₁) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth.
Solution:
a) decreases
b) decreases
c) mass of the body
d) more

Question 3.
Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ?
Solution:
Here, T_e = 1 year; T_p = [latex]\frac{T_c}{2}=\frac{1}{2}[/latex] year; r_e = 1
A.U.; r_p = ?
Using Kepler’s third law, we have
r_p = r_e[latex]\left(\frac{T_p}{T_e}\right)^{2 / 3}[/latex] = [latex]1\left(\frac{1 / 2}{1}\right)^{2 / 3}[/latex]
= 0.63 AU

Question 4.
Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10⁸m. Show that the mass of Jupiter is about-one-thousandth that of the sun.
Solution:
For a satellite of Jupiter, orbital period,
T₁ = 1.769 days = 1.769 × 24 × 60 × 60 s
Radius of the orbit of satellite,
r₁ = 4.22 × 10⁸ m
mass of Jupiter, M₁ is given by M₁
= [latex]\frac{4 \pi^2 \times\left(4.22 \times 10^8\right)^3}{G \times(1.769 \times 24 \times 60 \times 60)^2}[/latex]
= [latex]\frac{4 \pi^2 r_1^3}{\mathrm{GT}_1^2}[/latex] ……………. (1)
We know that the orbital period of earth around the sun,
T = 1 year = 365.25 × 24 × 60 × 60 s
Oribital radius, r = 1 A.U = 1.496 × 10¹¹ m

Question 5.
Let us assume that our galaxy consists of 2.5 × 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be 105 ly.
Solution:
Here, r = 50,000 ly =50,000 × 9.46 × 10¹⁵m
= 4.73 × 10²⁰m.
M = 2.5 × 10¹¹ solar, mass = 2.5 × 10¹¹ × (2 × 10³⁰) kg
= 5.0 × 10⁴¹ kg
We know that, M = [latex]\frac{4 \pi^2 r^3}{\mathrm{GT}^2}[/latex]
or T = [latex]\left(\frac{4 \pi^2 r^3}{G M}\right)^{1 / 2}[/latex]
= [latex]\left[\frac{4 \times(22 / 7)^2 \times\left(4.73 \times 10^{20}\right)^3}{\left(6.67 \times 10^{11}\right) \times\left(5.0 \times 10^{41}\right)}\right]^{1 / 2}[/latex]
= 1.12 × 10¹⁶S.

Question 6.
Choose the correct alternative :
a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potentia! energy.
b) The energy required to launch an orbiting .satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.
Solution:
a) Kinetic energy
b) Less.

Question 7.
Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the Ideation from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched ?
Solution:
The escape velocity is independent of mass of the body and the direction of projection it depends upon the gravitational potential at the point from where the body is launched. Since this potential depends slightly on the latitude and height of the point, therefore, the escape velocity depends slightly on these factors.

Question 8.
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit ? Neglect any mass loss of the comet when it comes very close to the Sun.
Solution:
A comet while going on elliptical orbit around the sun has constant angular momentum and total energy at all locations but other quantities vary with locations.

Question 9.
Which of the following symptoms is likely to afflict an astronaut in space
(a) swollen feet,
(b) swollen face,
(c) headache,
(d) orientational problem.
Solution:
a) We know that the legs carry the weight of the body in the normal position due to gravity pull. The astronaut in space is in weightless state. Hence, swollen feet may not affect his working.

b) In the conditions of weightless, the face of the astronaut is expected to get more supply. Due to it, the astronaut may develop swollen face. As eyes, ears, nose, mouth etc. are all embedded in the face, hence, swollen face may affect to great extent the seeing / hearing / eating / smelling capabilities of the astronaut in space.

c) Headache is due to metal strain it will persist whether a person is an astronaut in space or he is on earth it means headache will have the same effect on the astronaut in space as on a person on earth.

d) Space also has orientation. We also have the frames of reference in space. Hence, orientational problem will affect the astronaut in space.

Question 10.
In the following two exercises, choose the correct answer from among the given ones : The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig) (i) a, (ii) b, (iii) c, (iv) 0.

Solution:
We know that the gravitational potential is constant at all points upside a spherical shell. Therefore, the gravitational potential gradient at all points inside the spherical shell is zero [i.e as v is constant, [latex]\frac{\mathrm{dv}}{\mathrm{dr}}[/latex] = 0].

Since gravitational intensity is equal to negative of the gravitational potential gradient, hence the gravitational intensity is zero at all points inside a hollow spherical shell. This indicates that the gravitational forces acting on a particle at any point inside a spherical shell, will be symmetrically placed. Therefore if we remove the upper hemispherical shell, the net gravitational forces acting on the particle at the centre Q or at some other point P will be acting downwards which will also be the direction of gravitational intensity it is so because, the gravitational intensity at a point is the gravitational force per unit mass at that point. Hence the gravitational intensity at the centre Q will be along c, i.e., option (iii) is correct.

Question 11.
For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g. .
Solution:
As per explanation given in the answer of Q. 10, the direction of gravitational intensity at P will be along e i.e., option (ii) is correct.

Question 12.
A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun = 2 × 10³⁰ kg, mass of the earth 6 × 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius 1.5 × 10¹¹ m).
Solution:
Here M_s = 2 × 10³⁰ kg ; M_c = 6 × 10²⁴ kg ; r = 1.5 × 10¹¹ m .
Let x be the distance of a point from the earth where gravitational forces on the rocket due to sun and earth become equal and opposite. Then distance of rocket from the sun
= (r – x). If m is the mass of rocket then
[latex]\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{m}}{(\mathrm{r}-\mathrm{x})^2}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{x}^2} \text { or } \frac{(\mathrm{r}-\mathrm{x})^2}{\mathrm{x}^2}=\frac{\mathrm{M}_{\mathrm{s}}}{\mathrm{M}_{\mathrm{e}}}[/latex]

Question 13.
How will you ‘weigh the sun1, that is estimate its mass ? The mean orbital radius of the earth around the sun is 1.5 × 10⁸ km.
Solution:
To estimate the mass of the sun, we require, the time period of revolution T of one of its planets (say the earth). Let M_s, M_e be the masses of sun and earth respectively and r be the mean orbital radius of the earth around the sun. The gravitational force acting on earth due to sum is
F = [latex]\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{M}_{\mathrm{e}}}{\mathrm{r}^2}[/latex]
Let, the earth be moving in circular orbit around the sun, with a uniform angular velocity ω, the centripetal force acting on earth is.
F¹ = M_erω² = M_er [latex]\frac{4 \pi^2}{T^2}[/latex]
As this centripetal force is provided by the gravitational pull of sun on earth, So
[latex]\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{M}_{\mathrm{e}}}{\mathrm{r}^2}=\mathrm{M}_{\mathrm{e}} \mathrm{r} \frac{4 \pi^2}{\mathrm{~T}^2} \text { or } \mathrm{M}_{\mathrm{s}}=\frac{4 \pi^2 \mathrm{r}^3}{G \mathrm{~T}^2}[/latex]
Knowing r and T, mass M_s of the sun can be estimated.
In this Question, we are given, r = 1.5 × 10⁸ km
= 1.5 × 10¹¹ m
T = 365 days = 365 × 24 × 60 × 60 s
∴ M_s = [latex]\frac{4 \times(22 / 7)^2 \times\left(1.5 \times 10^{11}\right)^3}{\left(6.67 \times 10^{-11}\right) \times(365 \times 24 \times 60 \times 60)^2}[/latex]
= 2 × 10³⁰ kg.

Question 14.
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 10⁸ km away from the sun ?
Solution:
Here, T_s = 29.5 T_e; R_e = 1.5 × 10⁸ km; R_s =?
Using the relation, [latex]\frac{\mathrm{T}_{\mathrm{s}}^2}{\mathrm{R}_{\mathrm{s}}^3}=\frac{\mathrm{T}_{\mathrm{e}}^2}{\mathrm{R}_{\mathrm{e}}^3}[/latex]
or R = Re [latex]\left(\frac{\mathrm{T}_{\mathrm{s}}}{\mathrm{T}_{\mathrm{e}}}\right)^{2 / 3}[/latex]
= 1.5 × 10⁸ [latex]\left(\frac{29.5 \mathrm{~T}_{\mathrm{e}}}{\mathrm{T}_{\mathrm{e}}}\right)^{2 / 3}[/latex]
= 1.43 × 10⁹ km.

Question 15.
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Solution:
Weight of the body = mg = 63N
At height h, the value of g is given by
g’ = [latex]\frac{g R^2}{(R+h)^2}=\frac{g R^2}{(R+R / 2)^2}[/latex] = 4/9 g
Gravitational force on body at height h is
F = mg’ = m × [latex]\frac{4}{9}[/latex] g = [latex]\frac{4}{9}[/latex] mg
= [latex]\frac{4}{9}[/latex] × 63 = 28N

Question 16.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ?
Solution:
wt. of body at a depth d = mg¹
= m × g [latex]\left(1-\frac{d}{R}\right)[/latex]
= 250 [latex]\left(1-\frac{R / 2}{R}\right)[/latex]
= 125 N

Question 17.
A rocket is fired vertically with a speed of 5 km s^-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^-11 N m² kg^-2.
Solution:
Let the rocket be fired with velocity v from the surface of earth and it reaches a height h from the surface of earth where its velocity becomes zero.
Total energy of rocket at the surface of energy

or h = [latex]\frac{\mathrm{Rv}^2}{2 \mathrm{gR}-\mathrm{v}^2}[/latex]
= [latex]\frac{\left(6.4 \times 10^6\right) \times\left(5 \times 10^3\right)^2}{2 \times 9.8 \times\left(6.4 \times 10^6\right)-\left(5 \times 10^3\right)^2}[/latex]
= 1.6 × 10⁶m

Question 18.
The escape speed of a projectile on the earth’s surface is 11.2 km s-1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth ? Ignore the presence of the sun and other planets.
Solution:
Here, v_e = 11.2 kms^-1, velocity of projection of the body v = 3ve. Let m be the mass of the projectile and v₀ be the velocity of the projectile when far away from the earth (i.e) out of gravitational field of earth) then from the law of conservation of energy
[latex]\frac{1}{2}[/latex] mv₀² = [latex]\frac{1}{2}[/latex] mv² – [latex]\frac{1}{2}[/latex] mv_e²
or v₀ = [latex]\sqrt{v^2-v_e^2}[/latex]
= [latex]\sqrt{(3 v e)^2-v_e^2}[/latex]
= [latex]\sqrt{8} v_e=\sqrt{8}[/latex] × 11.2 = 31.68 kms^-1

Question 19.
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence ? Mass of the satellite = 200 kg; mass of the earth = 6.0 × 1024 kg; radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^-11 N m² kg^-2.
Solution:
Total energy of orbiting satellite at a hight h.
= – [latex]\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}+\frac{1}{2} \mathrm{mv}^2[/latex]
= – [latex]\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}+\frac{1}{2} m \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}[/latex]
= [latex]\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}[/latex]
energy expended to rocket the satellite out of the earth’s gravitational field.
= – (total energy of orbiting satellite)
= [latex]\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}[/latex]
= [latex]\frac{\left(6.67 \times 10^{-11}\right) \times\left(6 \times 10^{24}\right) \times 200}{2\left(6.4 \times 10^6+4 \times 10^5\right)}[/latex]
= 5.9 × 10⁹J

Question 20.
Two stars each of one solar mass (= 2 × 10^30< kg) are approaching each other for a head on collision. When they are a distance i09 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10^4< km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Solution:
Here, mass of each star, M = 2 × 10^30< kg
initial distance between two stars, r = 10^9<
km = 10^12< m.
initial potential energy of the system = – [latex]\frac{\text { GMM }}{r}[/latex]
Total K.E. of the stars = [latex]\frac{1}{2}[/latex] mv^2< + [latex]\frac{1}{2}[/latex] mv^2<
= Mv^2<
Where v is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres, r^1< = 2R.
∴ Final potential energy of two starts = [latex]\frac{-\mathrm{GMM}}{2 \mathrm{R}}[/latex]
since gain in K.E. is at the cost of loss in P.E

Question 21.
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium ? If so, is the equilibrium stable or unstable ?
Solution:
Gravitational field at the mid – point of the line joining the centres of the two spheres.
= [latex]\frac{\mathrm{GM}}{(r / 2)^2}(-\hat{r})+\frac{\mathrm{GM}}{(r / 2)^2} \hat{r}=0[/latex]
Gravitational potential at the mid point of the list joining the centres of the two spheres is
v = [latex]\frac{-\mathrm{GM}}{r / 2}+\left(\frac{-\mathrm{GM}}{r / 2}\right)=\frac{-4 \mathrm{GM}}{r}[/latex]
[latex]\frac{-4 \times 6.67 \times 10^{-11} \times 100}{1.0}[/latex] = -2.7 × 10^-8< J/kg
As the effective force on the body placed at mid-point is zero, so the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its initial position of equilibrium. Hence, the body is in unstable equilibrium.

Question 22.
As you have learnt in the text, a geo-stationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 × 10²⁴ kg, radius = 6400 km.
Solution:
Gravitational potential at height h from the surface of earth is
v = [latex]\frac{-\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}[/latex]
= [latex]\frac{-6.67 \times 10^{-11} \times\left(6 \times 10^{24}\right)}{\left(6.4 \times 10^6+36 \times 10^6\right)}[/latex]
= -9.4 × 10⁶ J/kg.

Question 23.
A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity ? (mass of the sun = 2 × 10^30< kg).
Solution:
The object will remain struck to the surface of star due to gravity, if the accerlation due to gravity is more than the centrifugal accerlation due to its rotation.
Accerlation due to gravity, g = [latex]\frac{\mathrm{GM}}{\mathrm{R}^2}[/latex]
= [latex]\frac{6.67 \times 10^{-11} \times 2.5 \times 2 \times 10^{30}}{(12000)^2}[/latex]
= 2.3 × 10¹² m/s²
centrifugal accerlation = rw²
= r(2πv)²
= 12000 (2π × 1.5)²
= 1.1 × 10⁶ ms^-2
since g > rω² , therefore the body will remain struck with the surface of star.

Question 24.
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun = 2 × 10³⁰ kg; mass of mars = 6.4 × 10^23< kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 10^8< km; G = 6.67 × 10^-11< N m² kg ² .
Solution:
Let R, be the radius of the orbit of mars and R be the radius of the mars. M be the mass of the sun and M’ be the mass of mars. If m is the mass of the space ship, then potential energy of space-ship due to gravitational attraction of the sun = [latex]\frac{-\mathrm{GMm}}{\mathrm{R}}[/latex]
potential energy of space – ship due to gravitational attraction of mars = – [latex]\frac{\mathrm{GM}^1 \mathrm{~m}}{\mathrm{R}^1}[/latex]
since K.E of space ship is zero, therefore total energy of spaceship
= [latex]\frac{-\mathrm{GMm}}{\mathrm{R}}[/latex] – [latex]\frac{\mathrm{GM}^1 \mathrm{~m}}{\mathrm{R}^1}[/latex]
= – Gm [latex]\left(\frac{M}{R}+\frac{M^1}{R^1}\right)[/latex]
∴ energy required to rocket out the spaceship from the solar system = – (total energy of space ship)

Question 25.
A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s^-1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it ? Mass of mars = 6.4 × 10^23< kg; radius of mars = 3395 km; G = 6.67 × 10^-11< N m² kg^-2.
Solution:
Let m = mass of the rocket, M = mass of the mars and
R = radius of mars. Let v be the initial velocity of rocket.
Initial K.E = [latex]\frac{1}{2}[/latex] mv²; Initial P.E = – [latex]\frac{-\mathrm{GMm}}{\mathrm{R}}[/latex]
Total initial energy = [latex]\frac{1}{2}[/latex] mv² – [latex]\frac{-\mathrm{GMm}}{\mathrm{R}}[/latex]
since 20% of K.E is lost, only 80% is left behind to reach the height. Therefore
Total energy available = [latex]\frac{80}{100} \times \frac{1}{2}[/latex] mv²
– [latex]\frac{-\mathrm{GMm}}{\mathrm{R}}[/latex] = 0.4 mv² – [latex]\frac{-\mathrm{GMm}}{\mathrm{R}}[/latex]
If the rocket reaches the higher point which is at a height h from the surface of Mars, its
K.E. is zero and P.E. = [latex]\frac{-\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}[/latex]
using principle of conservation of energy, we have
0.4 mv² – [latex]\frac{\mathrm{GMm}}{\mathrm{R}}=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}[/latex]
or [latex]\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}=\frac{\mathrm{GM}}{\mathrm{R}}[/latex] – 0.4 v²

Textual Examples

Question 1.
Let the speed of the planet at the perihelion P in Fig. be υ_p and the Sun- planet distance SP be r_p. Relate {r_p, υ_p} to the corresponding quantities at the aphelion {r_A, υ_A}. Will the planet take equal times to traverse BAC and CPB ?

(a) An ellipse traced out by a planet around the sun. The colsest point is P and the farthest point is A. P is called the perihelion and A the aphelion. The semimajor axis (a) is half the distance AP
Answer:
The magnitude of the angular momentum at P is L_p = m_p r_p υ_p. Similarly, L_A = m_p r_A υ_A. From angular momentum conservation
m_p r_p υ_p = m_p r_A υ_A
or [latex]\frac{v_p}{v_A}=\frac{r_A}{r_p}[/latex]

Question 2.
Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. (a) What is the force acting on a mass,2m placed at the centroid O of the triangle ? (b) What is the force if the mass at the vertex A is doubled ?
Take AO = BO = CO = 1 m (see Fig)

Three equal masses are placed at the three vertices of the ∆ABC. A mass 2m is placed at the centroid O.
Answer:
(a) The angle between OC and the positive x- axix is 30° and so is the angle between OB and the negative x-axis. The individual forces a vector notation are

From the principle of superposition and the law of vector addition, the resultant gravitational force F_R on (2m) at O is
F_R = F_OA + F_OB + F_OC
F_R = 2Gm² [latex]\hat{\mathrm{j}}[/latex] + 2Gm² [latex]-\hat{\mathrm{i}}[/latex] cos 30° – [latex]\hat{\mathrm{j}}[/latex] sin 30°) + 2Gm² ([latex]\hat{\mathrm{i}}[/latex] cos 30° – [latex]\hat{\mathrm{j}}[/latex] sin 30°) = 0
Alternatively, one expects on the basis of symmetry that the resultant force ought to be zero.

(b) By symmetry the x-component of the force cancels out. The y-component survives.
F_R = 4Gm² [latex]\hat{\mathrm{j}}[/latex] – 2Gm² [latex]\hat{\mathrm{j}}[/latex] = 2Gm² [latex]\hat{\mathrm{j}}[/latex]

Question 3.
Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square.
Answer:
We have four mass pairs at distance l and two diagonal pairs at distance [latex]\sqrt{2}[/latex]1 Hence,

= [latex]\frac{2 \mathrm{Gm}}{1}\left(2+\frac{1}{\sqrt{2}}\right)[/latex] = -5.41 [latex]\frac{\mathrm{Gm}^2}{l}[/latex]
The gravitational potential U(r) at the centre of the square (r = [latex]\sqrt{2}[/latex] l / 2) is
U(r) = [latex]-4 \sqrt{2} \frac{\mathrm{Gm}}{\mathrm{l}}[/latex]

Question 4.
Two uniform solid spheres of equal radii R, but mass M and 4 M have a centre to centre separation 6 R, as shown in Fig. The two spheres are held fixed A projeetile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.

Answer:
If ON = r, we have
[latex]\frac{\mathrm{GMm}}{\mathrm{r}^2}=\frac{4 \mathrm{GMm}}{\left(6 \mathrm{R}-\mathrm{r}^2\right)}[/latex]
(6R – r)² = 4r²
6R – r = ±2r
r = 2R or – 6R.
The neutral point r = -6R does not concern us in this example. Thus ON = r = 2R.
Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface of M is
E_i = [latex]\frac{1}{2} \mathrm{~m} v^2-\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{4 \mathrm{GMm}}{5 \mathrm{R}}[/latex]
The mechanical energy at N is purely potential.
E_N = [latex]-\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{4 \mathrm{GMm}}{4 \mathrm{R}}[/latex]
From the principle of conservation of mechanical energy
[latex]\frac{1}{2} v^2-\frac{G M}{R}-\frac{4 G M}{5 R}=-\frac{G M}{2 R}-\frac{G M}{R}[/latex]
υ² = [latex]\frac{2 G M}{R}\left(\frac{4}{5}-\frac{1}{2}\right)[/latex]
υ² = [latex]\left(\frac{3 \mathrm{GM}}{5 R}\right)^{1 / 2}[/latex]

Question 5.
The planet Mars has two moons, phobos and delmos. (i) phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 × 10³ km. Calculate the mass of Mars, (ii) Assume that Earth and Mars move in circular orbits around the sun, with the Martian orbits being 1.52 times the orbital radius of the earth. What is the length of the Martain year in days ?
Answer:
(i) We employ T² = K (R_E + h)³ (where K = 4π² / GM_E) with the Earth’s mass replaced by the Martian mass M_m

(ii) Once again Kepler’s third law comes to our aid,
[latex]\frac{T_M^2}{T_E^2}=\frac{R_{M S}^3}{R_{E S}^3}[/latex]
Where R_MS is the Mars-Sun distance and R_ES is the Earth-Sun distance.
∴ T_M = (1.52)^3/2 × 365
= 684 days
For example. the ratio of the semi-minor to semi-major axis for our Earth is, b/a = 0.99986.

Question 6.
Weighing the Earth : You are given the following data g = 9.81 ms² R_E = 6.37 x106 m the distance to the moon R = 3.4 × 10⁸ m and the time period of the moons revolution is 27.3 days. Obtain the mass of the Earth M_E in two different ways.
Answer:
(1) From g = [latex]\frac{F}{m}=\frac{G M_E}{R_E^2}[/latex]
M_E = [latex]\frac{g R_E^2}{G}[/latex]
= [latex]\frac{9.81 \times\left(6.37 \times 10^6\right)^2}{6.67 \times 10^{-11}}[/latex]
= 5.97 × 10²⁴kg. (by Method – 1)

(2) The moon is a satellite of the Earth. From the derivation of Kepler’s third law

= 6.02 × 10²⁴kg (by Method – 2)
Both methods yield almost the same answer the difference between them being less than 1%.

Question 7.
Express the constant k T² = K (R_E + h)² where K = 4π²/GM_E of in days and kilometres. Given k = 10^-13 s² m^-3. The moon is at a distance of 3.84 × 10⁵ km from the earth. Obtain its time period of revolution in days.
Answer:
Given
k = 10^-13 s² m^-3 (d = day)
= 10^-13 [latex]\left[\frac{1}{(24 \times 60 \times 60)^2} d^2\right][/latex]
[latex]\left[\frac{1}{(1 / 1000)^3 \mathrm{~km}^3}\right][/latex] = 1.33 × 10^-14 d² km^-3
Using T² = K (R_E + h)³ (where k = 4π²/ GM_E) and the given value of k the time period of the moon is
T² = (1.33 × 10^-14) (3.84 × 10⁵)³
T = 27.3 d

Question 8.
A 400 kg satellite is in a circular orbit of radius 2R_E about the Earth. How much energy is required to transfer it to a circular orbit of radius 4R_E? What are the changes in the kinetic and potential energies ?
Answer:
Initially,
E₁ = [latex][/latex]
While finally
E_f = [latex][/latex]
The change in the total energy is
∆E = E_f – E_i
= [latex]\frac{\mathrm{GM}_E \mathrm{~m}}{8 R_E}=\left(\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}^2}\right) \frac{\mathrm{mR} \mathrm{R}_{\mathrm{E}}}{8}[/latex]
∆E = [latex]\frac{\mathrm{gm} \mathrm{R}_{\mathrm{E}}}{8}=\frac{9.81 \times 400 \times 6.37 \times 10^6}{8}[/latex]
= 3.13 × 10⁹J
The kinetic enegy is reduced and it mimics ∆E, namely, ∆K = K_f – K_i = -3.13 × 10⁹ J.
The change in potential energy is twice the change in the total energy, namely
∆V = V_f – V_i= -6.25 × 10⁹ J

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 8th Lesson Oscillations Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 8th Lesson Oscillations

Very Short Answer Questions

Question 1.
Give two examples of periodic motion which are not oscillatory.
Answer:

1. The motion of planets around the sun.
2. The motion of an electron round the nucleus.

Question 2.
The displacement in S.H.M. is given by y = a sin (20t + 4). What is the displacement when it is increased by 2π/ω ?
Answer:
The displacement in S.H.M. is y = a sin (20t + 4)
The time period T = [latex]\frac{2 \pi}{\omega}[/latex] is increased, the displacement of the particle remains the same.

Question 3.
A girl is swinging seated in a swing. What is the effect on the frequency of oscillation if she stands ?
Answer:
Frequency (n) = [latex]\frac{1}{2 \pi} \sqrt{\frac{g}{l}}[/latex], n ∝ [latex]\frac{1}{\sqrt{l}}[/latex]
A girl swinging in standing position location of centre of mass shifts upwards l decreases, frequency of oscillation increases.

Question 4.
The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change, if the water begins to drain out of the hollow sphere ?
Answer:
The period of the pendulum is same, when the bob is hollow (or) completely filled with water. As water flows out from the bob, the centre of gravity of the bob lowered. The pendulum length increases. Hence time period also increases. When the bob becomes empty, again centre of gravity shifts upwards. The pendulum length decreases. The time period also decreases.

Question 5.
The bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of aluminum ?
Answer:
Time period (T) = [latex]2 \pi \sqrt{\frac{l}{g}}[/latex], Time period is independent of mass of the bob.
Hence wooden bob is replaced by an identical aluminium bob, Time period remains constant.

Question 6.
Will a pendulum clock gain or lose time when taken to the top of a mountain.?
Answer:
T ∝ [latex]\frac{1}{\sqrt{g}}[/latex], At the mountain top, the value of g decreases, hence time period increases i.e. the pendulum will take longer time to complete one vibration. Hence pendulum clock will loose time on the mountain top.

Question 7.
A pendulum dock gives correct time at the equator. Will it gain or lose time if it is taken to the poles ? If so, why ?
Answer:
Time period (T) = [latex]2 \pi \sqrt{\frac{l}{g}}[/latex]
g value at poles is greater than at equator.
If it is taken to the poles g value increases, time period decreases.
So pendulum clock gains time.

Question 8.
What fraction of the total energy is K.E when the displacement is one half of a amplitude of a particle executing S.H.M.
Answer:
Total energy (E) = [latex]\frac{1}{2}[/latex] mω² A²
Given y = [latex]\frac{\mathrm{A}}{2}[/latex], Kinetic energy = [latex]\frac{1}{2}[/latex]mω²(A² – y²) = [latex]\frac{1}{2}[/latex] mω²[latex]\left(A^2-\frac{A^2}{4}\right)[/latex] = [latex]\frac{3}{4}[/latex] × [latex]\frac{1}{2}[/latex]mω²A²
K.E = [latex]\frac{3}{4}[/latex] × E
∴ [latex]\frac{K \cdot E}{E}[/latex] = [latex]\frac{3}{4}[/latex]

Question 9.
What happens to the energy of a simple harmonic oscillator if its amplitude is doubled ?
Answer:
Total energy (E) = [latex]\frac{1}{2}[/latex]mω²A²
Given Amplitude A is doubled
E’ = [latex]\frac{1}{2}[/latex]mω²(2A)²
E’ = 4 × [latex]\frac{1}{2}[/latex]mω²A²
E’ = 4E
∴ Energy becomes four times.

Question 10.
Can a simple pendulum be used in an artificial satellite? (T.S. Mar. 16)
Answer:
No, in an artificial satellite acceleration due to gravity is zero. Hence we cannot use simple pendulum in an artificial satellite.

Short Answer Questions

Question 1.
Define simple harmonic motion. Give two examples.
Answer:
Simple harmonic motion : “A body is said to be in simple harmonic motion, if it moves to and fro along a straight line, about its mean position such that, at any point its acceleration is proportional to its displacement but opposite in direction and is directed always towards the mean position”.
w ∝ -x

If a is the acceleration of the body at any given displacement x from the mean position, the time for the body to be in S.H.M.
Displacement of a particle in S.H.M is given by
x(t) = A cos (ωt + ϕ)

Example:

1. Motion of a simple pendulum.
2. Motion of mass attached to a spring.
3. Motion of atoms in solids.
4. Cork floating on water.

Question 2.
Present graphically the variations of displacement, velocity and acceleration with time for a particle in S.H.M.
Answer:
Let us put ϕ = 0 and write the expressions for x(t),
υ(t) and a(t).
x(t) = A cos ωt. υ(t) = -Aωsinωt
a(t) = -ω²A cos ωt. The corresponding plots are shown in figure. All quantities varies sinusoidally with time.

x(t) varies between – A to A; υ(t) varies from -ωA to ωA and a(t) varies from – ω²A to ω²A with respect to displacement plot, velocity plot has a phase difference of [latex]\frac{\pi}{2}[/latex] and acceleration plot has a phase difference of π.

Question 3.
What is phase ? Discuss the phase relations between displacement, velocity and acceleration in simple harmonic motion.
Answer:
Phase : The phase of a particle executing S.H.M. at any instant is defined as its state (or) condition as regards to its position and direction of motion at that instant.

1. Displacement : x = A cos (ωt – ϕ), (ωt – ϕ) is called phase and ϕ is epoch.
2. Velocity : V = -Aω sin (ωt – ϕ), Here also (ωt – ϕ) is phase angle
3. Acceleration : a = -ω²A cos (ωt – ϕ), Here also (ωt – ϕ) is phase angle.

Phase difference between displacement and velocity = [latex]\frac{\pi}{2}[/latex]
Phase difference between velocity and acceleration = [latex]\frac{\pi}{2}[/latex]
Phase difference between displacement and acceleration = π.

Question 4.
Obtain an equation for the frequency of oscillation of spring of force constant k to which a mass m is attached. Answer:
Let us consider a spring suspended vertically from a rigid support and loaded with a mass m. If it is now pulled down and released, it executes vertical oscillations about mean position.
Restoring force is directly proportional to the displacement, but oppositely directed

F ∝ -y
F = -ky —–> (1)
Where k is force constant
Ma = -ky (∴ F = Ma)
a = [latex]-\left(\frac{K}{M}\right) y[/latex] —-> (2)
Since K and M are constant. We can write a ∝ -y
ie Acceleration is directly proportional to the displacement but oppositely directed. Hence oscillations of a loaded spring is S.H.M.
Comparing eq. (2) with a = -ω²y
ω² = [latex]\frac{K}{M}[/latex] ; ω = [latex]\sqrt{\frac{\mathrm{K}}{\mathrm{M}}}[/latex]
T = [latex]\frac{2 \pi}{\omega}[/latex] ; T = [latex]2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}[/latex]
Frequency of oscillation (n) = [latex]\frac{1}{2 \pi} \sqrt{\frac{K}{M}}[/latex] —> (3)

Question 5.
Derive expressions for the kinetic energy and potential energy of a simple harmonic oscillator.
Answer:
Kinetic energy of simple harmonic oscillator : The velocity of a particle in S.H.M. is given by
v = [latex]\omega \sqrt{A^2-y^2}[/latex] ∴ Kinetic energy = [latex]\frac{1}{2}[/latex] mv² = [latex]\frac{1}{2}[/latex] mω²(A² – y²)
When y = 0, (K.E)_Max = [latex]\frac{1}{2}[/latex] mω²A² (Mean position)
When y = A, (K.E)_Min = 0 (Extreme position)
K.E is maximum at mean position and minimum at extreme position.

Potential energy of simple harmonic oscillator: When the displacement of a particle executing simple Harmonic oscillations increases, the restoring force also increases. The restoring force is in the opposite direction to the displacement. Therefore work is done in moving through the displacement, against restoring force. If F is the restoring force at the displacement y.
The average force against which work is done = [latex]\frac{\mathrm{O}+\mathrm{F}}{2}[/latex] = [latex]\frac{F}{2}[/latex]

∴ Workdone on the particle for the displacement (y) = Average force × displacement F
i.e. w = [latex]\frac{\mathrm{F}}{2}[/latex] × y
w = [latex]\frac{\max y}{2}[/latex] —- (1) (∵ F = ma)
But acceleration of a particle in S.H.M. is given by
a = ω²y —– (2)
Using eq’s (1) and (2). we get
The work done (w) = [latex]\frac{1}{2} m \omega^2 y^2[/latex] mw2y2
This work done is stored in the form of P.E.
∴ P.E. = [latex]\frac{1}{2} m \omega^2 y^2[/latex] —– (3)
If y = 0, (P.E)_Min = 0 (At mean position)
If y = A, (P.E)_Max = [latex]\frac{1}{2} m \omega^2 A^2[/latex] (At extreme position).
∴ P.E. is maximum at extreme position and minimum at mean position.

Question 6.
How does the energy of a simple pendulum vary as it moves from one extreme position to the other during its oscillations?
Answer:
The total energy associated with a particle executing S.H.M. at any point is the sum of potential energy and kinetic energy at that point.

Total energy (E) = K.E + P.E
K.E = [latex]\frac{1}{2} m \omega^2\left(A^2-y^2\right)[/latex]
P.E. = [latex]\frac{1}{2} m \omega^2 y^2[/latex]
T.E. = [latex]\frac{1}{2}[/latex]mω²(A² – y²) + [latex]\frac{1}{2}[/latex]mω²y² = [latex]\frac{1}{2}[/latex]mω²A²
At mean position y = 0, P.E = 0
and (K.E)_Max = [latex]\frac{1}{2} m \omega^2 A^2[/latex]
∴ T.E. = 0 + [latex]\frac{1}{2} m \omega^2 A^2[/latex] = [latex]\frac{1}{2} m \omega^2 A^2[/latex]
At extreme position y = A, K.E = 0 and P.E = [latex]\frac{1}{2} m \omega^2 A^2[/latex]
∴ (P.E)_Max = [latex]\frac{1}{2} m \omega^2 A^2[/latex]
From mean position to extreme position K.E. is to be converted into P.E.

Question 7.
Derive the expressions for displacement, velocity and acceleration of a particle executes S.H.M.
Answer:
Consider a particle P moves on the circumference of a circle of radius A with uniform angular velocity ω. Let PN be the perpendicular drawn to the diameter yy’ from P.
As P moves on the circumference of the circle, N moves on the diameter yy’ to and fro about the centre O.

Velocity : The velocity of a partcile executing SHM is given by
v = [latex]\frac{d y}{d t}[/latex] = [latex]\frac{\mathrm{d}}{\mathrm{dt}}[/latex] (A sin ωt)
= Aω cos ωt = Aω[latex]\sqrt{1-\sin ^2 \omega t}[/latex]
v = [latex]A \omega \sqrt{1-\left(\frac{y}{A}\right)^2}[/latex] (∵ sin ωt = [latex]\frac{y}{A}[/latex])
v = ω[latex]\omega \sqrt{A^2-y^2}[/latex] —- (2)

Acceleration : As the rate of change of velocity gives acceleration of the particle executing S.H.M is given by
a = [latex]\frac{d v}{d t}[/latex] = [latex]\frac{\mathrm{d}}{\mathrm{dt}}[/latex](Aω cos ωt) = -Aω² sin ωt
∴ a = -ω²y —- (3)

Long Answer Questions

Question 1.
Define simple harmonic motion. Show that the motion of (point) projection of a particle performing uniform circular motion, on any diameter, is simple harmonic. (AP – Mar. ‘18; TS – Mar. ‘16)
Answer:
Simple harmonic motion : A body is said to be in simple harmonic motion, if it moves to and fro along a straight line, about its mean position such that, at any point its acceleration is proportional to its displacement but opposite in direction and directed always towards the mean position.

Show that the projection of uniform circular motion on any diameter is simple harmonic : Consider a particle P moving on the circumference of a circle of radius A with uniform angular velocity ω. Let O be the centre of the circle. XX’ and YY’ are two mutually perpendicular diameters of the circle as shown in the figure. Let PN be drawn perpendicular to the diameter YY’ from P. As P moves on the circumference of the circle, N moves on the diameter YY’ to and fro about the centre O. Let us consider the position of N at any time t, after leaving the point ‘O’, during its motion. The corresponding angular displacement of the particle P is

Hence acceleration is directly proportional to the displacement and opposite direction. Hence motion of N is simple harmonic.

Question 2.
Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is seconds pendulum ? (TS – Mar. ’18, ’17, ’15, ’14, ’13; AP – Mar. ’17, ’16, ’15, ’14, ’13)
Answer:

1. Consider simple pendulum, a small bob of mass m tied to an inextensible mass less string of length L and other end of the string is fixed from a rigid support.
2. Once the bob is slightly displaced and released, it begins to oscillate about mean position.
3. Let θ be the angular displacement and T be the tension in the string.
4. The forces acting on the bob are (a) tension T along the string (b) weight mg acts vertically downwards.
5. The force mg can be resolved into two components (1) mg cos θ along the PA and (2) mg sin θ acts along PB.
   
6. From figure at point P,
   T = mg cos θ —– (1)
7. The force mg sin θ will provide the restoring torque, which tends to bring the bob back to its mean position O.
8. The restoring torque is given by
   [latex]\tau[/latex] = Restoring force × ⊥^lar distance
   [latex]\tau[/latex] = -mg sin θ × L —– (2)
   Here negative sign shows that the torque acts to reduce θ.
   Then sin θ is replaced by θ i.e., sin θ ≈ θ
   x = -mg L θ —– (3) (∵ sin θ = θ – [latex]\frac{\theta^3}{3 !}[/latex] + [latex]\frac{\theta^3}{5 !}[/latex] …..)
9. From equation (3), we note that [latex]\tau[/latex] ∝ θ. and This [latex]\tau[/latex] will bring the bob back towards its equilibrium position.
   So, if the bob is left free, it will execute angular simple harmonic motion.
   Comparing equation (3), with the equation [latex]\tau[/latex] = -kθ, we have Spring factor, k = mgL.
10. Here inertia factor = Moment of inertia of the bob about the point of suspension = mL2
11. In S.H.M, Time period (T) = [latex]2 \pi \sqrt{\frac{\text { Inertia factor }}{\text { Spring factor }}}[/latex]
    T = [latex]2 \pi \sqrt{\frac{m L^2}{m g L}}[/latex]
    T = [latex]2 \pi \sqrt{\frac{L}{g}}[/latex] — (4)
    Seconds pendulum : A pendulum whose time period is 2 seconds is called seconds pendulum.
    T = 2 seconds.

Question 3.
Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path.
Answer:
Kinetic energy : The velocity of a particle in S.H.M is given by v = ω[latex]\sqrt{A^2-y^2}[/latex]
∴ Kinetic energy = [latex]\frac{1}{2} \mathrm{mv}^2[/latex] = [latex]\frac{1}{2} m \omega^2\left(A^2-y^2\right)[/latex] —— (1)
We know that y = A sin ωt
K.E = [latex]\frac{1}{2}[/latex]mω²A²[1 – sin²ωt) —— (2)
When y = 0, (K.E)_Max = -[latex]\frac{1}{2}[/latex]mω²A² (Mean position)
When y = A, (K.E)_Min = 0 (At extreme position)
∴ K.E is maximum at mean position and minimum at extreme position.

Potential energy : When the displacement of a particle executing simple harmonic oscillations increase, the restoring force also increases. The restoring force is in the opposite direction to the displacement. Therefore work is done in moving through the displacement, against restoring force. If F is restoring force at the displacement y.
The average force against which work is done = [latex]\frac{\mathrm{O}+\mathrm{F}}{2}[/latex] = [latex]\frac{F}{2}[/latex]

∴ Workdone on the particle for the displacement (y) = Average force × displacement.
i.e., w = [latex]\frac{1}{2}[/latex] × y
w = [latex]\frac{\max y}{2}[/latex] —– (3) (∵ F = ma)
But acceleration of a particle in S.H.M is given by
a = -ω²y —– (4)
Using equations (3) and (4), we get
The work done (W) = [latex]\frac{1}{2}[/latex]mω²y²
This work done is stored in the form of P.E
∴ P.E = [latex]\frac{1}{2}[/latex]mω²y² —– (5)
∴ P.E = [latex]\frac{1}{2}[/latex]mω²A² sin²ωt —– (6) (∵ y = A sin ωt)
If y = 0, (P.E)_Min = 0 (At mean position)
If y = A, (P.E)_Max = – [latex]\frac{1}{2}[/latex]mω²A² (At extreme position)
∴ P.E. is maximum at extreme position and minimum at mean position.

Total energy (E) : The total energy associated with a particle executing S.H.M at any point is the sum of RE and K.E at that point.

Total energy (E) = K.E + P.E
K.E. = [latex]\frac{1}{2}[/latex]mω²y²
∴ T.E. = [latex]\frac{1}{2}[/latex]mω² (A² – y²) + [latex]\frac{1}{2}[/latex]mω²y² = [latex]\frac{1}{2}[/latex]mω²A²
At mean position y = 0, RE = 0, (KE)_Max = – [latex]\frac{1}{2}[/latex]mω²A²
∴ T.E = K.E + P.E
T.E = [latex]\frac{1}{2}[/latex]mω²A² + 0 = [latex]\frac{1}{2}[/latex]mω²A²
At extreme position, y = A, K.E = 0 and,
(P.E)_Max = [latex]\frac{1}{2}[/latex]mω²A².
∴ T.E = K.E + P.E

T.E. = O + [latex]\frac{1}{2}[/latex]mω²A² = [latex]\frac{1}{2}[/latex]mω²A²
From mean position to extreme position K.E is to be converted into RE.

Problems

Question 1.
The bob of a pendulum is made of a hollow brass sphere. What happens to the time period of the pendulum, if the bob is filled with water completely ? Why?
Solution:
Time period (T) = [latex]2 \pi \sqrt{\frac{1}{g}}[/latex]
The period of the pendulum is same. When the bob is hollow (or) completely filled with water. As water flows out from the bob, the centre of gravity of the bob lowers.

The pendulum length increases. Hence time period also increases. When the bob becomes empty, again centre of gravity shifts upwards. The pendulum length decreases. The time period also decreases.

Question 2.
Two identical springs of force constant “k” are joined one at the end of the other (in series). Find the effective force constant of the combination.
Solution:
k₁ = k₂ = k
If two springs are connected in series

Question 3.
What are the physical quantities having maximum value at the mean position in SHM ?
Solution:

1. Velocity, V_max = Aω
2. Kinetic energy, (K.E)_Max = [latex]\frac{1}{2}[/latex]mω²A².

Question 4.
A particle executes SHM such that, the maximum velocity during the oscillation is numerically equal to half the maximum acceleration. What is the time period ?
Solution:
Given V_Max = [latex]\frac{1}{2}[/latex]a_Max
Aω = [latex]\frac{1}{2}[/latex]ω²
ω = 2
T = [latex]\frac{2 \pi}{\omega}[/latex] = [latex]\frac{2 \pi}{2}[/latex] = π sec.

Question 5.
A mass of 2 kg attached to a spring of force constant 260 Nm^-1 makes 100 oscillations. What is the time taken ?
Solution:
m = 2 kg, k = 260 N/m
T = [latex]2 \pi \sqrt{\frac{m}{k}}[/latex] = 2 × 3.14[latex]\sqrt{\frac{2}{260}}[/latex] = 0.5508sec.
∴ Time for 100 oscillations = 100 × 0.5508
= 55.08 sec.

Question 6.
A simple pendulum in a stationery lift has time period T. What would be the effect on the time period when the lift

(i) moves up with uniform velocity
(ii) moves down with uniform velocity
(iii) moves up with uniform acceleration a
(iv) moves down with uniform acceleration ‘a’
(v) begins to fall freely under gravity ?

Solution:

i) When the lift moves up with uniform velocity
T = [latex]2 \pi \sqrt{\frac{1}{g}}[/latex]
No change in time period.

ii) When the lift moves down with uniform velocity. No change in the time period.

iii) When the lift moves up with acceleration.
T = [latex]2 \pi \sqrt{\frac{1}{g+a}}[/latex]
Time period decreases.

iv) When the lift moves down with acceleration.
T = 2π[latex]\sqrt{\frac{1}{g-a}}[/latex]
Time period increases.

v) Lift falls freely, a = g
T = 2π[latex]\sqrt{\frac{l}{g-g}}[/latex] = 2π[latex]\sqrt{\frac{1}{0}}[/latex] = ∝
Time period becomes infinity.

Question 7.
A particle executing SHM has amplitude of 4 cm and its acceleration at a distance of 1 cm from the mean position is 3 cm s^-2. What will is velocity be when it is at a distance of 2 cm from its mean position ?
Solution:
A = 4 cm, x₁ = 1 cm, a = 3 cm/s²
a = ω²x₁
3 = ω² × ₁
ω = [latex]\sqrt{3}[/latex]
Velocity v = ω[latex]\sqrt{A^2-x_2^2}[/latex]
(∵ x2 = 2 cm)
v = [latex]\sqrt{3} \sqrt{4^2-2^2}[/latex]
v = [latex]\sqrt{3} \times \sqrt{12}[/latex]
v = [latex]\sqrt{36}[/latex] = 6 cm/s.

Question 8.
A simple harmonic oscillator has a time period of 2s. What will be the change in the phase 0.25 s after leaving the mean position ?
Solution:
T = 2 sec
t = 0.25 sec
sin ωt = sin[latex]\left(\frac{2 \pi}{T}\right) t[/latex]
ϕ = ωt = [latex]\frac{2 \pi}{T}[/latex] × t
= [latex]\frac{2 \pi}{2}[/latex] × 0.25
ϕ = [latex]\frac{\pi}{4}[/latex]

Question 9.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. find the acceleration and velocity of the body when the displacement is
(a) 5 cm.
(b) 3 cm.
(c) 0 cm.
Solution:
A = 5 cm = 5 × 10^-2 m
T = 0.2 sec.

i) y = 5 cm = 5 × 10^-2 m
ω = [latex]\frac{2 \pi}{T}[/latex] = [latex]\frac{2 \pi}{0.2}[/latex] = 10π
Acceleration (a) = -ω²2y = -(10π)² × 5 × 10^-2
a = -5π² m/s²
Velocity (v) = ω[latex]\sqrt{A^2-y^2}[/latex]
= 10π[latex]\sqrt{\left(5 \times 10^{-2}\right)^2-\left(5 \times 10^{-2}\right)^2}[/latex]

ii) y = 3 cm = 3 × 10^-2 m
Acceleration (a) = -ω²y = -(10π)² × 3 × 10^-2
= -3π² m/s²
Velocity (v) = ω[latex]\sqrt{A^2-y^2}[/latex]
= 10π[latex]\sqrt{\left(5 \times 10^{-2}\right)^2-\left(3 \times 10^{-2}\right)^2}[/latex]
= 10π[latex]\sqrt{25-9}[/latex] × 10^-2
v = 0.4π m/s.

iii) y = 0 cm
a = -ω²y = -(10π)² × 0
Velocity (v) = ω[latex]\sqrt{A^2-y^2}[/latex]
= 10π[latex]\sqrt{\left(5 \times 10^{-2}\right)^2-0}[/latex]
= 0.5π m/s

Question 10.
The mass and radius of a planet are double that of the earth. If the time period of a simple pendulum on the earth is T, find the time period on the planet.
Solution:
g = [latex]\frac{\mathrm{GM}}{\mathrm{R}^2}[/latex]
g ∝ [latex]\frac{M}{R^2}[/latex]

Question 11.
Calculate the change in the length of a simple pendulum of length 1 m, when its period of oscillation changes from 2 s to 1.5 s. (T.S. Mar. ’18)
Solution:

Question 12.
A freely falling body takes 2 seconds to reach the ground on a plane, when it is dropped from a height of 8 m. If the period of a simple pendulum is seconds on the planet, calculate the length of the pendulum.
Solution:
u = 0, t = 2 sec, s = h = 8 m
s = ut + [latex]\frac{1}{2}[/latex]at
s = 0 × t + [latex]\frac{1}{2}[/latex] × g × 2²
g = 4m/s²
T = [latex]2 \pi \sqrt{\frac{1}{g}}[/latex]
π = [latex]2 \pi \sqrt{\frac{1}{4}}[/latex]
l = 1 m = 100 cm.

Question 13.
The period of a simple pendulum is found to increase by 50% when the length of the pendulum is increased by 0.6 m. Calculate the initial length and the initial period of oscillation at a place where g = 9.8 m s^-2.
Solution:

Question 14.
A clock regulated by a second’s pendulum keeps correct time. During summer the length of the pendulum increases to 1.02 m. How much will the clock gain or lose in one day ?
Solution:
T = 2π[latex]\sqrt{\frac{1}{g}}[/latex]
T ∝ [latex]\sqrt{l}[/latex]
[latex]\frac{d T}{T}[/latex] = [latex]\frac{1}{2} \frac{\mathrm{d} l}{l}[/latex]
T = 2 sec, l = [latex]\frac{\mathrm{g}}{\pi^2}[/latex] = 0.9927
dl = 1.02 – 0.9927 = 0.0273
[latex]\frac{d T}{2}[/latex] = [latex]\frac{1}{2}\left(\frac{0.0273}{0.9927}\right)[/latex]
dT = [latex]\frac{0.0273}{0.9927}[/latex]
No. of oscillations performed per day by seconds pendulum = [latex]\frac{\text { One day }}{2 \text { Sec. }}[/latex]
= [latex]\frac{86,400}{2}[/latex] = 43,200.

The gain (or) loss of time per day = No. of oscillations in one day to the change in time for one oscillation = 43,200 × [latex]\frac{0.0273}{0.9927}[/latex]
= 1180 sec.

Question 15.
The time period of a body suspended from a spring is T. What will be the new time period if the spring is cut into two equal parts and the mass is suspended
(i) from one part
(ii) simultaneously from both the parts ?
Answer:
T = [latex]2 \pi \sqrt{\frac{m}{k}}[/latex]
i) Spring is cut into two parts, k’ = 2k
T’ = [latex]2 \pi \sqrt{\frac{m}{k^{\prime}}}[/latex] = [latex]\frac{T}{\sqrt{2}}[/latex]
ii) When the mass is suspended from both the parts
T = [latex]2 \pi \sqrt{\frac{m}{2 k+2 k}}[/latex] = [latex]2 \pi \sqrt{\frac{m}{4 k}}[/latex] = [latex]\frac{T}{2}[/latex]

Additional Problems

Question 1.
Which of the following examples represent periodic motion ?
a) A swimmer completing one (return) trip from one bank of a river to the other and back. .
b) A freely suspended bar magnet displaced from its N – S direction and released.
c) A hydrogen molecule rotating about its centre of mass.
d) An arrow released from a bow.
Solution:
a) It is not a periodic motion. Though the motion of a swimmer is to and fro but will not have a definite period.
b) It is a periodic motion because a freely suspended magnet if once displaced from N-S direction and let it go, it oscillation about this position. Hence it is simple harmonic motion also.
c) It is also a periodic motion.
d) It is not a periodic motion.

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion ?
a) The rotation of earth about its axis.
b) Motion of an oscillating mercury column in a U-tube.
c) Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
d) General vibrations of a polyatomic molecule about its equilibrium position.
Solution:
a) It is periodic but not SHM because it is not to and fro motion about a fixed point.
b) It is SHM
c) It is SHM
d) It is a periodic but not SHM. A polyatomic gas molecule has a number of natural frequencies and its general motion is the resultant of S.H.M’s of a number of different frequencies. The resultant motion is periodic but not SHM.

Question 3.
Fig. depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion ? What is the period of motion (in case of periodic motion)?

Solution:
a) Does not represent periodic motion, as the motion neither repeats nor comes to mean position.
b) Represents periodic motion with period equal to 2s.
c) Does not represent periodic motion, because it is not identically repeated.
d) Represents periodic motion with periodic equal to 2s.

Question 4.
Which of the following functions of time represent
(a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion ? Give period for each case of periodic motion (ω is any positive constant)
a) sin ωt – cos ωt
b) sin³ ωt
c) 3 cos (π/4 – 2ωt)
d) cos ωt + cos 3ωt + cos 5 ωt
e) exp (-ω²t²)
f) 1 + ωt + ω²t².
Solution:
The function will represent a periodic motion, if it is identically repeated after a fixed interval of time and will represent SHM. If it can be written uniquely in the form of a
cos [latex]\left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}+\phi\right)[/latex] or a
sin [latex]\left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}+\phi\right)[/latex], where T is the time period.

a) sin ωt – cos ωt
= [latex]\sqrt{2}[/latex]([latex]\frac{1}{\sqrt{2}}[/latex]sin ωt – [latex]\frac{1}{\sqrt{2}}[/latex]cos ωt)
= [latex]\sqrt{2}[/latex](sin ωt cos [latex]\frac{\pi}{4}[/latex] – cos ωt sin [latex]\frac{\pi}{4}[/latex])
= [latex]\sqrt{2}[/latex](ωt[latex]\frac{\pi}{4}[/latex])
It is a S.H.M and its period is 2π/ω.

b) sin³ ωt = [latex]\frac{1}{4}[/latex](3sin ωt – sin 3ωt)
Here each term sin ωt and sin 3ωt individually represents S.H.M. But (ii) which is the out come of the super position of two S.H.Ms will only be periodic but not S.H.M. its time period is 2π/ω.
(∵ cos (-θ) = cos θ).

c) 3 cos [latex]\left(\frac{\pi}{4}-2 \omega t\right)[/latex] = 3 cos [latex]\left(2 \omega t-\frac{\pi}{4}\right)[/latex]
Clearly it represents S.H.M. and its time period is 2π/2ω.

d) cos ωt + cos 3ωt + cos 5ωt it represents the periodic but not S.H.M its time period is 2π/ω.

e) [latex]\mathrm{e}^{-\omega^2 t^2}[/latex] it is an exponential function which never repeats itself. Therefore it represents non-periodic motion.

f) 1 + ωt + ω²t² also represents non periodic motion.

Question 5.
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
a) at the end A
b) at end B
c) at the mid-point of AB going towards A
d) at 2 cm away from B going towards A
e) at 3 cm away from A going towards B and
f) at 4 cm away from B going towards A.
Solution:

Refer figure here A and B represent the two extreme positions of a S.H.M. For velocity, the direction from A to B is taken as positive for acceleration and the force, the direction is taken positive if directed along AP and negative if directed along BP.
a) At the end A, the artice executing S.H.M is momentarily at rest being its extreme position of motion, hence its velocity is zero, acceleration is +ve because directed along AP. Force is also +ve since the force is directed towards AP i.e + ve direction.
b) At the end B, velocity is zero. Here acceleration and force are negative as they are directed along BP i.e. along negative direction.
c) At the mid point AB going towards A, the particle is at its mean position P, with a tendency to move along PA i.e. -ve direction. Hence velocity is -ve both acceleration and force are zero.
d) At 2 cm away from B going towards A, the particle is at Q, with a tendency to move along QP, which is a negative direction there velocity, acceleration and force are all -ve.
e) At 3 cm away from A going towards B, the particle is at R with a tendency to move along RP, which is positive direction, there, velocity, acceleration and force are all +ve.
f) At 4 cm away from A going towards A, the particle is at S, with a tendency to move along SA which is negative direction for velocity. Therefore velocity is negative but acceleration is directed towards mean position i.e., along SP, hence +ve. Similarly force is also +ve.

Question 6.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion ?
a) a = 0.7x
b) a = -200x²
c) a = -10x
d) a = 100x³
Solution:
In S.H.M acceleration a is related to displacement by the related of the form a = -kx which is for relation (c).

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function.
x(t) = A cos (ωt + ϕ).
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is πs^-1. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Solution:
Here, at t = 0, x = 1 cm and v = ω cm s^-1, ϕ = ?; ω = πs^-1
Given x = A cos (ωt + ϕ)
1 = A cos (π × 0 + ϕ) or
= A cos ϕ —– (i)
Velocity, v = [latex]\frac{d x}{d t}[/latex] = – Aω(sin ωt + ϕ)
∴ ω = -Aω sin (π × 0 + ϕ) or 1 = – A sin ϕ
or A sin ϕ = -1 —— (ii)
Squaring and adding (i) and (ii)
A²(cos² ϕ + sin² ϕ) = 1 + 1 = 2 or A² = 2 or A = [latex]\sqrt{2}[/latex] cm
Dividing (ii) by (i), we get
tan ϕ = -1 or ϕ = [latex]\frac{3 \pi}{4}[/latex] or [latex]\frac{7 \pi}{4}[/latex]
For, x = B sin (ωt + α) —— (iii)
At t = 0, x = 1, so,
1 = B sin (ω × 0 + α) = B sin α —– (iv)
Differentiating (iii), w.r.t, t we have dx
velocity, v = [latex]\frac{d x}{d t}[/latex] = Bω cos (ωt + α)
Applying initial conditions i.e. at t = 0, v = ω
ω = Bωcos (π × 0 + α)
or 1 = B cos α —– (v)
Squaring and adding (iv) and (v) we get
B²sin²α + B² cos² α = 1² + 1² = 2 or B² = 2 or B = [latex]\sqrt{2}[/latex] cm
Dividing (iv) by (v), we have
[latex]\frac{B \sin \alpha}{B \cos \alpha}[/latex] = [latex]\frac{1}{1}[/latex] or tan α = 1 or α = [latex]\frac{\pi}{4}[/latex] or [latex]\frac{5 \pi}{4}[/latex].

Question 8.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ?
Solution:
Here. m = 50 kg, max. extension.
y = 20 – 0 = 20 cm = 0.2 m; T = 0.65
Max. Force, F = mg = 5 × 9.8 N
K = [latex]\frac{F}{y}[/latex] = [latex]\frac{50 \times 9.8}{0.2}[/latex] = 2450 Nm^-1
As T = [latex]2 \pi \sqrt{\frac{m}{k}}[/latex]
m = [latex]\frac{T^2 k}{4 \pi^2}[/latex]
= [latex]\frac{(0.6)^2 \times 2450}{4 \times(3.14)^2}[/latex]
= 22.36 kg
∴ Weight of body = mg = 22.36 × 9.8
= 219.1 N
= 22.36 kgf

Question 9.
A spring having with a spring constant 1200 Nm^-1 is mounted on a horizontal table as shown in fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine
(i) the frequency of oscillations,
(ii) maximum acceleration of the mass, and
(iii) the maximum speed of the mass.
Solution:
Here, k = 1200 Nm^-1, m = 3.0 kg, a = 2.0 cm = 0.02 m
a)
Frequency,
υ = [latex]\frac{1}{T}[/latex] = [latex]\frac{1}{2 \pi} \sqrt{\frac{k}{m}}[/latex]
= [latex]\frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}}[/latex]
= 3.2s^-1

b) Acceleration A = ω²y = [latex]\frac{k}{m}[/latex]y
Acceleration will be maximum when y is maximum i.e. y = a
Max. acceleration, A_max = [latex]\frac{k_a}{m}[/latex] = [latex]\frac{1200 \times 0.02}{3}[/latex]
= 8 ms^-2

c) Max. speed of the mass will be when it is passing through the mean position, which is given by
V_max = aω = a[latex]\sqrt{\frac{k}{m}}[/latex]
= 0.02 × [latex]\sqrt{\frac{1200}{3}}[/latex] = 0.4 ms^-1

Question 10.
In exercise, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
a) at the mean position
b) at the maximum stretched position and
c) at the maximum compressed position
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase ?
Solution:
Here, a = 2.0 cm; ω = [latex]\sqrt{\frac{k}{m}}[/latex] = [latex]\sqrt{\frac{1200}{3}}[/latex] = 20s^-1

a) As time is noted from the mean position, hence using
x = a sin ωt, we have x = 2 sin 20t

b) At maximum stretched position, the body is the extreme right position, with an initial phase of [latex]\frac{\pi}{2}[/latex] rad. Then
x = a sin [latex]\left(\omega t+\frac{\pi}{2}\right)[/latex]
= a cos ωt = 2 cos 20t

c) At maximum compressed position, the body is at the extreme left position, with an initial phase of [latex]\frac{3 \pi}{2}[/latex] rad.
Then x = a sin (ωt + [latex]\frac{3 \pi}{2}[/latex])
= -a cos ωt
= -2 cos 20t

Question 11.
Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple ‘ harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Solution:
In Fig. (a) T = 2s ; a = 3 cm
At t = 0, OP makes an angle [latex]\frac{\pi}{2}[/latex] with X-axis
i.e., ϕ = [latex]\frac{\pi}{2}[/latex] radian.
While moving clockwise, here ϕ = +[latex]\frac{\pi}{2}[/latex]. Thus the X-projection of OP at time t will give us the equation of S.H.M. given by
x = a cos [latex]\left(\frac{2 \pi t}{T}+\phi\right)[/latex]
= 3cos [latex]\left(\frac{2 \pi t}{2}+\frac{\pi}{2}\right)[/latex] or x = -3
sin πt (x is in cm)
In Fig. (b) T = 4s ; a = 2m
At t = 0, OP makes an angle π with the positive direction of X-axis i.e., ϕ = π. While moving anticlockwise, here ϕ = +π.
Thus the X-projection of OP at time t will give us the equation of S.H.M. as
x = a cos [latex]\left(\frac{2 \pi t}{T}+\phi\right)[/latex]
= 2 cos [latex]\left(\frac{2 \pi t}{4}+\pi\right)[/latex]
= -2 cos [latex]\left(\frac{\pi}{2} t\right)[/latex] (x is in m)

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case : (x is in cm and t is in s).
a) x = -2 sin (3t + π/3)
b) x = cos (π/6 – t)
c) x = 3 sin (2πt + π/4)
d) x = 2 cos πt.
Solution:
If we express each function of the form
x = a cos (cot + ϕ) —— (i)
Where ϕ is the initial phase i.e., ϕ represents the angle which the initial radius vector of the particle makes with the -l-ve direction of X-axis.

a) x = -2 sin (3t + [latex]\frac{\pi}{3}[/latex]) = 2 cos ([latex]\frac{\pi}{2}[/latex] + 3t + [latex]\frac{\pi}{3}[/latex])
or x = 2 cos (3t + [latex]\frac{5 \pi}{6}[/latex])
Comparing it with equation (i), we note
that a = 2, ω = 3 and ϕ = [latex]\frac{5 \pi}{6}[/latex]
Hence, the reference circle will be shown in Fig. (a).

b) x = cos[latex]\left(\frac{\pi}{6}-t\right)[/latex] = cos [latex]\left(t-\frac{\pi}{6}\right)[/latex]
(∵ cos (-θ) = cos θ))
Comparing it with equation (i), we note
that a = 1, ω and ϕ = [latex]\frac{\pi}{6}[/latex]
The reference circle will be as shown in Fig. (b).

c) x = 3 sin [latex]\left(2 \pi t+\frac{\pi}{4}\right)[/latex]
= 3 cos [latex]\left(2 \pi t+\frac{3 \pi}{2}+\frac{\pi}{4}\right)[/latex]
Comparing it with equation (i), we note that a = 3, ω = 2π and ϕ = [latex]\frac{3 \pi}{2}[/latex] + [latex]\frac{\pi}{4}[/latex] = [latex]\frac{7 \pi}{4}[/latex]
The reference circle will be as in Fig. (c).

d) x = 2 cos πt
Comparing it with equation (i), we note
that a = 2, = π and ϕ = 0.
The reference circle will be as shown Fig. (d)

Question 13.
Figures (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass at either end. Each end of the spring in Fig. (b) is stretched by the same force F.

a) What is the maximum extension of the spring in two cases ?
b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ?
Solution:

a) The maximum extension of the spring in both the cases will be = [latex]\frac{F}{K}[/latex], where K is the spring constant of the spring used.
b) In Fig. (a), if x is the extension in the spring, when mass m is returning to its mean posi-tion after being released free, then restoring force on the mass is F = -Kx i.e., F ∝ x.
As this F is directed towards mean position of the mass, hence the mass attached to the spring will execute SHM. Here, spring factor = spring constant
= K
Inertia factor = mass of the given mass
= m

In Fig.(b). we have a two body system of spring constant K and reduced mass,
μ = [latex]\frac{m \times m}{m+m}[/latex] = [latex]\frac{m}{2}[/latex]
Here, inertia factor = [latex]\frac{m}{2}[/latex]
and spring factor = K
∴ Time period, T = 2π[latex]\sqrt{\frac{(m / 2)}{k}}[/latex]
= 2π[latex]\sqrt{\frac{m}{2 K}}[/latex]

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min. what is its maximum speed ?
Solution:
Given a = [latex]\frac{1}{2}[/latex]m; ω = 200 rev/min ;
V_max = aω
= [latex]\frac{1}{2}[/latex] × 200
= 100 m/min.

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 ms^-2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 ms^-2).
Solution:

Question 16.
Answer the following questions :
a) Time period of a particle in SHM depends on the force constant k and mass m of the particle :
T = [latex]2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}[/latex]. A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum ?
b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than [latex]2 \pi \sqrt{\frac{l}{g}}[/latex]. Think of a qualitative argument to appreciate this result.
c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ?
d) What is the frequency of oscillation of a simple pendulum mounted in cabin that is freely falling under gravity ?
Solution:
a) For a simple pendulum, force constant or spring factor K is proportional to mass m, therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.

b) The effective restoring force acting on the bob of simple pendulum in displaced position is
F = -mg sin θ. When θ is small, sin θ = θ. Then the expression for time period of simple pendulum is given by T = [latex]2 \pi \sqrt{\mu \mathrm{g}}[/latex] When θ is large sin θ < θ, if the restoring force mg sin θ is replaced by mgθ, this amounts to effective reduction in the value of ‘g’ for large angles and hence an increase in the value of time period T.

c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.

d) We know that gravity disappears for a man under free fall, so frequency is zero.

Question 17.
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ?
Solution:
Centripetal acceleration, a_c = [latex]\frac{v^2}{R}[/latex], it is acting horizontally.
Acceleration due to gravity = g acting vertically downwards.
Effective acceleration due to gravity
g’ = [latex]\sqrt{g^2+\frac{v^4}{R^2}}[/latex]
∴ Time period, T = 2π[latex]\sqrt{\frac{1}{g}}[/latex]
= 2π[latex]\sqrt{\frac{1}{g^2+v^4 / R^2}}[/latex]

Question 18.
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ₁. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period.
T = [latex]2 \pi \sqrt{\frac{h \rho}{\rho_1 g}}[/latex]
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Solution:
Mass of the cylinder (m) = volume × density
= Ahρ ——- (1)
F₁ = weight of the liquid displaced by the length l of the cylinder
= (Al)ρ₁g —— (2)
Weight of the cylinder = mg —– (3)
In equilibrium position, mg = Alρ₁g
m = Alρ₁ —– (4)
F₂ = A(l + y)ρ₁g —– (5)

Restoring force (F) = -(F₂ – mg)
= -[A(l + y)ρ₁g – Alρ₁g]
F = Ayρ₁g = -(Aρ₁g)y —— (6)
In S.H.M. F = -Ky —– (7)
From eqs. (6) & (7),
spring factor (K) = Aρ₁g ——(8)
Inertia factor, m = Ahρ ——- (9)

Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Solution:
Consider a liquid of density ρ contained in a vertical U-tube of cross-sectional area A. Total length of the liquid column from P to P₁ is L.

Mass (m) = LAρ
PQ = y, P₁Q₁ = y, QQ₁ = 2y
Restoring force (F) = -(A2y)ρg
= -(2Aρg)y —– (1)
F ∝ -y
Hence oscillations in U-tube is S.H.M.

Question 20.
An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal (see Fig.).

Solution:
Consider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m filled smoothly in the neck at position C. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount P, so that the ball is depressed to position D, where CD = y.
There will be decrease in volume and hence increases in pressure of air inside the chamber. The decrease in volume of the air inside the chamber.

Here, negative sign shows that the increase in pressure will decrease the volume of air in the chamber.
Now, ρ = [latex]\frac{-E A y}{V}[/latex]
Due to this excess pressure, the restoring force acting on the ball is
F = P × A = [latex]\frac{-E A y}{V} \cdot A[/latex]
= [latex]\frac{-E A^2}{V} y[/latex] —- (1)
Since F ∝ y and negative sign shows that the force is directed towards equilibrium position, if the applied increased pressure is removed from the ball, the ball with start executing linear SHM in the neck of chamber with C as mean position.
In a S.H.M., the restoring force,
F = -Ky —- (2)
Comparing (1) and (2), we have spring factor.
K = [latex]\frac{E A^2}{V}[/latex]
Here, inertia factor = mass of ball = m
Inertia factor Spring factor

Note : If the ball oscillates in the neck of chamber under isothermal conditions, then E = P = Pressure of air inside the chamber, when ball is at equilibrium position. If the ball oscillates in the neck of chamber under adiabatic conditions, then E = υP, where v = [latex]\frac{c_p}{c_v}[/latex].

Question 21.
You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
Solution:
Here, M = 3000 kg ; x = 0.15 cm ; if K is the spring constant of each spring, then spring constant of 4 springs in parallel to support the whole mass is, K = 4 K.
4 kx = Mg
k = [latex]\frac{M g}{4 x}[/latex]
= [latex]\frac{3000 \times 10}{4 \times 0.15}[/latex]
= 5 × 10⁴ N/m.

b) If m is the mass supported by each spring, then m = [latex]\frac{3000}{4}[/latex] = 750 kg.

Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Solution:
Consider a particle of mass m be executing S.H.M. with period T. The displacement of the particle at an instant t, when time period is noted from the mean position is given by
Y = a sin ωt
∴ Velocity 1 V = [latex]\frac{\mathrm{dy}}{\mathrm{d} \mathrm{t}}[/latex] = aω cos ωt
K.E., E_K = [latex]\frac{1}{2}[/latex]mv² = [latex]\frac{1}{2}[/latex]ma²ω² cos² ωt

Average P.E. over one cycle is

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to is centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = -α θ, where J is the restoring couple and θ the angle of twist).
Solution:
Here, m = 10kg; R = 15 cm = 0.15m; T = 1.55, ∝ = ?
Moment of inertia of disc,
I = [latex]\frac{1}{2} m R^2[/latex]
= [latex]\frac{1}{2}[/latex] × 10 × (0.15)² kgm²
Now T = [latex]2 \pi \sqrt{\frac{1}{\alpha}}[/latex]
so, α = [latex]\frac{4 \pi^2 1}{T^2}[/latex]
= 4 × [latex]\left(\frac{22}{7}\right)^2[/latex] × [latex]\frac{1}{2}[/latex] × [latex]\frac{10 \times(0.15)^2}{(1.5)^2}[/latex]
= 1.97Nm/rad.

Question 24.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is
(a) 5 cm
(b) 3 cm
(c) 0 cm.
Solution:
Here, r = 5 cm = 0.05 m ; T = 0.25 ;
ω = [latex]\frac{2 \pi}{\mathrm{T}}[/latex] = [latex]\frac{2 \pi}{0.2}[/latex]
= 10p rad/s
When displacement is y, then acceleration
A = -ω²y
Velocity, v = ω[latex]\sqrt{r^2-y^2}[/latex]

Case (a) : When y = x cm = 0.05 m
A = -(10π)² × 0.05
= -5π² m/s²
V = 10π[latex]\sqrt{(0.05)^2-(0.05)^2}[/latex] = 0

Case (b) : When y = 3 cm = 0.03 m
A = -(10π)² × 0.03
= -3π² m/s²
V= 10π × [latex]\sqrt{(0.05)^2-(0.03)^2}[/latex]
= 10π × 0.04
= 0.4π m/s

Case (c): When y = 0,
A = -(10π)² × 0 = 0
V= 10π[latex]\sqrt{(0.05)^2-(0)^2}[/latex]
= 10π × 0.05
= 0.5π m/s.

Question 25.
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the centre with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x₀ and V₀.
[Hint : Start with the equation x = a cos (ωt + θ) and note that the initial velocity is negative.]
Solution:
x = A cos (ωt + θ) dx
Velocity, [latex]\frac{\mathrm{dx}}{\mathrm{dt}}[/latex] = -Aω sin (ωt + θ) dt
When t = 0, x = x₀, and [latex]\frac{\mathrm{dx}}{\mathrm{dt}}[/latex] = -V₀
∴ x₀ = A cos θ
-V₀ = -Aω sin θ or A sin θ = [latex]\frac{V_0}{\omega}[/latex]
Squaring and adding- (i) and (ii), we get
A²(sin² θ + cos² θ) = [latex]\left(\frac{v_0^2}{\omega^2}\right)+x_0^2[/latex]
A = [latex]\left[\frac{v_0^2}{\omega^2}+x_0^2\right]^{1 / 2}[/latex]

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 = [latex]\frac{1}{2}[/latex] I ω² = [latex]\frac{1}{2} \frac{\mathrm{L}^2}{1}[/latex] , E oc [latex]\frac{1}{l}[/latex] (∵ 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 = [latex]I\left(\frac{2 \pi}{T}\right)[/latex]= 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 [latex]\overrightarrow{r_1}, \overrightarrow{r_2}, \overrightarrow{r_3} \ldots \ldots \overrightarrow{r_n}[/latex] 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 [latex]\vec{C}=\vec{A} \times \vec{B}[/latex]. 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.
∴ [latex]|\vec{C}|[/latex] = AB sin θ. and [latex]\vec{C}[/latex] = AB sin θ [latex]\hat{n}[/latex]. Where, [latex]\hat{n}[/latex] is the unit vector perpendicular to the plane containing the vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex]
Example: 1) Torque is cross product of position vector and Force. i.e., [latex]\overrightarrow{\mathrm{T}}=\vec{r} \times \vec{F}[/latex]
2) Angular momentum is cross product of position vector and momentum
i.e., [latex]\vec{L}=\vec{r} \times \vec{p}[/latex]

Properties:

1. Cross product does not obey commutative law. But its magnitude obeys commutative low.
   [latex]\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}|[/latex]
2. It obeys distributive law [latex]\vec{A} \times(\vec{B} \times \vec{C})=\vec{A} \times \vec{B}+\vec{A} \times \vec{C}[/latex]
3. The magnitude of cross product of two vectors which are parallel is zero.
   Since θ = 0; [latex]|\vec{A} \times \vec{B}|[/latex] = AB sin 0° = 0
4. For perpendicular vectors, θ = 90°, [latex]|\vec{A} \times \vec{B}|[/latex] = AB sin 90° [latex]|\hat{n}|[/latex]= 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 = [latex]\frac{\mathrm{d} \theta}{\mathrm{dt}}[/latex]
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 = [latex]\frac{\mathrm{d} \omega}{\mathrm{dt}}[/latex]
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 = [latex]\frac{\mathrm{dL}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{I} \omega)}{\mathrm{dt}}[/latex]
Where I = MR² = moment of inertia of a body
τ = I [latex]\frac{\mathrm{d} \omega}{\mathrm{dt}}[/latex] [∵ I = constant]
But [latex]\frac{\mathrm{d} \omega}{\mathrm{dt}}[/latex] 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 = [latex]\left(\frac{\omega_f+\omega_t}{2}\right) t[/latex] [∵ like v = [latex]\left(\frac{v_1+v_2}{2}\right) t[/latex]]
3. θ = ω_it + [latex]\frac{1}{2}[/latex] α t² [∵ like s = ut + [latex]\frac{1}{2}[/latex] 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 = [latex]\frac{1}{2}[/latex] Mv² + [latex]\frac{1}{2}[/latex] Iω²
= [latex]\frac{1}{2}[/latex] MR² ω² + [latex]\frac{1}{2}[/latex] Mk²ω²
E = [latex]\frac{1}{2}[/latex] M ω² (R² + K²)
where k is radius of gyration
E = [latex]\frac{1}{2}[/latex] Mv² (1 + [latex]\frac{K^2}{R^2}[/latex]) [∵ ω = [latex]\frac{V}{R}[/latex]]

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 = [latex]\sqrt{\frac{I}{M}}[/latex]
Where M = mass of disk
I = M.I of disk
The disk is cut into two halves about AB, when each,
Mass M’ = [latex]\frac{\mathrm{M}}{\mathrm{2}}[/latex] and each M.I, I = [latex]\frac{1}{2}[/latex]
Therefore Radius of gyration of each piece is
K’ = [latex]\sqrt{\frac{I^{\prime}}{M^{\prime}}}=\sqrt{\frac{\left(\frac{I}{2}\right)}{\left(\frac{M}{2}\right)}}=\sqrt{\frac{I}{M}}[/latex] = 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 = [latex]\frac{\mathrm{MR}_{\mathrm{r}}^2}{2}[/latex]
For a thin flat circular dist, M.I., I_d = [latex]\frac{\mathrm{MR}_{\mathrm{d}}^2}{2}[/latex]
Given that I_r = I_d
[latex]\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}[/latex]
∴ [latex]\frac{R_r}{R_d}=\frac{1}{\sqrt{2}} \text { or } R_r: R_d=1: \sqrt{2}[/latex]

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 = [latex]\frac{\mathrm{dL}}{\mathrm{dt}}[/latex]
If τ = 0 ⇒ [latex]\frac{\mathrm{dL}}{\mathrm{dt}}[/latex] = 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 = [latex]\frac{\tau}{\mathrm{l}}=\frac{12}{0.48}[/latex] = 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 = [latex]\frac{\mu \mathrm{g}}{\mathrm{r}}=\sqrt{\frac{0.8 \times 9.8}{0.1}}=\sqrt{8 \times 9.8}[/latex]
= 8.854 rad/s
w = 2πn
∴ frequency n = [latex]\frac{\omega}{2 \pi}=\frac{8.854}{2 \times 3.14}[/latex] = 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 = [latex]\frac{\mathrm{mL}^2}{12}=\frac{5.1 \times 1^2}{12}[/latex]
= 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 = [latex]\frac{\mathrm{MR}^2}{4}[/latex] + MR²
I₁ = [latex]\frac{5\mathrm{MR}^2}{4}[/latex]
Similarly I₂ = [latex]\frac{5\mathrm{MR}^2}{4}[/latex]
∴ I = [latex]\frac{10 \mathrm{MR}^2}{4}=\frac{10 \times 1 \times\left(20 \times 10^{-2}\right)^2}{4}[/latex] = 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 = [latex]\frac{2}{5}[/latex] ma²
I₃ = [latex]\frac{2}{5}[/latex] ma²
I₄ = mb²
∴ Moment of inertia of the system
I = I₁ + I₂ + I₃ + I₄
= mb² + [latex]\frac{2}{5}[/latex] ma² + [latex]\frac{2}{5}[/latex] ma² + mb²
I = [latex]\frac{4}{5}[/latex] 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 = [latex]\frac{10 \times 33}{5}[/latex] = 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 × [latex]\frac{\mathrm{N}}{\mathrm{t}}[/latex] = 2n × [latex]\frac{60}{60}[/latex] rad /s = 2π rad/s
The M.I of disc about parallel axis passing through a point on its circumferance
I = [latex]\frac{\mathrm{MR}^2}{2}[/latex] + MR² = [latex]\frac{3}{2}[/latex] MR²
∴ Kinetic energy = [latex]\frac{1}{2}[/latex] I ω²
= [latex]\frac{1}{2}[/latex] × [latex]\frac{3}{2}[/latex] MR² ω² = [latex]\frac{3}{4}[/latex] × 5 × (1)² × (2π)²
= [latex]\frac{3}{4}[/latex] × 5 × 4π² = 15 × ([latex]\frac{22}{7}[/latex]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;
[latex]\vec{V}_1=\vec{V}_2=\vec{V}[/latex]
momentum of 1^st particle p₁ = [latex]\mathrm{m} \vec{\mathrm{V}}[/latex]
∴ p₁ = [latex]\mathrm{m} \vec{\mathrm{V}}[/latex]
momentum of 2^nd particle [latex]\vec{\mathrm{P}}_2=\mathrm{mV}[/latex]

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

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 = [latex]\frac{60}{60}[/latex] = 1Hz
Initial angular velocity
ω₁ = 2π n₁ = 2π rad /sec
Final frequency n₂ = [latex]\frac{180}{60}[/latex] = 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 = [latex]\frac{1}{2}[/latex] I ω₂² – [latex]\frac{1}{2}[/latex] I ω₁²
100 = [latex]\frac{1}{2}[/latex] I [(6π)² – (2π)²]
100 = [latex]\frac{1}{2}[/latex] I (32π)²
I = [latex]\frac{200}{32 \pi^2}[/latex] = 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 = [latex]\frac{45.085}{36.5}[/latex]
= 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 [latex]\vec{\mathrm{a}}[/latex] represented by [latex]\vec{\mathrm{OP}}[/latex] and [latex]\vec{\mathrm{b}}[/latex] be represented by [latex]\vec{\mathrm{OQ}}[/latex].
Let ∠POQ = θ

∴ area of ∆ OPQ = [latex]\frac{1}{2}|\vec{a} \times \vec{b}|[/latex], 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 [latex]\hat{n}[/latex] is unit vector along [latex]\overrightarrow{\mathrm{OA}}[/latex] perpendicular to the plane containing [latex][/latex] and [latex]\vec{\mathrm{b}}[/latex]. Now [latex]\vec{\mathrm{b}}[/latex] = (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 [latex]\vec{\mathrm{r}}[/latex] and linear momentum vector [latex]\vec{\mathrm{p}}[/latex] 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
[latex]\vec{L_B}=m \vec{v} \times d+m \vec{v} \times 0=m \vec{v} d[/latex]
Let us consider any other point (on AB, where AC = x)
∴ Vector angular momentum of the two particle system about c is
[latex]\vec{L_c}=m \vec{v}(x)+m \vec{v}(d-x)=m \vec{v} d[/latex]
Clearly, [latex]\vec{\mathrm{L}_{\mathrm{A}}}=\vec{\mathrm{L}_{\mathrm{B}}}=\vec{\mathrm{L}_C}[/latex]
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) [latex]\frac{T_1}{T_2}=\frac{\sin \theta_2}{\sin \theta_1}=\frac{\sin 53.1^{\circ}}{\sin 36.9^{\circ}}[/latex]
= [latex]\frac{0.7407}{0.5477}[/latex] = 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)
[latex]\frac{R_1}{R_2}=\frac{0.75}{1.05}=\frac{5}{7}[/latex]
Putting in (i)
[latex]\frac{5}{7}[/latex] R₂ + R₂ = 1800 × 9.8
R₂ = [latex]\frac{7 \times 1800 \times 9.8}{12}[/latex]
= 10290 N
R₁ = [latex]\frac{5}{7}[/latex] R₂
= [latex]\frac{5}{7}[/latex] × 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 = [latex]\frac{2}{5}[/latex]MR²
Applying theorem of parallel axes,
Moment of inertia of sphere about a tangent to the sphere = [latex]\frac{2}{5}[/latex]MR² + M(R)²
= [latex]\frac{7}{5}[/latex]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 = [latex]\frac{1}{4}[/latex]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 × [latex]\frac{1}{4}[/latex]MR² = [latex]\frac{1}{2}[/latex]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
= [latex]\frac{1}{2}[/latex]R² + MR² = [latex]\frac{3}{2}[/latex] 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 = [latex]\frac{2}{5}[/latex] MR²
Torque applied, I = I₁ α₁ = I₂ α₂
[latex]\frac{\alpha_2}{\alpha_1}=\frac{l_1}{l_2}=\frac{M^2}{\frac{2}{5} M R^2}=\frac{5}{2}[/latex]
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
= [latex]\frac{\mathrm{MR}^2}{2}=\frac{20 \times(0.25)^2}{2}[/latex] = 0.625 kg/m²
K.E of rotation = [latex]\frac{1}{2}[/latex] Iω² = [latex]\frac{1}{2}[/latex] × 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 = [latex]\frac{2}{5}[/latex]I, Intial moment of Inertia
As no external torque acts in the process, therefore
Iω = constant
i.e. I₂ω₂ = I₁ω₁
ω₂ = [latex]\frac{I_1}{I_2}[/latex] ω₁ = [latex]\frac{5}{2}[/latex] × 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 = [latex]\frac{1}{2}[/latex]I₂ω₂²
Intial K.E of rotation, E₁ = [latex]\frac{1}{2}[/latex]I₂ω₁²
[latex]\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[/latex]
= [latex]\frac{2}{5} \times\left(\frac{100}{40}\right)^2=\frac{5}{2}[/latex] = 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 = [latex]\frac{\tau}{\mathrm{I}}=\frac{12}{0.48}[/latex] = 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[latex]\left(\frac{R}{2}\right)^2 \times m=\frac{\pi R^2}{4} m=\frac{M}{4}[/latex]
In fig, mass M is connected at O and mass
M’ is concentrated at O’, where OO’ = [latex]\frac{\mathrm{R}}{\mathrm{2}}[/latex].
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’ = [latex]\frac{\mathrm{-M}}{\mathrm{4}}[/latex] at O’. If x is the distance of centre of 4 mass (p) of the remaining part, then
x = [latex]\frac{M \times O-M^{\prime} \times \frac{R}{2}}{M-M^{\prime}}[/latex]
= [latex]\frac{\frac{-M}{4} \times \frac{R}{2}}{M-\frac{M}{4}}=\frac{-M R}{8} \times \frac{4}{3 M}=\frac{-R}{6}[/latex]
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 = [latex]\frac{10 \times 33}{5}[/latex]
= 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
[latex]\frac{1}{2}[/latex] mv² + [latex]\frac{1}{2}[/latex] Iω² = mgh
As I = [latex]\frac{2}{5}[/latex] mr²; [latex]\frac{1}{2}[/latex] mv² + [latex]\frac{1}{2}[/latex] ([latex]\frac{2}{5}[/latex] mr²)ω² = mgh
as rω = v, [latex]\frac{1}{2}[/latex] mv² + [latex]\frac{1}{5}[/latex] mv² = mgh
v = [latex]\sqrt{\frac{10}{7} \mathrm{gh}} .[/latex]
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 = [latex]\frac{1}{2}[/latex] mv² + [latex]\frac{1}{2}[/latex] Iω²
= [latex]\frac{1}{2}[/latex] mv² + [latex]\frac{1}{2}[/latex] (MR)²ω² = [latex]\frac{1}{2}[/latex] mv² + [latex]\frac{1}{2}[/latex] 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 [latex]\frac{\mathrm{m}}{\mathrm{2}}[/latex] 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 [latex]\frac{1}{2}[/latex] mv² + [latex]\frac{1}{2}[/latex] Iω² = mgh
[latex]\frac{1}{2}[/latex] mv² + [latex]\frac{1}{2}[/latex] ([latex]\frac{1}{2}[/latex] mr²) ω² = mgh
[latex]\frac{3}{4}[/latex] mv² = mgh
h = [latex]\frac{3 v^2}{4 g}=\frac{3 \times 5^2}{4 \times 9.8}[/latex] = 1.913 m
If s is the distance up the inclined plane, then as sin θ = [latex]\frac{\mathrm{h}}{\mathrm{s}^2}[/latex]
s = [latex]\frac{\mathrm{h}}{\sin \theta}=\frac{1.913}{\sin 30^{\circ}}[/latex] = 3.826 m
Time taken to return to the bottom
t = [latex]\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}}}[/latex] = 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 = [latex]\frac{l_1}{l_2} \omega_1=\frac{15.7 \times 30}{8.0}[/latex] = 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 × [latex]\frac{1}{2}[/latex] = 2.5
Also, L = [latex]\frac{M L^2}{3}=\frac{12 \times 1.0^2}{3}[/latex] = 4kgm
As L = Lω
∴ ω = [latex]\frac{\mathrm{L}}{1}=\frac{2.5}{4}[/latex] = 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 = [latex]\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}[/latex]

b) Initial K.E of two disc E₁ = [latex]\frac{1}{2}[/latex] I₁ω₁² + [latex]\frac{1}{2}[/latex] I₂ω₂²
Final K.E of the system E₂ = [latex]\frac{1}{2}[/latex] (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 = [latex]\frac{2 g h}{\left(1+k^2 / R^2\right.}[/latex] 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 . [latex]\frac{1}{2}[/latex] mv² + [latex]\frac{1}{2}[/latex] Iω² = mgh
[latex]\frac{1}{2}[/latex] mv² + [latex]\frac{1}{2}[/latex] (mk²) ω² = mgh
As w = [latex]\frac{\mathrm{V}}{\mathrm{R}}[/latex]
∴ [latex]\frac{1}{2}[/latex]mv² + [latex]\frac{1}{2}[/latex]m [latex]\frac{k^2}{R^2}[/latex]v² = mgh
or mv² (1 + [latex]\frac{k^2}{R^2}[/latex]) = mgh
v² = [latex]\frac{2 g h}{\left(1+\frac{K^2}{R^2}\right)}[/latex]

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 = ([latex]\frac{\mathrm{R}}{\mathrm{2}}[/latex] 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 = [latex]\frac{\mu_k \mathrm{mgR}}{\mathrm{I}}[/latex]
ω = ω₀ + αt
∴ ω = ω₀ – [latex]\frac{\mu_k \mathrm{mgRt}}{\mathrm{I}}[/latex]
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 = [latex]\frac{2}{3}[/latex]g sin θ = [latex]\frac{2}{3}[/latex] × 9.8 sin 30° = [latex]\frac{9.8}{3}[/latex] m/s²
a) Force of friction, f = mg sin θ – ma = m(g sin θ – α) = 10(9.8 sin 30° – [latex]\frac{9.8}{3}[/latex]) = 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 = [latex]\frac{1}{3}[/latex] 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 = [latex]\mathrm{P}_{\mathrm{i}}^{-}+\mathrm{m}_{\mathrm{i}} \mathrm{V}[/latex]
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 [latex]\frac{d l^{\prime}}{d t}=\sum r_i^{\prime} \frac{d p^{\prime}}{d t}[/latex]
Further, show that [latex]\frac{\mathrm{dL}^{\mathrm{l}}}{\mathrm{dt}}[/latex] = 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 [latex]\overrightarrow{r_1}, \overrightarrow{r_2}, \overrightarrow{r_i}[/latex] w.r.t origin ‘O’. The position vector of C.M. say [latex]\overrightarrow{\mathrm{OP}}[/latex]
i.e., [latex]\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}[/latex]
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. [latex]\frac{1}{2}[/latex] m₁v₁² + [latex]\frac{1}{2}[/latex] m₂v₂² + …..
= [[latex]\frac{1}{2}[/latex] m₁v₁² + [latex]\frac{1}{2}[/latex]m₂v₂ +……] + [latex]\frac{1}{2}[/latex]Mv²
[latex]\frac{1}{2}[/latex] m_iv_i² = [latex]\frac{1}{2}[/latex]m_iv_i² + [latex]\frac{1}{2}[/latex] mv²
∴ k = k’ + [latex]\frac{1}{2}[/latex]
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. [latex]\sqrt{3}[/latex] ). 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[latex]\sqrt{2}[/latex] 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[latex]\sqrt{2}[/latex] 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 [latex]\sqrt{2}[/latex] = 196.0/4 [latex]\sqrt{2}[/latex]
= 34.6 N
From (ii) F = F₁ = 34.6 N
F₂ = [latex]\sqrt{F^2+N^2}[/latex] = 199.0 N

Question 5.
Find the scalar and vector products of two vectors
a = [latex](3 \vec{i}-4 \vec{j}+5 \vec{k})[/latex] and b = [latex](-2 \vec{i}+\vec{j}-3 \vec{k})[/latex]
Solution:
a.b = ([latex](3 \vec{i}-4 \vec{j}+5 \vec{k})[/latex]) . ([latex](-2 \vec{i}+\vec{j}-3 \vec{k})[/latex])
= -6 – 4 – 15 = -25
a × b = [latex]\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}[/latex]
Note b × a = [latex]7 \vec{i}-\vec{j}-5 \vec{k}[/latex]

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
= [latex]\frac{2 \pi \times \text { angular speed in rev } / \mathrm{min}}{60 \mathrm{~s} / \mathrm{min}}[/latex]
= [latex]\frac{2 \pi \times 1200}{60} \mathrm{rad} / \mathrm{s}[/latex]
= 40π rad/ s
Similarly ω = final angular speed in rad /s
= [latex]\frac{2 \pi \times 3120}{60} \mathrm{rad} / \mathrm{s}[/latex]
= 2π × 52 rad/s = 104 π rad/s
∴ Angular acceleration
α = [latex]\frac{\omega-\omega_0}{t}[/latex] = 4π rad/s²
The angular acceleration of the engine = 4π rad/s²

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

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

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’ = [latex]\mathrm{M} \frac{l^2}{12}+\mathrm{M}\left(\frac{l}{2}\right)^2=\frac{\mathrm{M} l^2}{3}[/latex]
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’ = [latex]2 \mathrm{M} \frac{4 l^2}{12} \times \frac{1}{2}=\frac{\mathrm{M} l^2}{3}[/latex]

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 = [latex]\frac{1}{2}[/latex] 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 × [latex]\frac{1}{2}[/latex] × 12.5 × 10.0 = 250 (rad/s)²
∴ K.E. gained = [latex]\frac{1}{2}[/latex] × 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 = [latex]\frac{1}{2} m v_{\mathrm{cm}}^2\left[1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right][/latex]
K = [latex]\frac{1}{2} m v^2 1+\left(\frac{K^2}{R^2}\right)[/latex] 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 = [latex]\sqrt{\frac{2 g h}{1+1}}=\sqrt{g h}[/latex]
For a solid cylinder K² = R²/0.2
v_disc = [latex]\sqrt{\frac{2 g h}{1+1 / 2}}=\sqrt{\frac{4 g h}{3}}[/latex]
For a solid sphere K² = 2R²/5
v_sphere = [latex]\sqrt{\frac{2 g h}{1+1 / 52}}=\sqrt{\frac{10 g h}{7}}[/latex]

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 = [latex]\vec{F} \cdot \vec{S}[/latex] = F S cos θ.
Unit: Joule

Power : The rate of doing work by a force is called power, i.e., P = [latex]\frac{\mathrm{W}}{\mathrm{t}}[/latex]
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) = [latex]\frac{\mathrm{P}^2}{2 \mathrm{~m}}[/latex]; 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 = [latex]\frac{\mathrm{h}}{2}[/latex]
e = [latex]\sqrt{\frac{\mathrm{h}_2}{\mathrm{~h}_1}}=\sqrt{\frac{\frac{\mathrm{h}}{2}}{\mathrm{~h}}}=\frac{1}{\sqrt{2}}[/latex]

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 = [latex]\frac{h\left(1+e^2\right)}{\left(1-e^2\right)}[/latex], 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 = [latex]\frac{1}{2}[/latex] mv² – [latex]\frac{1}{2}[/latex] mu²; [latex]\frac{1}{2}[/latex] mu² = [latex]\frac{\mathrm{P}^2}{2 \mathrm{~m}}[/latex]
∴ W = F.S = [latex]\frac{\mathrm{P}^2}{2 \mathrm{~m}}[/latex]
∴ S ∝ [latex]\frac{1}{m}[/latex]
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.
[latex]\frac{1}{2}[/latex]m₁u₁² + [latex]\frac{1}{2}[/latex]m₂u₂² = [latex]\frac{1}{2}[/latex] m₁v₁² + [latex]\frac{1}{2}[/latex]m₂v₂²
m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
m₁(u₁² – v₁²) = m₂(v₂² – u₂²) ……………….. (2)
[latex]\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)}[/latex]
[latex]\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)}[/latex]
(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.
⇒ [latex]\frac{1}{2}[/latex] m₁ (u₁cosα)² + [latex]\frac{1}{2}[/latex]m₂ (u₂cosβ)² = [latex]\frac{1}{2}[/latex] m₁ (v₁cosθ)² + [latex]\frac{1}{2}[/latex]m₂(v₂cos126,Φ)²
⇒ m₁ [(u₁cosα)² – (v₁cosθ)² ] = m₂ [(v₂cosΦ)² – (u₂cosβ)²] …………….. (5)
[latex]\frac{(5)}{(4)}[/latex] ⇒ 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 = [latex]\left(\frac{m_1-m_2}{m_1+m_2}\right)[/latex] u₁ cosα + [latex]\left(\frac{2 m_u u_2}{m_1+m_2}\right)[/latex] cosβ ……………… (8)
Sub-equation (6) in equation (3), we get
and v₂ cosΦ = [latex]\left(\frac{m_2-m_1}{m_1+m_2}\right)[/latex] u₂ cosβ + [latex]\left(\frac{2 m_1 u_1}{m_1+m_2}\right)[/latex] 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 = [latex]\sqrt{2 g h}[/latex] …………….. (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 = [latex]\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}}[/latex]
∴ v₁ = -e[latex]\sqrt{2 g h}[/latex] ……………… (2)
(-) indicates that the sphere rebounds
∴ The height (h₁) attained by the sphere after first rebound is
h₁ = [latex]\frac{v_1^2}{2 g}=\frac{(e \sqrt{2 g h})^2}{2 g}[/latex] = e²h
h₁ = (e²)¹ h ………………… (3)
Similarly we can that velocity attained by the sphere after second rebound.
v₂ = -e² [latex]\sqrt{2 g h}[/latex] …………………. (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 [latex]\sqrt{2 g h}[/latex]
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 [latex]\frac{\mathrm{m}}{2}[/latex], we have 11
[latex]\frac{1}{2}[/latex] mv² – [latex]\frac{1}{2}[/latex] 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 [latex]\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}[/latex]
Once again multiplying both sides by [latex]\frac{\mathrm{m}}{2}[/latex], we obtain
[latex]\frac{1}{2}[/latex] mv² – [latex]\frac{1}{2}[/latex] mu² = m [latex]\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}[/latex] – [latex]\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{d}}[/latex] …………………. (3)
The above equation provides a motivation for the definitions of work and kinetec energy.
In equation (3), [latex]\frac{1}{2}[/latex] mv² – [latex]\frac{1}{2}[/latex] mu² = k_f – k_i
Where k_i, k_f are initial and final kinetic energies and [latex]\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{d}}[/latex] = 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 = [latex]\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[/latex] ………………… (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 = [latex]\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)}[/latex] 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. = [latex]\frac{1}{2}[/latex] 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) = [latex]\frac{1}{2}[/latex] mv_B² = [latex]\frac{1}{2}[/latex] 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 = [latex]\sqrt{2 \mathrm{gH}}[/latex]
Kinetic energy of the body (k) = [latex]\frac{1}{2}[/latex] mv_c²
= [latex]\frac{1}{2}[/latex] 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 = [latex]\sqrt{5 r g}=\sqrt{5 \times 5 \times 10^{-2} \times 10}[/latex]
= 5 × [latex]\sqrt{10^{-1}}[/latex] 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 = [latex]\frac{\text { K.E of n} \text { bullets }}{t}=\frac{\frac{1}{2} \mathrm{mnV}^2}{t}[/latex]
= [latex]\frac{\frac{1}{2} \times 5 \times 10^{-3} \times 360 \times 600 \times 600}{60}[/latex]
∴ 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 = [latex]\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}}[/latex]
= [latex]\frac{600}{60}\left(9.8 \times 25+\frac{50 \times 50}{2}\right)[/latex]
= 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 = [latex]\int_{x=0}^{x=4} F d x[/latex]
= [latex]\int_0^4(20+5 x) d x[/latex]
= [latex][20 x]_0^4+\left[5 \frac{x^2}{2}\right]_0^4[/latex]
= 20 × 4 + 5 × [latex]\frac{16}{2}[/latex]
= 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 × [latex]\frac{3}{5}[/latex] = 600 × x
⇒ x = [latex]\frac{15}{600}[/latex]
= 0.025 cm.

Question 7.
A force F = -[latex]\frac{K}{x^2}[/latex](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 = [latex]\frac{+2 b-b}{2}=\frac{b}{2}[/latex]
⇒ x = -a to x = 2a
W = [latex]\int_{-a}^{+2 a} F d x=F[x]_a^{+2}=\frac{b}{2}[2 a-(-a)]=\frac{3 a b}{2}[/latex]

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 = [latex]\sqrt{800}[/latex]
And v₁ = [latex]\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 20}=\sqrt{400}[/latex]
Since floor is at rest u₂ = 0; v₂ = 0
e = [latex]\frac{v_2-v_1}{u_1-u_2}=\frac{0-\sqrt{400}}{800-0}=-\frac{1}{\sqrt{2}}[/latex]
∴ e = [latex]\frac{1}{\sqrt{2}}[/latex] […. -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 [latex]\frac{1}{\sqrt{2}}[/latex]. What is the total distance travelled by the ball before it ceases to rebound ?
Solution:
Given h = 10m; e = [latex]\frac{1}{\sqrt{2}}[/latex]
d = h[latex]\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][/latex]
∴ 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 = [latex]\vec{F} \cdot \vec{S}[/latex] = FS cosθ where θ is smaller angle between force j and displacement [latex]\vec{S}[/latex].
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 = -[latex]\frac{F}{m}=\frac{7}{2}[/latex] = 3.5 m/s²
force of friction, f = µR = µmg = 0.1 × 2 × 9.8 = 1.96 N
Retardation produced by friction
a₂ = -[latex]\frac{F}{m}=\frac{-1.96}{2}[/latex] = 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 + [latex]\frac{1}{2}[/latex] at²
= 0 + [latex]\frac{1}{2}[/latex] × 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 = [latex]\frac{1}{2}[/latex] mv² = [latex]\frac{1}{2}[/latex] × 2 × (25.2)²
= 635J
Initial K.E = [latex]\frac{1}{2}[/latex]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 = [latex]\frac{1}{2}[/latex] mu² + [latex]\frac{1}{2}[/latex] kx²
The particles turns back at the instant, when its velocity becomes zero i.e. u = 0.
∴ E = 0 + [latex]\frac{1}{2}[/latex] kx² as E = 1 Joule and K = [latex]\frac{1}{2}[/latex] N/m
∴ 1 = [latex]\frac{1}{2}[/latex] × [latex]\frac{1}{2}[/latex] 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 + [latex]\frac{1}{2}[/latex] 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) [latex]\frac{\mathrm{L}^2}{\mathrm{~T}^3}[/latex] = 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 = [latex]-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}[/latex] N where [latex]\hat{i}, \hat{j}, \hat{k}[/latex] 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 = [latex]-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}[/latex]N
[latex]\overrightarrow{\mathrm{S}}=4 \hat{\mathrm{k}}[/latex] (∵ 4m distance is along z-axis)
W = 2; As W = [latex]\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{S}}[/latex]
∴ W = [latex](-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{k}})=12 \hat{\mathrm{k}} \cdot \hat{\mathrm{k}}[/latex] = 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 = [latex]\frac{1}{2}[/latex] m_ev_e² = 10 × 1.6 × 10^-16 ……………… (i)
E₂ = [latex]\frac{1}{2}[/latex] m_pv_p² = 100 × 1.6 × 10^-16 ……………… (ii)
Divide (i) by (ii) [latex]\frac{m_v v_v^2}{m_p v_p^2}=\frac{1}{10}[/latex]
[latex]\frac{v_{\mathrm{e}}}{v_\rho}=\sqrt{\frac{1}{10}} \frac{m \rho}{m_e}[/latex]
= [latex]\sqrt{\frac{1.67 \times 10^{-27}}{10 \times 9.11 \times 10^{-31}}}=10^2 \sqrt{\frac{1.67}{91.1}}[/latex] = 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 = [latex]\frac{500}{2}[/latex] = 250 m
Density of water, ρ = 10³ kg/m³
Mass of rain drop
= Volume of drop × Density
m = [latex]\frac{4}{3}[/latex]πr³ × ρ = [latex]\frac{4}{3}[/latex] × [latex]\frac{22}{7}[/latex] (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 = [latex]\frac{1}{2}[/latex] mv² = [latex]\frac{1}{2}[/latex] × 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 + [latex]\frac{1}{2}[/latex] M(0)²
E₂ = 2 × 10⁴ mJ
Which is the K.E. of the molecule before collision.
[E₁ = [latex]\frac{1}{2}[/latex]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 = [latex]\frac{\mathrm{w}}{\mathrm{t}}=\frac{\mathrm{mgh}}{\mathrm{t}}[/latex]
= [latex]\frac{30 \times 10^3 \times 9.8 \times 40}{900}[/latex] = 10370 watt.
If P_i is input power (required), then as η = [latex]\frac{P_0}{P_i}[/latex]
∴ P_i =[latex]\frac{P_0}{\eta}=\frac{10370}{\frac{30}{100}}[/latex] = 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 =
[latex]\frac{1}{2}[/latex]mV² + 0 = [latex]\frac{1}{2}[/latex] mV²
After collision K.E of the system is
Case I, E₁ = [latex]\frac{1}{2}[/latex] (2m) (V/2)2 = [latex]\frac{1}{4}[/latex] mV²
Case II, E₂ = [latex]\frac{1}{2}[/latex] mV²
Case III, E₃ = – (3m) (V/3)² = [latex]\frac{1}{6}[/latex]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 = [latex]\frac{1}{2}[/latex] mv²
= [latex]\frac{95}{100}[/latex]mgh
V = [latex]\sqrt{\frac{95}{100} \times 2 \mathrm{gh}}=\sqrt{\frac{19}{20} \times 2 \times 9.8 \times 1.5}[/latex]
= 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. = [latex]\frac{1}{2}[/latex] m
(v₂² – v₁²), W = [latex]\frac{1}{2}[/latex] × 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 = [latex]\frac{1}{2}[/latex] mv² = [latex]\frac{1}{2}[/latex] (AVρt)
v² = [latex]\frac{1}{2}[/latex] av³ρt.

c) Electrical energy produced = [latex]\frac{25}{100}[/latex]
K.E of air = [latex]\frac{1}{4}[/latex] × [latex]\frac{1}{2}[/latex] × Av³ρt
Electrical power = [latex]\frac{1}{8} \frac{A v^3 \rho t}{t}=\frac{1}{8} A v^3 \rho t[/latex]
P = [latex]\frac{1}{8}[/latex] × 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 × [latex]\frac{20}{100}[/latex] = 0.76 × 10⁷ J/kg
∴ Fat used up by the dieter
= [latex]\frac{1 \mathrm{~kg}}{0.76 \times 10^7}[/latex] = 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 = [latex]\frac{20}{100}[/latex]
(200A) = 8KW = 40A = 8000 (watt)
Therefore, A = [latex]\frac{8000}{40}[/latex] = 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 = [latex]\frac{m_1 u_1}{m_1+m_2}=\frac{0.012 \times 70}{0.012+0.4}=\frac{0.84}{0.412}[/latex]
= 2.04 m/s.
Let the block rise to a height h.
RE of combination = K.E. of the combination
(m₁ + m₂)gh = [latex]\frac{1}{2}[/latex] (m₁ + m₂)v²
h = [latex]\frac{v^2}{2 g}=\frac{2.04 \times 2.04}{2 \times 9.8}[/latex] = 0.212m.
For calculating heat produced. We calculate energy lost (W). where
W = intial K.E of bullet – final K.E of combination
= [latex]\frac{1}{2}[/latex] m₁u₁² [latex]\frac{1}{2}[/latex] (m₁ + m₂)v²
= [latex]\frac{1}{2}[/latex] × 0.012 (70)² – [latex]\frac{1}{2}[/latex] (0.412) (2.04)²
W = 29.4 – 0.86 = 28.54 Joule.
∴ heat produd. H = [latex]\frac{W}{j}=\frac{28.54}{4.2}[/latex] = 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 = [latex]\frac{1}{2}[/latex] 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 = [latex]\frac{2280}{220}[/latex] = 10.36 ms^-1
Time taken by the child to run a distance of 10m over the trolley, t = [latex]\frac{10 \mathrm{~m}}{4 \mathrm{~ms}^{-1}}[/latex] = 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) ∝ [latex]\frac{\mathrm{1}}{\mathrm{r}}[/latex].
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 = [latex](3 \bar{i}+4 \bar{J}-5 \bar{k})[/latex] unit and displacement d = [latex](3 \bar{i}+4 \bar{J}-5 \bar{k})[/latex] 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 = [latex]\frac{16}{\sqrt{50} \sqrt{50}}=\frac{16}{50}[/latex] = 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 = [latex]\frac{1}{2}[/latex]mv² – 0
= [latex]\frac{1}{2}[/latex] × 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,
[latex]\frac{1}{2}[/latex] mv²_f = 100 J
v_f = [latex]\frac{\sqrt{2 \times 100 \mathrm{~J}}}{0.05 \mathrm{~kg}}[/latex] = 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 + [latex]\frac{1}{2}[/latex](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 = [latex]\frac{1}{2} {\mathrm{mv}_0^2}[/latex] ………………… (1)
T_A – mg = [latex]\frac{\mathrm{mv}_0^2}{\mathrm{~L}}[/latex] [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 = – [latex]\frac{1}{2} {\mathrm{mv}_c^2}[/latex] + 2mgL ………………… (2) .
mg = [latex]\frac{\mathrm{mv}_c^2}{\mathrm{~L}}[/latex] [Newtons Second law] ……………… (3)
where v_c is the speed at C. From equations (2) & (3)
E = [latex]\frac{5}{2}[/latex]mgL
Equating this to the energy at A
[latex]\frac{5}{2}[/latex] mgL = [latex]\frac{m}{2} v_0^2[/latex] or, v₀ = [latex]\sqrt{5 \mathrm{gL}}[/latex]

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

(iii) The ratio of the kinetic energies at Band C is :
[latex]\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}[/latex]
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 = [latex]\frac{1}{2}[/latex]mv² = [latex]\frac{1}{2}[/latex] × 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 = [latex]\frac{1}{2}[/latex] 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 – [latex]\frac{1}{2}[/latex]mv²
The work done by the net force is
W = [latex]\frac{1}{2}[/latex]Kx_m² – μm g x_m
Equating we have
[latex]\frac{1}{2}[/latex]mv² = [latex]\frac{1}{2}[/latex]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 = [latex]\frac{-\mu m g+\left[\mu^2 m^2 g^2+m k v^2\right]^{1 / 2}}{k}[/latex]
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 [latex]\frac{10^{-20} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}[/latex] ; 0.06 eV
Note 0.1 eV = 100 meV (100 millielectron volt).

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

(c) The average human consumption in a day
is [latex]\frac{10^7 \mathrm{~J}}{4.2 \times 10^3 \mathrm{~J} / \mathrm{kcal}}[/latex]; 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 [latex]{ }_{92}^{235} \mathrm{U}[/latex] 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 = [latex]\frac{1}{2}[/latex] m₁ v_1i²
While its final kinetic energy from
V_1f = [latex]\frac{\left(m_1-m_2\right)}{m_1+m_2}[/latex] v_1i²
K_1f = [latex]\frac{1}{2}[/latex]m₁ v₁²_f = [latex]\frac{1}{2}[/latex]m₁ [latex]\left(\frac{m_1-m_2}{m_1+m_2}\right)^2[/latex] v_1i²
The fractional kinetic energy lost is
f₁ = [latex]\frac{K_{1 f}}{k_{1i}}\left(\frac{m_1-m_2}{m_1+m_2}\right)^2[/latex]
while the fractional kinetic energy gained by the moderating nuclei K_2f/ K_1i is f₂ = 1 – f₁
elastic collision = [latex]\frac{4 m_1 m_2}{\left(m_1+m_2\right)^2}[/latex]
One canalso verify this result by substituting from
v_2f = [latex]\frac{2 m_1 v_1i}{m_1+m_2}[/latex]
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 = [latex]\frac{\text { Force }}{\text { Mass }}[/latex]
∴ 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 = [latex]\frac{\mathrm{mu}}{\mathrm{M}}[/latex]
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 = [latex]\frac{\mathrm{mu}}{\mathrm{M}}[/latex]
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 = [latex]m \frac{(v-u)}{t} t[/latex]
= (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
[latex]\frac{\mathrm{dp}}{\mathrm{dx}}[/latex] ∝ F, where F = external force
(or)F = K[latex]\frac{\mathrm{dp}}{\mathrm{dx}}[/latex] ………………. (2)
From equations (1) and (2) we have
F = [latex]K \frac{d(m v)}{d t}=K \cdot m \frac{d v}{d t}[/latex]
= Kma …………………. (3)
Since the rate of change of velocity [latex]\frac{\mathrm{dv}}{\mathrm{dx}}[/latex] 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 = [latex]\frac{\mathrm{dp}}{\mathrm{dx}}[/latex] = 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 [latex]\frac{(2)}{(1)} \Rightarrow \frac{f_s}{N}=\frac{m g \sin \theta}{m g{con} \theta}[/latex]
= [latex]\frac{f_s}{N}[/latex] = 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 = [latex]\frac{F}{m}=\frac{30}{4}[/latex] = 7.5 ms^-2
µ = [latex]\frac{\mathrm{a}}{\mathrm{g}}=\frac{7.5}{10}=\frac{3}{4}[/latex]
∴ Contact force = Frictional force
= µ mg
= [latex]\frac{3}{4}[/latex] × 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 = [latex]\frac{\mathrm{dp}}{\mathrm{dt}}[/latex] = [latex]\frac{\mathrm{d}}{\mathrm{dt}}[/latex](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 = [latex]m \frac{(v-u)}{t} \Rightarrow 5=\frac{10 \times 2}{t}[/latex]
∴ 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 = [latex]\sqrt{2 g h}[/latex]

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

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 [latex]\left(\frac{v-u}{t}\right)[/latex] = 3[latex]\left(\frac{3.5-2}{25}\right)[/latex]
= [latex]\frac{3 \times 15}{25}=\frac{0.9}{5}[/latex] = 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 = [latex]\frac{g}{3}[/latex]
Apparent weight W = m(g + a)
= m(g + [latex]\frac{g}{3}[/latex]) = [latex]\frac{4mg}{3}[/latex]
⇒ mg = [latex]\frac{3 W}{4}[/latex] …………….. (1)
When the lift is moving down, a = [latex]\frac{g}{2}[/latex]
Apparent weight W¹ = m(g – a)
= m(g – [latex]\frac{g}{2}[/latex]) = [latex]\frac{mg}{2}[/latex]
∴ |W¹| = [latex]\frac{\left[\frac{3 W}{4}\right]}{2}=\frac{3 W}{8}[/latex]

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 = [latex]\frac{a}{g}=\frac{1.5}{9.8}[/latex] = 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 + [latex]\frac{1}{2}[/latex] gt²
= 10 × 2 × [latex]\frac{1}{2}[/latex] × 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 + [latex]\frac{1}{2}[/latex] gt²
= 10 × 2 × [latex]\frac{1}{2}[/latex] × 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 = [latex]\left(\frac{\dot{m}_1-m_2}{m_1+m_2}\right) g=\left(\frac{6-4}{6+4}\right) \times 10[/latex] = 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 [latex]\frac{\sqrt{3}}{2}[/latex].
(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 = [latex]\frac{\sqrt{3}}{2}[/latex]
i) F_down = mg (sinθ – µ cosθ)
= 2 × 9.8 (sin30° – [latex]\frac{\sqrt{3}}{2}[/latex]cos30°)
= 2 × 98 [[latex]\frac{1}{2}[/latex] – [latex]\frac{\sqrt{3}}{2}[/latex] × [latex]\frac{\sqrt{3}}{2}[/latex]] = 49 N

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

Question 10.
A block is placed on a ramp of parabolic shape given by the equation g = [latex]\frac{x^2}{20}[/latex], 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 = [latex]\frac{\mathrm{d} y}{\mathrm{~d} x}[/latex])

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 = [latex]\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[/latex]
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 = [latex]\frac{\mu^2}{2 \mu g}=\frac{0.4 \times 0.4}{2 \times 0.2 \times 9.8}[/latex] = 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 = [latex]\frac{F}{\left(m_A+m_B\right)}[/latex]
= [latex]\frac{30}{10+5}[/latex] = 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 – [latex]\frac{\mathrm{m} v^2}{l}[/latex],
iii) T + [latex]\frac{\mathrm{m} v^2}{l}[/latex],
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 = [latex]\frac{F}{m}=\frac{-50}{20}[/latex] = -2.5 m/s²
From v = u + at
0 = 15 – 2.5t
t = [latex]\frac{15}{2.5}[/latex] = 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 = [latex]\frac{m(v-u)}{t}=\frac{3.0(3.5-2.0)}{25}[/latex]
= 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 = [latex]\frac{F}{m}=\frac{10}{5}[/latex] = 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 = [latex]\frac{m(v-u)}{t}=\frac{465(0-10)}{4}[/latex] = -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 = [latex]\frac{F}{m}=\frac{-8.0}{0.40}[/latex] = -20 m/s²
for 0 ≤ t ≤ 30s.
i) At t = -5s, x = Ut = 10 × (-5) = -50 m

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

iii) At t = 100s, The problem is divided into two parts, upto 30s, there is force/acc.
∴ from x₁ = Ut + [latex]\frac{1}{2}[/latex] at²
= 10 × 30 + [latex]\frac{1}{2}[/latex] (-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 = [latex]\sqrt{v_x^2+v_y^2}=\sqrt{20^2+(9.8)^2}[/latex]
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 = [latex]\frac{v_y}{v_x}=\frac{9.8}{20}[/latex] = 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 = [latex]\frac{3}{4}[/latex]
= 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 = [latex]\frac{F}{m_1+m_2}=\frac{500}{10+20}=\frac{50}{3}[/latex] m/s²
a) When force is applied on heavier block,
T = m₁ a = 10 × [latex]\frac{50}{3}[/latex] N
T = 166.66 N

b) When force is applied on lighter block,
T = m₂a = 20 × [latex]\frac{50}{3}[/latex] 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 = [latex]\frac{\left(m_1-m_2\right) g}{m_1+m_2}[/latex]
a = [latex]\frac{(12-8) 9.8}{12+8}=\frac{39.2}{20}[/latex]
= 1.96 m/s²
Again
T = [latex]\frac{2 m_1 m_2 g}{m_1+m_2}=\frac{2 \times 12 \times 8 \times 9.8}{(12+8)}[/latex] = 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 [latex]\vec{v}_1, \vec{v}_2[/latex] be their respective velocities. Therefore total linear momentum after disintegration = [latex]m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}[/latex]. 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.
[latex]m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}[/latex] (Or) [latex]\vec{v}_2=\frac{-m_1 \vec{v}_1}{m_2}[/latex]
Negative sign shows that [latex]\overrightarrow{v_1}[/latex] and [latex]\overrightarrow{v_2}[/latex] 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 = [latex]\frac{-\mathrm{mv}}{\mathrm{M}}=\frac{-0.02 \times 80}{100}[/latex] = 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 = [latex]\frac{40}{60}[/latex], rps = [latex]\frac{2}{3}[/latex], T = ?
T = mrw² = mr(2 π)² = 4 π²m²
T = 4 × [latex]\left(\frac{22}{7}\right)^2[/latex] × 0.25 × 1.5 × [latex]\left(\frac{2}{3}\right)^2[/latex] = 6.6N
If T_max = 200 N. then from
T_max = [latex]\frac{\mathrm{mv}_{\max }^2}{r}[/latex]
[latex]v_{\max }^2=\frac{T_{\max } \times r}{m}=\frac{200 \times 1.5}{0.25}[/latex] = 1200
v_max = [latex]\sqrt{1200}[/latex] = 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 = [latex]\frac{(2-0) \mathrm{cm}}{(2-0) \mathrm{s}}[/latex] = 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 = [latex]\frac{\text { Change in linear momentum }}{\text { Time }}[/latex]
∴ F = [latex]\frac{m \times v}{t}=\frac{150 \times 15}{1}[/latex] = 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 =[latex]\frac{720 \times 1000}{60 \times 60}[/latex] = 200 ms^-1
g = 9.8 ms^-2
From tan θ = [latex]\frac{\mathrm{v}^2}{\mathrm{rg}}[/latex]
v² = rg tan θ
r = [latex]\frac{v^2}{g \tan \theta}=\frac{(200)^2}{9.8 \times \tan 15^{\circ}}[/latex]
= 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 =[latex]\frac{54 \times 1000}{60 \times 60}[/latex] = 15 m/s
g = 9.8 ms^-2
As tan θ = [latex]\frac{v^2}{r g}=\frac{15 \times 15}{30 \times 9.8}[/latex] = 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 = [latex]\frac{F^1}{m_1+m_2}=\frac{177.95}{5+10}[/latex]
= 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 = [latex]\frac{p}{m}=\frac{21.2}{40}[/latex] = 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 + [latex]\frac{1}{2}[/latex] at²
5 = 0 × t + [latex]\frac{1}{2}[/latex] × 0.53 t²
t = [latex]\frac{\sqrt{5 \times 2}}{0.53}[/latex] = 4.345
If the truck travels a distance x during this time, then again from
S = ut + [latex]\frac{1}{2}[/latex] at²
x = 0 × 4.34 + [latex]\frac{1}{2}[/latex] × 2(4.34)² = 18.84 m

Question 37.
A disc revolves with a speed of 33[latex]\frac{1}{3}[/latex] 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 [latex]\frac{m v^2}{r}[/latex] 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 = [latex]\frac{4}{100}[/latex] m
n = 33 [latex]\frac{1}{2}[/latex] rev/min = [latex]\frac{100}{3 \times 60}[/latex] rev/sec
w = 2πn = 2π × [latex]\frac{100}{180}[/latex] = 3.49 S^-1
∴ rw² = [latex]\frac{4}{100}[/latex] × (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 = [latex]\frac{m v^2}{r}[/latex],
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 = [latex]\frac{m v^2}{r}[/latex] or
v = [latex]\sqrt{\mathrm{rg}}=\sqrt{25 \times 10}[/latex]
= 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 = [latex]\frac{200}{60}[/latex] 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 = [latex]\sqrt{\frac{g}{\mu \mathrm{r}}}=\sqrt{\frac{10}{0.15 \times 3}}[/latex]
= 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 ≤ [latex]\sqrt{g / R}[/latex] . 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 = [latex]\frac{\mathrm{g}}{\mathrm{R} \omega^2}[/latex] ……………….. (3)
As |cos θ| ≤ 1, therefore, bead will remain at its lower most point for
[latex]\frac{\mathrm{g}}{\mathrm{R} \omega^2}[/latex] ≤ 1 or w ≤ [latex]\sqrt{\frac{g}{R}}[/latex]
When ω = [latex]\sqrt{\frac{2 g}{R}}[/latex], from (iii),
cos θ = [latex]\frac{g}{R}\left(\frac{R}{2 g}\right)=\frac{1}{2}[/latex]
∴ 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 = [latex]\frac{-u^2}{2 s}=\frac{-90 \times 90}{2 \times 0.6}[/latex] 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 + [latex]\frac{1}{2}[/latex]at². Find the force acting on the particle.
Answer:
We know, y = ut + [latex]\frac{1}{2}[/latex] gt²
Now, υ = [latex]\frac{\mathrm{dy}}{\mathrm{dt}}[/latex]u + gt
acceleration, a = [latex]\frac{\mathrm{dv}}{\mathrm{dt}}[/latex] = g
Then the force is given by F = [latex]\frac{\mathrm{dp}}{\mathrm{dt}}[/latex] = 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°) = [latex]\frac{2}{\sqrt{3}}[/latex] ≈ 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 = [latex]\frac{5}{6}[/latex] or
θ = tan^-1 ([latex]\frac{5}{6}[/latex]) = 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 = [latex]\frac{22}{23}[/latex] 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 = [latex]\sqrt{\mu_s R_g}[/latex] 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[latex]\left(\frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)^{1 / 2}[/latex] = 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.