Chapter 06: Work, Energy and Power
Physics XI : Complete NCERT Exercise Solutions
Practice these examination-oriented questions to master the concepts of rectilinear motion, velocity, acceleration, and kinematic equations. Each question includes a detailed solution to enhance your understanding.
Question 6.1
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:
Answer & Explanation:
(a) Positive – Force applied by man is upward and displacement is upward.
(b) Negative – Gravitational force acts downward, displacement upward.
(c) Negative – Friction opposes motion, so force opposite to displacement.
(d) Positive – Applied force is in direction of displacement.
(e) Negative – Resistive force opposes motion, reducing kinetic energy.
(b) Negative – Gravitational force acts downward, displacement upward.
(c) Negative – Friction opposes motion, so force opposite to displacement.
(d) Positive – Applied force is in direction of displacement.
(e) Negative – Resistive force opposes motion, reducing kinetic energy.
Question 6.2
A body of mass 2 kg 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
Answer & Explanation:
Given: m = 2 kg, u = 0, F_app = 7 N, μ = 0.1, t = 10 s
Step 1: Acceleration
Friction f = μ mg = 0.1 × 2 × 9.8 ≈ 1.96 N
Net force F_net = 7 − 1.96 = 5.04 N
a = F_net / m = 5.04 / 2 = 2.52 m/s²
Step 2: Displacement in 10 s
s = ut + ½ at² = 0 + 0.5 × 2.52 × 100 = 126 m
(a) Work by applied force: W_app = F_app × s = 7 × 126 = 882 J
(b) Work by friction: W_fric = −f × s = −1.96 × 126 ≈ −247 J
(c) Work by net force: W_net = F_net × s = 5.04 × 126 ≈ 635 J
(d) Change in KE: ΔK = W_net = 635 J (by Work-Energy Theorem)
Interpretation: The net work equals change in kinetic energy.
Step 1: Acceleration
Friction f = μ mg = 0.1 × 2 × 9.8 ≈ 1.96 N
Net force F_net = 7 − 1.96 = 5.04 N
a = F_net / m = 5.04 / 2 = 2.52 m/s²
Step 2: Displacement in 10 s
s = ut + ½ at² = 0 + 0.5 × 2.52 × 100 = 126 m
(a) Work by applied force: W_app = F_app × s = 7 × 126 = 882 J
(b) Work by friction: W_fric = −f × s = −1.96 × 126 ≈ −247 J
(c) Work by net force: W_net = F_net × s = 5.04 × 126 ≈ 635 J
(d) Change in KE: ΔK = W_net = 635 J (by Work-Energy Theorem)
Interpretation: The net work equals change in kinetic energy.
Question 6.3
Given in Fig. 6.11 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.
Answer & Explanation:
For Fig. 6.11(a):
Particle cannot be found where total energy E < V(x). That is in regions where potential curve is above E.
Minimum total energy = lowest point of potential well.
For Fig. 6.11(b):
Particle confined between turning points where V(x) = E.
Minimum energy = minimum of V(x).
For Fig. 6.11(c):
Particle cannot exist where V(x) > E.
Minimum energy = 0 (if well is symmetric).
For Fig. 6.11(d):
Particle free in region where E > V(x), forbidden where E < V(x).
Minimum energy = min(V(x)).
Particle cannot be found where total energy E < V(x). That is in regions where potential curve is above E.
Minimum total energy = lowest point of potential well.
For Fig. 6.11(b):
Particle confined between turning points where V(x) = E.
Minimum energy = minimum of V(x).
For Fig. 6.11(c):
Particle cannot exist where V(x) > E.
Minimum energy = 0 (if well is symmetric).
For Fig. 6.11(d):
Particle free in region where E > V(x), forbidden where E < V(x).
Minimum energy = min(V(x)).
Question 6.4
The potential energy function for a particle executing linear simple harmonic motion is given by \( V(x) = kx^2/2 \), 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. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
Answer & Explanation:
Total energy E = 1 J, k = 0.5 N/m.
At turning point, kinetic energy = 0, so E = V(x).
V(x) = ½ k x² = ½ × 0.5 × x² = 0.25 x²
Set E = V(x): 1 = 0.25 x² → x² = 4 → x = ±2 m.
Thus, particle turns back at x = ±2 m.
At turning point, kinetic energy = 0, so E = V(x).
V(x) = ½ k x² = ½ × 0.5 × x² = 0.25 x²
Set E = V(x): 1 = 0.25 x² → x² = 4 → x = ±2 m.
Thus, particle turns back at x = ±2 m.
Question 6.5
Answer the following :
Answer & Explanation:
(a) At the expense of the rocket’s kinetic energy (converted to heat due to friction).
(b) Gravitational force is conservative; work over closed path is zero.
(c) As satellite loses energy, orbit shrinks; by conservation of angular momentum, speed increases.
(d) In case (i), force is vertical but displacement horizontal → work = 0. In case (ii), force does work against gravity → work = mgh. So work is greater in (ii).
(b) Gravitational force is conservative; work over closed path is zero.
(c) As satellite loses energy, orbit shrinks; by conservation of angular momentum, speed increases.
(d) In case (i), force is vertical but displacement horizontal → work = 0. In case (ii), force does work against gravity → work = mgh. So work is greater in (ii).
Question 6.6
Underline the correct alternative :
Answer & Explanation:
(a) decreases (W = −ΔV, positive W means ΔV negative)
(b) kinetic energy (work against friction reduces KE)
(c) external force (Newton's second law for system)
(d) total linear momentum (momentum conserved in all collisions)
(b) kinetic energy (work against friction reduces KE)
(c) external force (Newton's second law for system)
(d) total linear momentum (momentum conserved in all collisions)
Question 6.7
State if each of the following statements is true or false. Give reasons for your answer.
Answer & Explanation:
(a) False – Total momentum and total energy conserved, not individually.
(b) True – Law of conservation of energy holds universally.
(c) False – True only for conservative forces.
(d) True – In inelastic collision, KE decreases (converted to other forms).
(b) True – Law of conservation of energy holds universally.
(c) False – True only for conservative forces.
(d) True – In inelastic collision, KE decreases (converted to other forms).
Question 6.8
Answer carefully, with reasons :
Answer & Explanation:
(a) No – During contact, KE is partly converted to elastic PE.
(b) Yes – Momentum conserved at every instant (no external force).
(c) Same: momentum conserved, KE not conserved during contact.
(d) Elastic – If force is conservative (depends only on separation), mechanical energy conserved → elastic collision.
(b) Yes – Momentum conserved at every instant (no external force).
(c) Same: momentum conserved, KE not conserved during contact.
(d) Elastic – If force is conservative (depends only on separation), mechanical energy conserved → elastic collision.
Question 6.9
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
Answer & Explanation:
Let a = constant, u = 0 → v = at, F = ma.
Power P = F v = ma × at = ma²t.
Thus P ∝ t → Option (ii) is correct.
Power P = F v = ma × at = ma²t.
Thus P ∝ t → Option (ii) is correct.
Question 6.10
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to
Answer & Explanation:
Constant power P = F v = m a v = constant.
Also a = dv/dt. Solving gives v ∝ t¹/², then s = ∫ v dt ∝ t³/².
So displacement ∝ t³/² → Option (iii) is correct.
Also a = dv/dt. Solving gives v ∝ t¹/², then s = ∫ v dt ∝ t³/².
So displacement ∝ t³/² → Option (iii) is correct.
Question 6.11
A body constrained to move along the z-axis of a coordinate system is subject to a constant force \( \mathbf{F} \) given by
\( \mathbf{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \, \text{N} \)
where \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
\( \mathbf{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \, \text{N} \)
where \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Answer & Explanation:
Displacement \( \mathbf{d} = 4\hat{k} \, \text{m} \) (along z-axis).
Force \( \mathbf{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \).
Work done \( W = \mathbf{F} \cdot \mathbf{d} \)
= \( (-\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (4\hat{k}) \)
= \( 3 \times 4 = 12 \, \text{J} \) (only k-component contributes).
Answer: 12 J
Force \( \mathbf{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \).
Work done \( W = \mathbf{F} \cdot \mathbf{d} \)
= \( (-\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (4\hat{k}) \)
= \( 3 \times 4 = 12 \, \text{J} \) (only k-component contributes).
Answer: 12 J
Question 6.12
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 & Explanation:
K = ½ mv² → v = √(2K/m).
For electron: K₁ = 10 keV = 10 × 10³ × 1.6×10⁻¹⁹ = 1.6×10⁻¹⁵ J
v_e = √(2 × 1.6×10⁻¹⁵ / 9.11×10⁻³¹) ≈ √(3.512×10¹⁵) ≈ 5.93×10⁷ m/s.
For proton: K₂ = 100 keV = 1.6×10⁻¹⁴ J
v_p = √(2 × 1.6×10⁻¹⁴ / 1.67×10⁻²⁷) ≈ √(1.916×10¹³) ≈ 4.38×10⁶ m/s.
Electron is faster.
Ratio v_e / v_p ≈ (5.93×10⁷) / (4.38×10⁶) ≈ 13.54.
For electron: K₁ = 10 keV = 10 × 10³ × 1.6×10⁻¹⁹ = 1.6×10⁻¹⁵ J
v_e = √(2 × 1.6×10⁻¹⁵ / 9.11×10⁻³¹) ≈ √(3.512×10¹⁵) ≈ 5.93×10⁷ m/s.
For proton: K₂ = 100 keV = 1.6×10⁻¹⁴ J
v_p = √(2 × 1.6×10⁻¹⁴ / 1.67×10⁻²⁷) ≈ √(1.916×10¹³) ≈ 4.38×10⁶ m/s.
Electron is faster.
Ratio v_e / v_p ≈ (5.93×10⁷) / (4.38×10⁶) ≈ 13.54.
Question 6.13
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration 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 gravitational 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 m/s?
Answer & Explanation:
Radius r = 2 mm = 2×10⁻³ m.
Volume V = 4/3 π r³ ≈ 3.35×10⁻⁸ m³.
Density of water ρ = 1000 kg/m³ → mass m = ρV ≈ 3.35×10⁻⁵ kg.
First half (250 m): W_g = mgh₁ = 3.35×10⁻⁵ × 9.8 × 250 ≈ 0.082 J.
Second half (250 m): W_g = mgh₂ = same = 0.082 J.
Total work by gravity = mgH = 3.35×10⁻⁵ × 9.8 × 500 ≈ 0.164 J.
Final KE = ½ m v² = 0.5 × 3.35×10⁻⁵ × 100 = 1.675×10⁻³ J.
By work-energy theorem:
W_gravity + W_resistive = ΔK = KE_f − 0
W_resistive = KE_f − W_gravity = 1.675×10⁻³ − 0.164 ≈ −0.162 J (negative).
Volume V = 4/3 π r³ ≈ 3.35×10⁻⁸ m³.
Density of water ρ = 1000 kg/m³ → mass m = ρV ≈ 3.35×10⁻⁵ kg.
First half (250 m): W_g = mgh₁ = 3.35×10⁻⁵ × 9.8 × 250 ≈ 0.082 J.
Second half (250 m): W_g = mgh₂ = same = 0.082 J.
Total work by gravity = mgH = 3.35×10⁻⁵ × 9.8 × 500 ≈ 0.164 J.
Final KE = ½ m v² = 0.5 × 3.35×10⁻⁵ × 100 = 1.675×10⁻³ J.
By work-energy theorem:
W_gravity + W_resistive = ΔK = KE_f − 0
W_resistive = KE_f − W_gravity = 1.675×10⁻³ − 0.164 ≈ −0.162 J (negative).
Question 6.14
A molecule in a gas container hits a horizontal wall with speed 200 m/s 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 & Explanation:
Momentum: For the molecule–wall system, horizontal momentum is conserved if wall is fixed and no external horizontal force.
But vertical momentum changes sign (normal direction). For wall+molecule system, total momentum conserved (wall gains negligible momentum).
Elasticity: Since speed is same before and after, kinetic energy conserved → elastic collision.
But vertical momentum changes sign (normal direction). For wall+molecule system, total momentum conserved (wall gains negligible momentum).
Elasticity: Since speed is same before and after, kinetic energy conserved → elastic collision.
Question 6.15
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 & Explanation:
Volume V = 30 m³, density ρ = 1000 kg/m³ → mass m = 30000 kg.
Height h = 40 m, time t = 15 min = 900 s.
Useful work output = mgh = 30000 × 9.8 × 40 = 1.176×10⁷ J.
Useful power output = Work / t = 1.176×10⁷ / 900 ≈ 13066.67 W.
Efficiency η = 0.3 = (Useful power output) / (Electric power input).
Electric power input = 13066.67 / 0.3 ≈ 43555.56 W ≈ 43.56 kW.
Height h = 40 m, time t = 15 min = 900 s.
Useful work output = mgh = 30000 × 9.8 × 40 = 1.176×10⁷ J.
Useful power output = Work / t = 1.176×10⁷ / 900 ≈ 13066.67 W.
Efficiency η = 0.3 = (Useful power output) / (Electric power input).
Electric power input = 13066.67 / 0.3 ≈ 43555.56 W ≈ 43.56 kW.
Question 6.16
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 is elastic, which of the following (Fig. 6.14) is a possible result after collision?
Answer & Explanation:
In elastic collision of equal masses:
When moving mass strikes stationary mass, they exchange velocities.
Here, incoming ball (mass m, speed V) hits first stationary ball.
After first collision: incoming ball stops, first ball moves with V.
Then first ball hits second stationary ball: first stops, second moves with V.
So final: Two balls stationary, one moves with V → Option (i) is correct.
When moving mass strikes stationary mass, they exchange velocities.
Here, incoming ball (mass m, speed V) hits first stationary ball.
After first collision: incoming ball stops, first ball moves with V.
Then first ball hits second stationary ball: first stops, second moves with V.
So final: Two balls stationary, one moves with V → Option (i) is correct.
Question 6.17
The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
Answer & Explanation:
For elastic collision between equal masses, velocities exchange.
Bob A comes down, gains speed v at bottom given by mgh = ½ mv².
Height h = L(1 − cos30°).
After collision, bob A comes to rest and bob B moves with speed v.
So bob A rises to zero height (does not rise after collision).
Bob A comes down, gains speed v at bottom given by mgh = ½ mv².
Height h = L(1 − cos30°).
After collision, bob A comes to rest and bob B moves with speed v.
So bob A rises to zero height (does not rise after collision).
Question 6.18
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 & Explanation:
Length L = 1.5 m.
Initial PE = mgh = mgL (since horizontal).
5% dissipated → 95% remains as KE at bottom.
0.95 mgL = ½ mv²
v² = 2 × 0.95 × gL = 1.9 × 9.8 × 1.5 ≈ 27.93
v ≈ 5.29 m/s.
Initial PE = mgh = mgL (since horizontal).
5% dissipated → 95% remains as KE at bottom.
0.95 mgL = ½ mv²
v² = 2 × 0.95 × gL = 1.9 × 9.8 × 1.5 ≈ 27.93
v ≈ 5.29 m/s.
Question 6.19
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on 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. What is the speed of the trolley after the entire sand bag is empty?
Answer & Explanation:
Initial total mass M = 300 + 25 = 325 kg.
Speed u = 27 km/h = 7.5 m/s.
Momentum = M u = 325 × 7.5 = 2437.5 kg·m/s.
Sand leaks vertically downward → no horizontal force → horizontal momentum conserved.
After sand gone, mass = 300 kg, momentum same.
v = Momentum / mass = 2437.5 / 300 = 8.125 m/s.
Speed increases because mass decreases.
Speed u = 27 km/h = 7.5 m/s.
Momentum = M u = 325 × 7.5 = 2437.5 kg·m/s.
Sand leaks vertically downward → no horizontal force → horizontal momentum conserved.
After sand gone, mass = 300 kg, momentum same.
v = Momentum / mass = 2437.5 / 300 = 8.125 m/s.
Speed increases because mass decreases.
Question 6.20
A body of mass 0.5 kg travels in a straight line with velocity \( v = a x^{3/2} \) where \( a = 5 \, \text{m}^{-1/2} \text{s}^{-1} \). What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Answer & Explanation:
Given v = a x^{3/2}, a = 5.
At x = 0, v = 0; at x = 2, v = 5 × (2)^{3/2} = 5 × 2.828 ≈ 14.14 m/s.
Work done by net force = Change in KE.
ΔK = ½ m (v_f² − v_i²) = 0.5 × 0.5 × (14.14² − 0) ≈ 0.25 × 200 = 50 J.
At x = 0, v = 0; at x = 2, v = 5 × (2)^{3/2} = 5 × 2.828 ≈ 14.14 m/s.
Work done by net force = Change in KE.
ΔK = ½ m (v_f² − v_i²) = 0.5 × 0.5 × (14.14² − 0) ≈ 0.25 × 200 = 50 J.
Question 6.21
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/m³. What is the electrical power produced?
Answer & Explanation:
(a) Volume per second = A v, mass per second = ρ A v.
In time t, mass m = ρ A v t.
(b) KE = ½ m v² = ½ ρ A v t × v² = ½ ρ A v³ t.
(c) A = 30 m², v = 36 km/h = 10 m/s, ρ = 1.2 kg/m³.
Wind power = KE per second = ½ ρ A v³ = 0.5 × 1.2 × 30 × 1000 = 18000 W.
Electrical power = 25% of 18000 = 4500 W = 4.5 kW.
In time t, mass m = ρ A v t.
(b) KE = ½ m v² = ½ ρ A v t × v² = ½ ρ A v³ t.
(c) A = 30 m², v = 36 km/h = 10 m/s, ρ = 1.2 kg/m³.
Wind power = KE per second = ½ ρ A v³ = 0.5 × 1.2 × 30 × 1000 = 18000 W.
Electrical power = 25% of 18000 = 4500 W = 4.5 kW.
Question 6.22
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each 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 × 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Answer & Explanation:
m = 10 kg, h = 0.5 m, n = 1000.
(a) Work per lift = mgh = 10 × 9.8 × 0.5 = 49 J.
Total work = 49 × 1000 = 49000 J.
(b) Fat energy per kg = 3.8×10⁷ J, efficiency 20% → usable = 0.2 × 3.8×10⁷ = 7.6×10⁶ J per kg.
Fat used = Total work / usable energy per kg = 49000 / 7.6×10⁶ ≈ 0.00645 kg = 6.45 g.
(a) Work per lift = mgh = 10 × 9.8 × 0.5 = 49 J.
Total work = 49 × 1000 = 49000 J.
(b) Fat energy per kg = 3.8×10⁷ J, efficiency 20% → usable = 0.2 × 3.8×10⁷ = 7.6×10⁶ J per kg.
Fat used = Total work / usable energy per kg = 49000 / 7.6×10⁶ ≈ 0.00645 kg = 6.45 g.
Question 6.23
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 & Explanation:
(a) Required power P = 8000 W.
Solar power per m² = 200 W, efficiency η = 0.2 → useful power per m² = 40 W.
Area needed = 8000 / 40 = 200 m².
(b) Typical house roof area ≈ 100–150 m². So 200 m² is larger than typical roof.
Solar power per m² = 200 W, efficiency η = 0.2 → useful power per m² = 40 W.
Area needed = 8000 / 40 = 200 m².
(b) Typical house roof area ≈ 100–150 m². So 200 m² is larger than typical roof.
Question 6.24
A bullet of mass 0.012 kg and horizontal speed 70 m/s 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 & Explanation:
Step 1: Conservation of momentum
m₁ = 0.012 kg, u₁ = 70 m/s, m₂ = 0.4 kg, u₂ = 0.
After collision, combined mass M = 0.412 kg, let velocity = V.
m₁u₁ + m₂u₂ = M V → 0.012 × 70 = 0.412 V → V = 0.84 / 0.412 ≈ 2.04 m/s.
Step 2: Height risen
KE of combined mass converts to PE: ½ M V² = M g h.
h = V²/(2g) = (2.04²)/(2×9.8) ≈ 4.1616/19.6 ≈ 0.212 m.
Step 3: Heat produced
Initial KE = ½ m₁ u₁² = 0.5 × 0.012 × 4900 = 29.4 J.
Final KE = ½ M V² = 0.5 × 0.412 × 4.1616 ≈ 0.857 J.
Heat = Initial KE − Final KE ≈ 29.4 − 0.857 ≈ 28.54 J.
m₁ = 0.012 kg, u₁ = 70 m/s, m₂ = 0.4 kg, u₂ = 0.
After collision, combined mass M = 0.412 kg, let velocity = V.
m₁u₁ + m₂u₂ = M V → 0.012 × 70 = 0.412 V → V = 0.84 / 0.412 ≈ 2.04 m/s.
Step 2: Height risen
KE of combined mass converts to PE: ½ M V² = M g h.
h = V²/(2g) = (2.04²)/(2×9.8) ≈ 4.1616/19.6 ≈ 0.212 m.
Step 3: Heat produced
Initial KE = ½ m₁ u₁² = 0.5 × 0.012 × 4900 = 29.4 J.
Final KE = ½ M V² = 0.5 × 0.412 × 4.1616 ≈ 0.857 J.
Heat = Initial KE − Final KE ≈ 29.4 − 0.857 ≈ 28.54 J.
Question 6.25
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ₁ = 30°, θ₂ = 60°, and h = 10 m, what are the speeds and times taken by the two stones?
Answer & Explanation:
Same speed? Yes – by conservation of energy: mgh = ½ mv² → v = √(2gh), independent of angle.
v = √(2×9.8×10) = √196 ≈ 14 m/s.
Same time? No – acceleration along incline a = g sinθ, distance s = h/sinθ.
Time t = √(2s/a) = √(2h/(g sin²θ)).
For θ₁ = 30°: t₁ = √(2×10/(9.8×0.25)) = √(20/2.45) ≈ √8.16 ≈ 2.86 s.
For θ₂ = 60°: t₂ = √(2×10/(9.8×0.75)) = √(20/7.35) ≈ √2.72 ≈ 1.65 s.
Steeper slope gives shorter time.
v = √(2×9.8×10) = √196 ≈ 14 m/s.
Same time? No – acceleration along incline a = g sinθ, distance s = h/sinθ.
Time t = √(2s/a) = √(2h/(g sin²θ)).
For θ₁ = 30°: t₁ = √(2×10/(9.8×0.25)) = √(20/2.45) ≈ √8.16 ≈ 2.86 s.
For θ₂ = 60°: t₂ = √(2×10/(9.8×0.75)) = √(20/7.35) ≈ √2.72 ≈ 1.65 s.
Steeper slope gives shorter time.
Question 6.26
A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N/m as shown in Fig. 6.17. 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 & Explanation:
m = 1 kg, k = 100 N/m, x = 0.1 m, incline angle θ not given (assume θ).
Loss in PE = gain in spring PE + work against friction.
m g (x sinθ) = ½ k x² + μ m g cosθ × x.
If θ = 30° (typical problem):
1×9.8×0.1×0.5 = 0.5×100×0.01 + μ×1×9.8×0.866×0.1
0.49 = 0.5 + 0.849 μ → μ = (0.49 − 0.5)/0.849 ≈ −0.0012 → friction negligible (≈0).
For θ such that sinθ =? Need θ given; general formula: μ = (mgx sinθ − ½ kx²)/(mgx cosθ).
Loss in PE = gain in spring PE + work against friction.
m g (x sinθ) = ½ k x² + μ m g cosθ × x.
If θ = 30° (typical problem):
1×9.8×0.1×0.5 = 0.5×100×0.01 + μ×1×9.8×0.866×0.1
0.49 = 0.5 + 0.849 μ → μ = (0.49 − 0.5)/0.849 ≈ −0.0012 → friction negligible (≈0).
For θ such that sinθ =? Need θ given; general formula: μ = (mgx sinθ − ½ kx²)/(mgx cosθ).
Question 6.27
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m/s. 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 & Explanation:
Elevator moving down with constant speed → inertial frame relative to ground.
Bolt falls from ceiling to floor (3 m) in elevator frame: initial relative velocity = 0, acceleration = g relative to elevator.
Time to fall t = √(2h/g) = √(2×3/9.8) ≈ 0.782 s.
Velocity just before impact relative to elevator v = gt ≈ 7.66 m/s.
In elevator frame, bolt's KE relative to elevator = ½ m v² = 0.5×0.3×58.7 ≈ 8.8 J.
This KE converts to heat on impact.
If elevator stationary, same height drop → same KE relative to elevator → same heat produced.
Answer: ≈ 8.8 J, same in both cases.
Bolt falls from ceiling to floor (3 m) in elevator frame: initial relative velocity = 0, acceleration = g relative to elevator.
Time to fall t = √(2h/g) = √(2×3/9.8) ≈ 0.782 s.
Velocity just before impact relative to elevator v = gt ≈ 7.66 m/s.
In elevator frame, bolt's KE relative to elevator = ½ m v² = 0.5×0.3×58.7 ≈ 8.8 J.
This KE converts to heat on impact.
If elevator stationary, same height drop → same KE relative to elevator → same heat produced.
Answer: ≈ 8.8 J, same in both cases.
Question 6.28
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 4 m/s relative to the trolley in a direction opposite to its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?
Answer & Explanation:
Step 1: Initial momentum
M = 200 kg, m = 20 kg, u_trolley = 36 km/h = 10 m/s, u_child = 10 m/s (same as trolley initially).
Total initial momentum = (200+20)×10 = 2200 kg·m/s.
Step 2: Child runs opposite
Child's velocity relative to ground = 10 − 4 = 6 m/s (opposite direction).
Let final trolley speed = V.
Momentum conservation: 2200 = 200V + 20×6 → 200V = 2200 − 120 = 2080 → V = 10.4 m/s.
Step 3: Distance moved by trolley
Time for child to run 10 m relative to trolley at 4 m/s: t = 10/4 = 2.5 s.
Trolley's average speed during this time = (10 + 10.4)/2 = 10.2 m/s.
Distance moved by trolley = 10.2 × 2.5 = 25.5 m.
M = 200 kg, m = 20 kg, u_trolley = 36 km/h = 10 m/s, u_child = 10 m/s (same as trolley initially).
Total initial momentum = (200+20)×10 = 2200 kg·m/s.
Step 2: Child runs opposite
Child's velocity relative to ground = 10 − 4 = 6 m/s (opposite direction).
Let final trolley speed = V.
Momentum conservation: 2200 = 200V + 20×6 → 200V = 2200 − 120 = 2080 → V = 10.4 m/s.
Step 3: Distance moved by trolley
Time for child to run 10 m relative to trolley at 4 m/s: t = 10/4 = 2.5 s.
Trolley's average speed during this time = (10 + 10.4)/2 = 10.2 m/s.
Distance moved by trolley = 10.2 × 2.5 = 25.5 m.
Question 6.29
Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.
Answer & Explanation:
In elastic collision, total mechanical energy conserved, and force is repulsive when r < 2R (R = radius).
Potential energy V(r) should:
1. Increase as r decreases below 2R (repulsive).
2. Tend to infinity as r → 0 (hard sphere).
3. Be constant for r > 2R (no interaction).
Curves that decrease as r decreases (attractive) or have finite value at r=0 cannot describe billiard balls.
Typically, the curve with a sharp rise near r = 2R and flat beyond is correct.
The incorrect ones are those with wells (attractive) or finite at r=0.
Potential energy V(r) should:
1. Increase as r decreases below 2R (repulsive).
2. Tend to infinity as r → 0 (hard sphere).
3. Be constant for r > 2R (no interaction).
Curves that decrease as r decreases (attractive) or have finite value at r=0 cannot describe billiard balls.
Typically, the curve with a sharp rise near r = 2R and flat beyond is correct.
The incorrect ones are those with wells (attractive) or finite at r=0.
Question 6.30
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 β-decay of a neutron or a nucleus (Fig. 6.19).
Answer & Explanation:
In two-body decay, conservation of energy and momentum uniquely determine energies of products.
Let neutron mass M, proton mass m_p, electron mass m_e.
Initial momentum = 0 → proton and electron must move oppositely with equal momentum p.
Energy conservation: M c² = √(p²c² + m_p²c⁴) + √(p²c² + m_e²c⁴).
This equation fixes p, hence fixes electron energy uniquely.
But in observed β-decay, electron energy is continuous → implies a third particle (neutrino) carries away variable energy → three-body decay.
Let neutron mass M, proton mass m_p, electron mass m_e.
Initial momentum = 0 → proton and electron must move oppositely with equal momentum p.
Energy conservation: M c² = √(p²c² + m_p²c⁴) + √(p²c² + m_e²c⁴).
This equation fixes p, hence fixes electron energy uniquely.
But in observed β-decay, electron energy is continuous → implies a third particle (neutrino) carries away variable energy → three-body decay.
📘 Exam Preparation Tip:
These exercise questions will help you understand energy transformations and conservation principles. You'll learn to calculate work done by constant and variable forces, apply the work-energy theorem, solve kinetic and potential energy problems, analyze elastic and inelastic collisions, and calculate power in different physical situations. Vital for energy-related concepts in advanced physics.