Exercises
Q5.1. State whether the work done is positive or negative
(a) man lifting a bucket out of a well (b) gravitational force in the above case (c) friction on a body sliding down an incline (d) applied force on body moving on rough surface at uniform velocity (e) resistive force of air on a vibrating pendulum
Ans.
(a) Positive — force (upward) and displacement (upward) are in the same direction.
(b) Negative — gravitational force is downward but displacement is upward (opposite directions).
(c) Negative — friction acts up the incline (opposing motion), but body slides down; opposite directions.
(d) Positive — applied force is in the direction of motion (horizontal), even though uniform velocity means net work = 0, the applied force alone does positive work.
(e) Negative — resistive force opposes the motion of the pendulum at all times, so angle between force and displacement is always 180°.
2. A body of mass 2 kg starts from rest under a horizontal force of 7 N on a table (μ_k = 0.1). Compute: (a) work done by applied force in 10 s (b) work done by friction (c) work done by net force (d) change in KE. (g = 10 m s⁻²)
Answer
Normal force N = mg = 2 × 10 = 20 N
Friction f = μN = 0.1 × 20 = 2 N
Net force = 7 − 2 = 5 N
Acceleration a = F_net/m = 5/2 = 2.5 m s⁻²
Displacement in 10 s: s = ½at² = ½ × 2.5 × 100 = 125 m
(a) Work by applied force = F × s = 7 × 125 = 875 J
(b) Work by friction = −f × s = −2 × 125 = −250 J
(c) Work by net force = F_net × s = 5 × 125 = 625 J
(d) ΔKE = ½mv² − 0 = ½m(at)² = ½ × 2 × (2.5 × 10)² = ½ × 2 × 625 = 625 J
Interpretation: Work done by net force = change in KE (work-energy theorem). Work by applied force = work against friction + increase in KE: 875 = 250 + 625 ✓
3. Potential energy functions in one dimension — specify regions where particle cannot be found, and minimum energy.
Answer
A particle cannot exist in regions where total energy E < potential energy V(x), because that would require negative kinetic energy (½mv² < 0, impossible).
(a) V(x) = ax² (parabola): If E is at some height, particle is confined between the two intersection points where V = E. Minimum total energy = 0 (at x = 0).
(b) V(x) has a finite well followed by a constant value: particle cannot be in regions where V > E. Minimum energy = V_min at the bottom of the well.
(c) V(x) is a step function: if E < step height, particle cannot cross into the high-V region.
General rule: Particle is confined to regions where E ≥ V(x). Where E < V(x), the particle cannot be found.
4. V(x) = kx²/2, k = 0.5 N m⁻¹. Show that a particle with total energy 1 J must turn back at x = ±2 m.
Answer
Total energy E = KE + V(x)
At turning points, KE = 0 → E = V(x)
1 = (0.5/2)x² = 0.25x²
x² = 1/0.25 = 4 → x = ±2 m
At x = ±2 m, all energy is potential, kinetic energy = 0
⇒ velocity = 0
⇒ particle must turn back.
For |x| > 2 m, V(x) > 1 J = E, which would require negative KE — impossible. Hence particle cannot go beyond ±2 m.
Q5.5. Answer the following
(a) Heat from burning rocket casing — at whose expense?
(b) Why is work by gravity over complete orbit of comet = 0?
(c) Why does satellite speed increase as it loses energy?
(d) Which case involves greater work — carrying a mass or pulling via pulley?
Ans.
(a) The heat energy comes at the rocket's expense. The kinetic energy of the rocket is converted to heat due to friction with the atmosphere. The rocket slows down slightly as a result.
(b) Gravitational force is a conservative force. Work done by a conservative force over any closed path is zero. A complete orbit returns the comet to the same position, so net work = 0.
(c) As satellite loses energy (mechanical energy E = −GMm/2r becomes more negative), it spirals inward to a smaller orbit. Since KE = GMm/2r, as r decreases, KE increases → speed increases. The loss in potential energy is greater than the gain in KE; the difference goes into heat from atmospheric resistance.
(d) In Fig 5.13(i), the man carries 15 kg horizontally; no work done against gravity (force perpendicular to horizontal displacement). In Fig 5.13(ii), man pulls rope to raise 15 kg → work is done against gravity. Work done in case (ii) is greater.
Q5.6. Underline the correct alternative
(a) When conservative force does positive work, PE increases/decreases/unchanged?
(b) Work against friction results in loss of KE/PE?
(c) Rate of change of total momentum is proportional to external force/sum of internal forces?
(d) In inelastic collision, what doesn't change — total KE/total linear momentum/total energy?
Ans.
(a) Decreases — when a conservative force (like gravity) does positive work, PE decreases (e.g., as a ball falls, gravitational PE decreases).
(b) Kinetic energy — friction acts opposite to motion, reducing the body's kinetic energy.
(c) External force — internal forces cancel in pairs; only external forces change total momentum.
(d) Total linear momentum — in any collision (elastic or inelastic), if no external force, total momentum is conserved. KE is not conserved in inelastic; total energy is not if heat is produced.
Q5.7. True or false
(a) In elastic collision, momentum and energy of each body is conserved
(b) Total energy of system always conserved
(c) Work done over closed loop is zero for every force
(d) In inelastic collision, final KE is always less than initial KE
Ans.
(a) False — the total momentum and total KE of the system are conserved; individual momenta and energies of each body change.
(b) False — total energy is conserved only when no external forces do work (isolated system). With external forces, total mechanical energy may not be conserved (though total energy including heat etc. is).
(c) False — work done over a closed loop is zero only for conservative forces (like gravity). For non-conservative forces (like friction), work done over a closed loop is non-zero.
(d) True — in an inelastic collision, some KE is lost to deformation, heat, sound, etc. So final KE < initial KE.
Q5.8. In elastic and inelastic collisions
(a) Is total KE conserved during contact in elastic collision?
(b) Is total linear momentum conserved during contact?
(c) Same for inelastic?
(d) If PE depends only on separation, is collision elastic or inelastic?
Ans.
(a) No — during contact, balls deform elastically, storing energy as elastic PE. Total KE is not conserved during contact (KE is partly converted to PE). It is restored to KE after contact ends.
(b) Yes — momentum is conserved at all times (no external horizontal force).
(c) For inelastic: Linear momentum is still conserved during contact. Total KE is not conserved (some permanently lost to heat/deformation). Even during contact, momentum is conserved.
(d) Elastic — if PE depends only on separation, the force is conservative. Conservative forces do zero work over a complete interaction (when balls separate to original distance), so total KE is fully recovered → elastic collision.
9. A body starts from rest with constant acceleration. Power delivered at time t is proportional to: (i) t^(1/2) (ii) t (iii) t^(3/2) (iv) t²
Answer
With constant acceleration a: v = at
Power P = F × v = ma × (at) = ma²t
P ∝ t → Answer: (ii) t
10. A body moves under constant power. Its displacement in time t is proportional to: (i) t^(1/2) (ii) t (iii) t^(3/2) (iv) t²
Answer
P = Fv = mav = constant → a = P/(mv)
v dv/dx = P/(mv)
⇒ mv dv = P dt
⇒ ½mv² = Pt
⇒ v = √(2Pt/m) ∝ t^(1/2)
Displacement: dx/dt = v ∝ t^(1/2)
⇒ x = ∫v dt ∝ ∫t^(1/2) dt ∝ t^(3/2)
Answer: (iii) t^(3/2)
11. Force F = (−î + 2ĵ + 3k̂) N acts on a body constrained to move along z-axis. Work done in 4 m displacement along z-axis?
Answer
Displacement d = 4k̂ m
W = F · d = (−î + 2ĵ + 3k̂) · (4k̂) = (−1 × 0) + (2 × 0) + (3 × 4) = 12 J
12. Electron with KE = 10 keV, proton with KE = 100 keV. Which is faster? Find ratio of speeds.
Answer
KE = ½mv² → v = √(2KE/m)
v_e = √(2 × 10 × 10³ × 1.6 × 10⁻¹⁹ / (9.11 × 10⁻³¹))
= √(3.2 × 10⁻¹⁵ / 9.11 × 10⁻³¹) = √(3.513 × 10¹⁵) ≈ 5.93 × 10⁷ m s⁻¹
v_p = √(2 × 100 × 10³ × 1.6 × 10⁻¹⁹ / (1.67 × 10⁻²⁷))
= √(3.2 × 10⁻¹⁴ / 1.67 × 10⁻²⁷) = √(1.916 × 10¹³) ≈ 1.384 × 10⁷ m s⁻¹
Electron is faster.
Ratio v_e/v_p = 5.93 × 10⁷ / 1.384 × 10⁷ ≈ 4.28
13. A rain drop of radius 2 mm falls from 500 m. Terminal speed reached at 250 m. Speed on reaching ground = 10 m s⁻¹. Find work by gravity in first and second half, and work by resistance over entire journey.
Answer
Mass of drop: ρ = 1000 kg m⁻³; r = 2 mm = 2 × 10⁻³ m
m = ρ × (4/3)πr³ = 1000 × (4/3)π × (2 × 10⁻³)³ = 1000 × (4/3)π × 8 × 10⁻⁹ = 3.35 × 10⁻⁵ kg
Work by gravity in first half (250 m): W₁ = mgh₁ = 3.35 × 10⁻⁵ × 10 × 250 = 8.375 × 10⁻² J ≈ 0.08 J
Work by gravity in second half (250 m): W₂ = mgh₂ = same = 0.08 J
Work done by gravity over entire journey = mgh = 3.35 × 10⁻⁵ × 10 × 500 = 0.1675 J
Final KE = ½mv² = ½ × 3.35 × 10⁻⁵ × 100 = 1.675 × 10⁻³ J
By work-energy theorem: W_gravity + W_resistance = ΔKE
(drop starts from rest) W_resistance = ΔKE − W_gravity = 1.675 × 10⁻³ − 0.1675 ≈ −0.166 J
14. A molecule hits a wall at 200 m s⁻¹ at 30° to the normal and rebounds at same speed. Is momentum conserved? Is it elastic?
Answer
The momentum of the molecule alone is not conserved (the wall exerts a force on it). However, the momentum of the system (molecule + wall/earth) is conserved.
Since the molecule rebounds with the same speed, KE of the molecule is unchanged. The kinetic energy is conserved
⇒ The collision is elastic.
15. A pump fills a 30 m³ tank at 40 m height in 15 min. Efficiency = 30%. Find electric power consumed.
Answer
Volume = 30 m³; density of water = 1000 kg m⁻³; mass = 30 × 1000 = 3 × 10⁴ kg
Work done by pump = mgh = 3 × 10⁴ × 10 × 40 = 1.2 × 10⁷ J
Time = 15 × 60 = 900 s
Useful power output = 1.2 × 10⁷ / 900 = 1.333 × 10⁴ W
Efficiency = useful output/electric input
⇒ Electric power = useful power / efficiency
= 1.333 × 10⁴ / 0.30 = 4.44 × 10⁴ W ≈ 44.4 kW
16. Two identical balls at rest on frictionless table are hit head-on by a same-mass ball at speed V (elastic collision). Which result is possible?
Answer
In a perfectly elastic head-on collision: for equal masses, the hitting ball stops and the struck ball moves with the hitting ball's speed.
With two balls together (effectively mass 2m at rest), after elastic collision with mass m at speed V:
By conservation: mV = mv₁ + 2mv₂ and energy: ½mV² = ½mv₁² + ½(2m)v₂²
Solving: v₁ = (m−2m)/(m+2m) × V = −V/3; v₂ = 2m/(m+2m) × V = 2V/3
But the correct physical result (considering each ball separately in a line) is that the incoming ball stops and the last ball moves at V. The result where the incoming ball stops and the two balls move together at V/2 would violate energy conservation. The correct result is: first ball stops, second ball moves at V (if balls are in a line).
17. Bob A released from 30° to vertical hits bob B (same mass) at rest. How high does A rise? (elastic collision)
Answer
In a perfectly elastic collision between equal masses, the striking body comes to rest and the struck body moves with the original velocity of the striker. (This is a standard result for elastic collision between equal masses.)
Therefore, bob A comes to rest after collision
⇒ A rises to height = 0 (it doesn't rise at all).
Bob B moves off with the velocity that A had just before impact.
18. Bob of pendulum released from horizontal. Length = 1.5 m. Speed at lowest point if 5% energy lost to air resistance?
Answer
Initial PE = mgl = m × 10 × 1.5 = 15m J (taking lowest point as reference)
5% is lost → 95% converts to KE
½mv² = 0.95 × mgl = 0.95 × 15m
v² = 2 × 0.95 × 15 = 28.5
v = √28.5 ≈ 5.34 m s⁻¹
19. A trolley (300 kg) + sandbag (25 kg) moves at 27 km h⁻¹. Sand leaks out at 0.05 kg s⁻¹. Speed after all sand gone?
Answer
The sand falls out vertically (no horizontal force). No external horizontal force acts on the trolley system.
By conservation of horizontal momentum:
Initial momentum = (300 + 25) × 27 km h⁻¹ = 325 × 7.5 m s⁻¹ = 2437.5 kg m s⁻¹
After all sand gone, mass = 300 kg.
Final momentum = 300 × v
Since no external horizontal force: 300 × v = 2437.5
⇒ v = 2437.5/300 = 8.125 m s⁻¹
But wait — as sand leaks vertically, there's no horizontal impulse. Horizontal momentum of trolley + remaining sand is conserved.
The sand that leaks carries away some horizontal momentum (it had the trolley's velocity when it leaked).
Actually, as sand leaks at the trolley's velocity, total horizontal momentum remains with the trolley+remaining sand system: v remains = 7.5 m s⁻¹ = 27 km h⁻¹ (unchanged)
20. Body of mass 0.5 kg travels with v = ax^(3/2) where a = 5 m^(−1/2) s⁻¹. Work done from x = 0 to x = 2 m?
Answer
Using work-energy theorem: W = ΔKE = ½m(v₂² − v₁²)
At x = 0: v₁ = a(0)^(3/2) = 0
At x = 2: v₂ = 5 × (2)^(3/2) = 5 × 2√2 = 10√2 m s⁻¹
W = ½ × 0.5 × ((10√2)² − 0) = ½ × 0.5 × 200 = 50 J
21. Windmill blades sweep area A. Wind velocity v perpendicular to area.
Answer
(a) Mass of air passing in time t? (b) KE of air? (c) Electrical power if A = 30 m², v = 36 km h⁻¹, ρ = 1.2 kg m⁻³, efficiency = 25%?
Answer
(a) Volume of air per unit time = A × v; mass in time t: m = ρAv × t
⇒ m = ρAvt
(b) KE = ½mv² = ½ρAvt × v² = ½ρAv³t
(c) v = 36 km h⁻¹ = 10 m s⁻¹; A = 30 m²; ρ = 1.2 kg m⁻³
Power in wind = KE/t = ½ρAv³ = ½ × 1.2 × 30 × (10)³ = ½ × 1.2 × 30 × 1000 = 18000 W
Electrical power = 25% × 18000 = 4500 W = 4.5 kW
22. A dieter lifts 10 kg to 0.5 m height 1000 times. PE lost on lowering is dissipated.
Answer
(a) Work done against gravity? (b) Fat used if 1 kg fat = 3.8 × 10⁷ J at 20% efficiency?
Answer
(a) Work against gravity per lift = mgh = 10 × 10 × 0.5 = 50 J
Total work = 1000 × 50 = 50,000 J = 5 × 10⁴ J
(b) Energy supplied by fat = mechanical work / efficiency = 5 × 10⁴ / 0.20 = 2.5 × 10⁵ J
Fat used = energy / energy per kg = 2.5 × 10⁵ / 3.8 × 10⁷ = 6.58 × 10⁻³ kg ≈ 6.6 g
23. A family uses 8 kW power. Solar energy = 200 W m⁻² at 20% efficiency.
Answer
(a) Area needed? (b) Compare to typical roof area.
Answer
(a) Useful power per m² = 20% × 200 = 40 W m⁻²
Area needed = 8000 / 40 = 200 m²
(b) A typical house roof area is about 50–100 m². The required area of 200 m² is about 2–4 times a typical roof area, suggesting solar power alone may not be sufficient for typical household needs without a much larger collection area.
Important Extra Questions
Q. State and prove the work-energy theorem for a constant force.
Ans. The work-energy theorem states that the net work done on a body equals the change in its kinetic energy: W_net = ΔKE = ½mv² − ½mu²
Proof: For constant force F, acceleration a = F/m
Using v² = u² + 2as → v² − u² = 2as → ½m(v² − u²) = mas = Fs = W
Therefore W_net = ½mv² − ½mu²
Q. Define potential energy. Derive expression for gravitational potential energy.
Ans. Potential energy is the energy stored in a body or system by virtue of its position or configuration. It is the work done against a conservative force to bring the body to its current configuration from a reference position.
Consider raising a mass m from height h₁ to h₂:
Work done against gravity = mg(h₂ − h₁) = stored as PE
Taking ground as reference (h₁ = 0): U = mgh
This is gravitational potential energy.
Q. What is the coefficient of restitution? Distinguish between elastic, inelastic, and perfectly inelastic collisions.
Answer
Coefficient of restitution e = (relative velocity of separation) / (relative velocity of approach)
Elastic collision: e = 1; both momentum and KE are conserved. No permanent deformation.
Inelastic collision: 0 < e < 1; momentum conserved but KE partially lost to deformation/heat.
Perfectly inelastic: e = 0; bodies stick together after collision; maximum KE loss; momentum still conserved.
Q. A block of mass 2 kg moving at 5 m s⁻¹ collides with a stationary block of mass 3 kg. They stick together. Find final velocity and loss in KE.
Ans.
By conservation of momentum: 2 × 5 + 3 × 0 = (2 + 3) × v
⇒ v = 10/5 = 2 m s⁻¹
Initial KE = ½ × 2 × 25 = 25 J
Final KE = ½ × 5 × 4 = 10 J
Loss in KE = 25 − 10 = 15 J
Q. Define power. A car of mass 1000 kg reaches 36 km h⁻¹ in 5 s from rest. Assuming no friction, find power of engine at the end of 5 s.
Ans.
Power = rate of doing work = work done / time = Force × velocity
v = 36 km h⁻¹ = 10 m s⁻¹; a = 10/5 = 2 m s⁻²
Force = ma = 1000 × 2 = 2000 N
Power at the end of 5 s = F × v = 2000 × 10 = 20,000 W = 20 kW