Questions & explanations
1. A 2 kg ball moving with velocity (3i + 4j) m/s collides with a stationary 1 kg ball. After the collision, the 2 kg ball moves with velocity (i + 2j) m/s. Assuming no external forces act, what is the velocity of the 1 kg ball after the collision?
- (4i + 4j) m/s
- (4i − 4j) m/s
- (i + j) m/s
- (5i + 2j) m/s
Answer: (4i + 4j) m/s
Using conservation of linear momentum in two dimensions: initial momentum = 2×(3i+4j) + 1×0 = (6i+8j) kg·m/s. Final momentum = 2×(i+2j) + 1×v_B = (2i+4j) + v_B. Equating: (2i+4j) + v_B = (6i+8j) ⇒ v_B = (4i+4j) m/s. Option (4i−4j) would give momentum (2i+4j)+(4i−4j)=(6i+0j), missing the j component; (i+j) gives (3i+5j); (5i+2j) gives (7i+6j). Only (4i+4j) satisfies conservation.
2. Which of the following is an example of a non-conservative force?
- Gravitational force
- Elastic spring force
- Frictional force
- Electrostatic force
Answer: Frictional force
A non-conservative force is one for which the work done depends on the path taken. Frictional force always opposes motion and dissipates energy as heat, making it path-dependent. Gravitational, elastic spring, and electrostatic forces are conservative because the work done by them is independent of the path and depends only on initial and final positions.
3. A ball is dropped from a height of 10 m onto a horizontal floor. If the coefficient of restitution between the ball and the floor is 0.6, what is the height to which the ball rebounds after the first impact?
- 3.6 m
- 6.0 m
- 1.8 m
- 0.36 m
Answer: 3.6 m
The coefficient of restitution e is defined as the square root of the ratio of rebound height to drop height: e = √(h'/h). Given e = 0.6 and h = 10 m, we get h' = e² × h = (0.6)² × 10 = 0.36 × 10 = 3.6 m. Option 6.0 m would correspond to e = √0.6 ≈ 0.775; 1.8 m would be e = √0.18 ≈ 0.424; 0.36 m would be e = √0.036 ≈ 0.19. Only 3.6 m matches e = 0.6.
4. A 2 kg block moving at 4 m/s collides head‑on with a stationary 3 kg block and they stick together. What is the kinetic energy lost during the collision?
- 9.6 J
- 4.8 J
- 16 J
- 24 J
Answer: 9.6 J
Momentum is conserved: m₁u₁ = (m₁ + m₂)v ⇒ v = (2×4)/(2+3) = 8/5 = 1.6 m/s. Initial kinetic energy = ½×2×(4)² = 16 J. Final kinetic energy = ½×(2+3)×(1.6)² = 0.5×5×2.56 = 6.4 J. Loss = 16 − 6.4 = 9.6 J. Option 4.8 J is half of the correct loss; 16 J is the initial KE; 24 J is greater than initial KE, impossible. Thus 9.6 J is correct.
5. A body moves with velocity v = (3i + 4j) m/s under the action of a constant force F = (2i + 1j) N. The instantaneous power delivered by the force is:
- A) 6 W
- B) 10 W
- C) 5 W
- D) 11 W
Answer: B) 10 W
Instantaneous power is given by the scalar product (dot product) of force and velocity vectors: P = F · v = (2)(3) + (1)(4) = 6 + 4 = 10 W. Option A (6 W) uses only the x-component, option C (5 W) incorrectly adds components after taking half, and option D (11 W) incorrectly adds all components without dot product.
6. A force does 500 J of work on a body in 10 s. What is the average power delivered to the body?
- A) 5 W
- B) 50 W
- C) 0.05 W
- D) 5000 W
Answer: B) 50 W
Average power is defined as total work done divided by total time taken. Here, work = 500 J and time = 10 s, so average power = 500 J / 10 s = 50 W. Option A (5 W) incorrectly divides by 100, option C (0.05 W) mistakenly multiplies time and work, and option D (5000 W) multiplies work by time instead of dividing.
7. An electric heater of power 1000 W is used for 3 hours per day. The energy consumed in kWh in 30 days is:
- A) 30 kWh
- B) 90 kWh
- C) 100 kWh
- D) 3 kWh
Answer: B) 90 kWh
Energy consumed per day = power × time = 1000 W × 3 h = 3000 Wh = 3 kWh. Over 30 days, total energy = 3 kWh/day × 30 days = 90 kWh. Option A (30 kWh) uses only one day's consumption, option C (100 kWh) incorrectly uses 1 h per day, and option D (3 kWh) is the consumption for one day only.
8. A block of mass m slides from rest down a frictionless incline of height h. At the bottom, it enters a vertical circular loop of radius R. What is the minimum value of h such that the block just completes the loop without falling off at the top?
- h = R
- h = 2R
- h = 5R/2
- h = 3R/2
Answer: h = 5R/2
To just complete the loop, the block must have at least the critical speed v = √(gR) at the top of the loop. Applying energy conservation from the start (height h) to the top of the loop (height 2R): mgh = mg(2R) + ½m(√(gR))² = 2mgR + ½mgR = (5/2)mgR. Hence h = 5R/2.
9. In a one-dimensional elastic collision, a moving ball of mass m collides head-on with a stationary ball of equal mass. What happens after the collision?
- The moving ball stops and the stationary ball moves with the same speed
- Both balls move with half the initial speed of the moving ball
- The moving ball reverses direction with its original speed
- The stationary ball remains at rest
Answer: The moving ball stops and the stationary ball moves with the same speed
When equal masses undergo a perfectly elastic head-on collision and one is initially at rest, the velocities are exchanged. The moving ball comes to rest, and the initially stationary ball moves away with the speed that the first ball had before collision.
10. A 2 kg ball moving at 3 m/s collides elastically head-on with a stationary 1 kg ball. What is the speed of the 1 kg ball after the collision?
- 1 m/s
- 2 m/s
- 3 m/s
- 4 m/s
Answer: 4 m/s
For an elastic collision, both momentum and kinetic energy are conserved. Using the formula for the velocity of the target (mass m₂) after collision when initially at rest: v₂' = (2m₁/(m₁+m₂)) u₁ = (2×2/(2+1)) × 3 = (4/3)×3 = 4 m/s.
11. A ball is dropped from rest from a height of 20 m. Assuming no air resistance, what is its speed just before hitting the ground? (Take g = 10 m/s²)
- 10 m/s
- 20 m/s
- 30 m/s
- 40 m/s
Answer: 20 m/s
Using the law of conservation of mechanical energy, the potential energy at the top (mgh) is completely converted into kinetic energy (½mv²) just before impact. Thus, mgh = ½mv² → v = √(2gh) = √(2 × 10 × 20) = √400 = 20 m/s.
12. In a perfectly inelastic collision, which of the following is true?
- Kinetic energy is conserved
- The colliding bodies separate after collision
- The colliding bodies stick together after collision
- Only mechanical energy is conserved
Answer: The colliding bodies stick together after collision
A perfectly inelastic collision is defined as one in which the colliding bodies stick together after impact, moving with a common velocity. Momentum is conserved, but kinetic energy is not conserved (maximum loss).