A boy of mass 52 kg jumps with a horizontal velocity of 2 m/s onto a stationary cart of mass 3 kg. The cart is fixed with frictionless wheels. Which one of the following would be the speed of the cart?
- (a)2.15 m/s
- (b)1.89 m/s
- (c)1.51 m/s
- (d)2.51 m/s
Answer
Why
Correct — B, 1.89 m/s. When the boy lands on the cart no external horizontal force acts, so horizontal momentum is conserved. Before the jump only the boy moves: momentum = 52 kg × 2 m/s = 104 kg·m/s. Afterwards the boy and cart move together as one mass of 52 + 3 = 55 kg at speed v. Setting 104 = 55 × v gives v = 104 ÷ 55 ≈ 1.89 m/s.
Why the others are wrong
- (a)2.15 m/s — This is greater than the boy's initial 2 m/s, which is impossible — adding the cart's mass can only slow the combined system, never speed it up.
- (c)1.51 m/s — This does not satisfy the momentum balance 104 = 55·v (which gives 1.89 m/s); it comes from using a wrong total mass.
- (d)2.51 m/s — Like (a), it exceeds the boy's original 2 m/s and so violates conservation of momentum once the extra mass is added.
Concept
Linear momentum (p = m·v) is conserved when no external force acts. When two bodies stick together and move as one — a perfectly inelastic collision — the total momentum before equals the combined mass times the common velocity afterwards.
Because the wheels are frictionless, nothing external pushes horizontally, so momentum is conserved. The combined mass (55 kg) is larger than the boy's mass, so the common speed must drop below the boy's original 2 m/s — which immediately rules out any option above 2 m/s.
Key facts
- Conservation of momentum: total m·v before = total m·v after when no external force acts.
- Boy + cart stick together — a perfectly inelastic collision, so they share one final velocity.
- 104 kg·m/s = 55 kg × v → v ≈ 1.89 m/s.
- Kinetic energy is not conserved in an inelastic collision, though momentum is.
- Before: boy 52 kg at 2 m/s, cart 3 kg at rest → total momentum = 104 kg·m/s
- Boy lands on cart — they stick and move together (inelastic collision)
- After: combined mass 55 kg moves at v → 55·v = 104
- v = 104 ÷ 55 ≈ 1.89 m/s
Momentum is conserved; the larger combined mass means the common speed drops below 2 m/s.
Study next
Common traps
- Adding the velocities instead of applying momentum conservation.
- Forgetting to add the cart's mass, or expecting the speed to rise after the extra mass joins.
A person or object landing on and sticking to a cart — apply conservation of momentum to find the common speed.
Related PYQs
A metallic bob X of mass m is released from position A. It collides elastically with another identical bob Y placed at rest at position B on a horizontal frictionless table. The angle AOB is 30°. How high does the bob X rise immediately after the collision?
- (a) To the same height as that of position A on the other side in the same trajectory
- (b) To half the height as that of position A on the other side along the same trajectory
- (c) The same height at position A
- (d) It stops at position B
Answer(d) It stops at position B
The companion collision problem — that one is a perfectly elastic collision between equal masses (X stops, Y moves off), while this one is inelastic (boy and cart move together). Both are solved with conservation of momentum.
Practice
- practice — not a real PYQ
A 4 kg trolley moving at 3 m/s collides with and sticks to a 2 kg trolley at rest. Their common speed is
- (a)1 m/s
- (b)2 m/s
- (c)3 m/s
- (d)1.5 m/s
Answer(b) 2 m/s — total momentum 12 kg·m/s ÷ combined mass 6 kg. - practice — not a real PYQ
A bullet embedding itself in a wooden block is an example of a collision that is
- (a)perfectly elastic
- (b)perfectly inelastic
- (c)partially elastic
- (d)frictionless
Answer(b) perfectly inelastic — the two move together afterwards, so kinetic energy is not conserved but momentum is.