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
Correct answer
Explanation
The correct answer is B, 1.89 m/s. By the law of conservation of momentum, the boy and the cart move together with the momentum the boy brought, so their speed is 104 ÷ 55, about 1.89 m/s.
With frictionless wheels, no outside horizontal force acts, so total momentum stays the same. Before the jump, the boy's momentum is 52 kg × 2 m/s = 104 kg m/s and the cart has none. After he lands, boy and cart move together with a total mass of 52 + 3 = 55 kg. So 55 × v = 104, which gives v = 1.89 m/s. The speed falls below 2 m/s because the same momentum is now shared by a larger mass.
A is wrong because 2.15 m/s is more than the boy's own speed, which would create momentum from nothing.
C is wrong because 1.51 m/s would mean momentum was lost, which cannot happen without an outside force.
D is wrong because 2.51 m/s is also faster than the boy, so it breaks the conservation of momentum.
Exam tip: for bodies that stick together, m1u1 + m2u2 = (m1 + m2)v.
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