WAEC Physics 2024 Paper

Concept:The total momentum before and after the jump is conserved because no external horizontal force acts on the boy–skateboard system.Explanation:Before the jump, both the boy and the skateboard are at rest.So, initial total momentum = $0$.Let the mass of the skateboard be $m$.The boy has mass $40 \, \text{kg}$ and moves with velocity $+1.5 \, \text{m/s}$.The skateboard moves in the opposite direction, so its velocity is $-20 \, \text{m/s}$.Using conservation of momentum:$$0 = (40)(1.5) + m(-20)$$$$0 = 60 - 20m$$$$20m = 60$$$$m = \frac{60}{20} = 3 \, \text{kg}$$Answer:The mass of the skateboard is $3 \, \text{kg}$, which is option C.