A 25-kg boy jumps off of a 130-kg raft.If the boy jumps off at a speed of 4.90 m/s relative to the lake,what is the speed of the raft relative to the lake? Ignore the effects of water resistance.
A)0.94 m/s
B)1.1 m/s
C)1.6 m/s
D)2.5 m/s
E)0.50 m/s

Question

Quizplus · Accepted Answer

The conservation of momentum principle states that the total momentum before an event must equal the total momentum after the event, assuming no external forces act on the system. Here, the system is the boy and the raft. Before the jump, both are stationary, so their total momentum is 0. When the boy jumps off the raft, his momentum is his mass times his velocity (25 kg * 4.90 m/s). The raft's momentum must be equal in magnitude and opposite in direction to conserve momentum. $$ \text{Momentum of boy} = \text{Momentum of raft} $$$$ 25 \times 4.90 = 130 \times V_{\text{raft}} $$$$ V_{\text{raft}} = \frac{25 \times 4.90}{130} = 0.94 \, \text{m/s} $$Therefore, the speed of the raft relative to the lake is 0.94 m/s.

A 25-Kg Boy Jumps Off of a 130-Kg Raft