A 75-kg swimmer dives horizontally off a 250-kg raft initially at rest.If the diver's speed immediately after leaving the raft is 2 m/s,what is the corresponding raft speed?
A) 0.2 m/s
B) 0.3 m/s
C) 0.6 m/s
D) 4.0 m/s

Question

Quizplus · Accepted Answer

According to the law of conservation of momentum, the initial momentum of the system (swimmer+raft) is zero since the raft is initially at rest. After the swimmer jumps, the momentum of the system is conserved. We can use the equation:
(mass of swimmer x velocity of swimmer) + (mass of raft x velocity of raft) = 0
Plugging in the numbers:
(75 kg x 2 m/s) + (250 kg x velocity of raft) = 0
Solving for velocity of raft:
Velocity of raft = -(75 x 2) / (250) = -0.6 m/s (negative sign indicating direction opposite to that of swimmer)
Taking the magnitude of the velocity, we get: 
Velocity of raft = 0.6 m/s 
Therefore, the answer is C.

A 75-Kg Swimmer Dives Horizontally Off a 250-Kg Raft Initially