A 74-kg swimmer dives horizontally off a 500-kg raft.The diver's speed immediately after leaving the raft is 4.0 m/s.A micro-sensor system attached to the edge of the raft measures the time interval during which the diver applies an impulse to the raft just prior to leaving the raft surface.If the time interval is read as 0.20 s,what is the magnitude of the average horizontal force by diver on the raft?

Question

Quizplus · Accepted Answer

We can use the principle of conservation of momentum to solve this problem. Let the initial momentum of the system (diver + raft) be zero, since they are at rest. After the diver jumps off with a speed of 4.0 m/s, the momentum of the diver is m*d, where d is the horizontal distance that the diver travels while pushing the raft. The momentum of the raft is M*(-v), where v is the velocity of the raft (equal in magnitude and opposite in direction to the velocity of the diver), and M is the mass of the raft.

Using the principle of conservation of momentum, we have:

m*d + M*(-v) = 0

Solving for v, we get:

v = m*d / M

The time interval during which the impulse is applied to the raft is 0.20 s. The impulse is equal to the change in momentum of the raft, and is given by:

I = M*(v - 0)

I = M*m*d / M

I = m*d

Finally, the average force applied by the diver on the raft is equal to the impulse divided by the time interval:

F = I / t

F = (m*d) / t

Substituting the given values, we get:

F = (74 kg)*(4.0 m/s) / 0.20 s

F = 1,480 N

Therefore, the answer is D) 1,480 N.

A 74-Kg Swimmer Dives Horizontally Off a 500-Kg Raft