Question

A 61.4kg swimmer is standing on a stationary 210kg floating raft. The swimmer then runs off the raft with the horizontal velocity of 6.81m/s relative to the shore. Find the recoil the raft would have relative to the shore.

Answer 1

The raft will recoil with a velocity of 1.99 m/s relative to the shore, calculated using the conservation of momentum.

Step-by-step explanation:

To find the recoil velocity of the raft relative to the shore, we must use the principle of conservation of momentum. Since the system of the swimmer and the raft is initially at rest, the total momentum of the system is zero. When the swimmer runs off the raft, the momentum of the swimmer will be equal and opposite to the momentum of the raft due to conservation of momentum.

The momentum of the swimmer (Pswimmer) is the product of the swimmer's mass (mswimmer) and velocity (vswimmer): Pswimmer = mswimmer × vswimmer.
The recoil velocity of the raft (vraft) can then be found by rearranging the equation for the momentum of the raft: Praft = mraft × vraft.

By using the values given for the mass of the swimmer (61.4 kg) and the velocity of the swimmer (6.81 m/s), and the mass of the raft (210 kg), we find:

Pswimmer = 61.4 kg × 6.81 m/s = 418.074 kg·m/s

Since Praft = -Pswimmer (the negative sign indicates the opposite direction),

418.074 kg·m/s = 210 kg × vraft

Rearrange to solve for vraft:

vraft = 418.074 kg·m/s / 210 kg = 1.99 m/s (to two decimal places).

The raft will recoil with a velocity of 1.99 m/s relative to the shore.