"A 75 kg swimmer dives horizontally off a 500 kg raft. If the diver's speed immediately after leaving the raft is 4 m/s, what is the corresponding raft speed?"

The speed of the raft after the swimmer jumps off is 0.6 m/s.

The given parameters;

Apply the principle of conservation of linear momentum, to determine the speed of the raft after the swimmer jumps off.

m₁u₁ = m₂u₂

Thus, the speed of the raft after the swimmer jumps off is 0.6 m/s.

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0.6 m/s

Explanation:

Let the raft speed be v.

Given:

Mass of the swimmer (m) = 75 kg

Mass of the raft (M) = 500 kg

Speed of swimmer after leaving the raft (u) = 4 m/s

Now, the given situation can be considered a problem of conservation of total momentum before and after collision.

Momentum is the product of mass and velocity.

Here, the swimmer and raft are the bodies in collision.

So, before the collision, both the bodies were at rest. So, initial momentum is 0. Now, from conservation of momentum, the final momentum of the system must be 0 after the collision. Therefore,

Final Momentum = 0

Plug in the given values and solve for 'v'. This gives,

The final velocity of the raft is -0.6 m/s. Now, speed is the magnitude of velocity.

Therefore, the corresponding raft speed is 0.6 m/s.

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