A 93 kg man holding a 0.653 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 11.3 m/s (relative to the ground) and then catches the ball after it rebounds from the wall. How fast is he moving after he catches the ball?

After catching the ball, the man will be moving in the opposite direction to the ball's initial velocity with a final velocity of approximately 0.0786 m/s.

Explanation:

After catching the ball, the man will be moving in the opposite direction to the ball's initial velocity. To find the man's final velocity, we can use the principle of conservation of momentum. The momentum of the man and the ball before the throw is equal to the momentum of the man and the ball after the catch.

First, we need to calculate the initial momentum of the man and the ball:

Initial momentum = mass * velocity

Man's initial momentum = 93 kg * 0 m/s (since he is stationary) = 0 kg·m/s

Ball's initial momentum = 0.653 kg * 11.3 m/s = 7.3669 kg·m/s

The total initial momentum is 0 kg·m/s + 7.3669 kg·m/s = 7.3669 kg·m/s.

to find the final velocity of the man after catching the ball, we can set up the equation:
0 kg·m/s + 7.3669 kg·m/s = (93 kg + 0.653 kg) * final velocity
Simplifying the equation:

7.3669 kg·m/s = 93.653 kg * final velocity

Final velocity = 7.3669 kg·m/s / 93.653 kg ≈ 0.0786 m/s

Therefore, the man is moving in the opposite direction of the thrown ball with a final velocity of approximately 0.0786 m/s.

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