Question

9. A 45.0 kg ice skater stands at rest on the ice. A friend tosses the skater a 5.0 kg ball. The skater and the ball then move backwards across the ice with a speed of 0.5 m/s. What was the speed of the ball at the moment just before the skater caught it?

Answer 1

The speed of the ball at the moment just before the skater caught it can be found using the conservation of momentum principle. According to the principle, the total momentum of the system before the collision (ball moving towards skater) is equal to the total momentum of the system after the collision (ball and skater moving backwards).

The momentum of the ball before the collision is m_ball * v_ball, where m_ball is the mass of the ball and v_ball is the speed of the ball. The momentum of the skater before the collision is 0, since she is at rest.

The momentum of the ball and skater after the collision is (m_ball + m_skater) * v_after, where v_after is the final speed of the ball and skater moving backwards.

So, we can set up the equation:

m_ball * v_ball = (m_ball + m_skater) * v_after

We know that the final speed of the ball and skater is 0.5 m/s and we know the masses of the ball and skater.

Solving for v_ball:

v_ball = (m_ball + m_skater) * v_after / m_ball

v_ball = (5.0 kg + 45.0 kg) * 0.5 m/s / 5.0 kg

v_ball = 9.0 m/s

Therefore, the speed of the ball at the moment just before the skater caught it was 9.0 m/s.