A 75-kg man is at rest on ice skates. A 0.20-kg ball is thrown to him. The ball is moving horizontally at \(25 \mathrm{m} / \mathrm{s}\) just before the man catches it. How fast is the man moving just after he catches the ball?

Before the man catches the ball: The man is initially at rest (initial velocity, \(v_1 = 0 \mathrm{m} / \mathrm{s}\)) and has mass \(m_1 = 75 \mathrm{kg}\). The ball is moving horizontally at initial velocity, \(v_2 = 25 \mathrm{m} / \mathrm{s}\), and has mass \(m_2 = 0.20 \mathrm{kg}\).

According to the principle of momentum conservation, the total momentum before the catch must be equal to the total momentum after the catch. The total momentum before the catch is \(m_1v_1 + m_2v_2\). After the catch, the man and the ball will have the same final velocity, represented as \(v_f\). Therefore, the total momentum after the catch is: \((m_1 + m_2) v_f\)

Equating the total initial and final momenta and solving for the final velocity \(v_f\), we get: \(m_1v_1 + m_2v_2 = (m_1 + m_2) v_f\) Substitute the values for initial velocities, and masses: \((75 \mathrm{kg} \times 0 \mathrm{m} / \mathrm{s}) + (0.20 \mathrm{kg} \times 25 \mathrm{m} / \mathrm{s}) = (75 \mathrm{kg} + 0.20 \mathrm{kg}) v_f\) Solve for \(v_f\): \(0 + (0.20 \mathrm{kg} \times 25 \mathrm{m} / \mathrm{s}) = 75.20 \mathrm{kg} \times v_f\) \(5 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} = 75.20 \mathrm{kg} \times v_f\) Now, divide both sides by \(75.20 \mathrm{kg}\) to get the final velocity: \(v_f = \frac{5 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}}{75.20 \mathrm{kg}}\) \(v_f \approx 0.066 \mathrm{m} / \mathrm{s}\)

The man's velocity just after catching the ball is approximately \(0.066 \mathrm{m} / \mathrm{s}\).