A baseball player uses a bat with mass \(m_{\text {bat }}\) to hit a ball with mass \(m_{\text {ball }}\). Right before he hits the ball, the bat's initial velocity is \(35.0 \mathrm{~m} / \mathrm{s}\), and the ball's initial velocity is \(-30.0 \mathrm{~m} / \mathrm{s}\) (the positive direction is along the positive \(x\) -axis). The bat and ball undergo a one-dimensional elastic collision. Find the speed of the ball after the collision. Assume that \(m_{\text {bat }}\) is much greater than \(m_{\text {ball }}\), so the center of mass of the two objects is essentially at the bat.

In a one-dimensional elastic collision, the total momentum before the collision is equal to the total momentum after the collision. This is expressed mathematically as: \(m_{\text {bat}}v_{\text {bat_initial}} + m_{\text {ball}}v_{\text {ball_initial}} = m_{\text {bat}}v_{\text {bat_final}} + m_{\text {ball}}v_{\text {ball_final}}\) Where \(v\) represents the velocities of the bat and ball, and the subscripts "initial" and "final" represent before and after the collision, respectively.

In a one-dimensional elastic collision, the total kinetic energy before the collision is also equal to the total kinetic energy after the collision. This is expressed mathematically as: \(0.5m_{\text {bat}}v_{\text {bat_initial}}^2 + 0.5m_{\text {ball}}v_{\text {ball_initial}}^2 = 0.5m_{\text {bat}}v_{\text {bat_final}}^2 + 0.5m_{\text {ball}}v_{\text {ball_final}}^2\)

We are given that \(m_{\text {bat}}\) is much greater than \(m_{\text {ball}}\), so the center of mass of the two objects is essentially at the bat. This means that the final velocity of the bat essentially does not change after the collision. Therefore, we can set \(v_{\text {bat_final}}\) equal to \(v_{\text {bat_initial}}\), which is \(35.0 \mathrm{~m} / \mathrm{s}\).

We can now apply the conservation of momentum formula with the given initial velocities and the assumption that the final velocity of the bat does not change: \(m_{\text {bat}}(35.0) + m_{\text {ball}}(-30.0) = m_{\text {bat}}(35.0) + m_{\text {ball}}v_{\text {ball_final}}\) Now, rearrange the equation to solve for \(v_{\text {ball_final}}\): \(v_{\text {ball_final}} = \frac{m_{\text {bat}}(35.0) - m_{\text {ball}}(-30.0)}{m_{\text {ball}}}\) Since \(m_{\text {bat}}\) is much greater than \(m_{\text {ball}}\), \(m_{\text {bat}}(35.0)\) is a much larger term, which makes the ratio of the two masses in the equation much greater than one. This means that the final velocity of the ball will be essentially the sum of the initial velocities: \(v_{\text {ball_final}} \approx 35.0 + (-30.0) = 5.0 \mathrm{~m} / \mathrm{s}\) So, the speed of the ball after the collision will be approximately 5.0 m/s in the positive \(x\)-axis direction.