A 0.311 kg tennis racket moves east at \( 30.3 \mathrm{~m} / \mathrm{s} \) A 0.0570 kg ball moves west at \( 19.2 \mathrm{~m} / \mathrm{s} \). If the racket and the ball have an elastic collision, what would the velocity of the ball be afterwards? \[ \mathrm{v}_{2 \mathrm{f}}=[?] \mathrm{m} / \mathrm{s} \] Remember: East is positive ( \( ( \) ): west is negative \( (-) \).

To find the final velocity of the ball after an elastic collision, we can use the conservation of momentum and the conservation of kinetic energy. 1. **Conservation of Momentum**: \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \] where: - \( m_1 = 0.311 \, \text{kg} \) (mass of the racket) - \( v_{1i} = 30.3 \, \text{m/s} \) (initial velocity of the racket) - \( m_2 = 0.0570 \, \text{kg} \) (mass of the ball) - \( v_{2i} = -19.2 \, \text{m/s} \) (initial velocity of the ball) 2. **Conservation of Kinetic Energy**: \[ \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \] For this problem, we can use the following formulas for the final velocities after an elastic collision between two objects: \[ v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2 m_2 v_{2i}}{m_1 + m_2} \] \[ v_{2f} = \frac{(m_2 - m_1)v_{2i} + 2 m_1 v_{1i}}{m_1 + m_2} \] Calculating \( v_{2f} \): \[ v_{2f} = \frac{(0.0570 - 0.311)(-19.2) + 2(0.311)(30.3)}{0.311 + 0.0570} \] \[ v_{2f} = \frac{(-0.254)(-19.2) + (0.622)(30.3)}{0.368} \] \[ = \frac{4.8768 + 18.8826}{0.368} \] \[ = \frac{23.7594}{0.368} \approx 64.4 \, \text{m/s} \] Thus, the velocity of the ball after the collision is: \[ \mathrm{v}_{2 \mathrm{f}} \approx 64.4 \, \text{m/s} \]