A 0.311 kg tennis racket moves east at \( 30.3 \mathrm{~m} / \mathrm{s} \). A 0.0570 kg ball moves east at \( 19.2 \mathrm{~m} / \mathrm{s} \). If the racket and the ball have an elastic collision, what would the velocity of the racket be afterwards? \[ v_{1 f}=[?] \mathrm{m} / \mathrm{s} \] Remember East is positive ( + ) : west is negative ( - ).

To determine the final velocity of the tennis racket (\( v_{1f} \)) after an elastic collision with the ball, we'll use the principles of **conservation of momentum** and **conservation of kinetic energy** applicable to elastic collisions. ### Given: - **Mass of the racket (\( m_1 \))**: 0.311 kg - **Initial velocity of the racket (\( v_{1i} \))**: +30.3 m/s (east) - **Mass of the ball (\( m_2 \))**: 0.0570 kg - **Initial velocity of the ball (\( v_{2i} \))**: +19.2 m/s (east) ### Formula for Final Velocities in Elastic Collisions: For two objects undergoing an elastic collision in one dimension, the final velocities (\( v_{1f} \) and \( v_{2f} \)) are given by: \[ v_{1f} = \frac{(m_1 - m_2)}{(m_1 + m_2)} v_{1i} + \frac{2m_2}{(m_1 + m_2)} v_{2i} \] \[ v_{2f} = \frac{2m_1}{(m_1 + m_2)} v_{1i} + \frac{(m_2 - m_1)}{(m_1 + m_2)} v_{2i} \] ### Calculating \( v_{1f} \): Plugging in the known values: \[ v_{1f} = \left( \frac{0.311 \, \text{kg} - 0.0570 \, \text{kg}}{0.311 \, \text{kg} + 0.0570 \, \text{kg}} \right) \times 30.3 \, \text{m/s} + \left( \frac{2 \times 0.0570 \, \text{kg}}{0.311 \, \text{kg} + 0.0570 \, \text{kg}} \right) \times 19.2 \, \text{m/s} \] \[ v_{1f} = \left( \frac{0.254 \, \text{kg}}{0.368 \, \text{kg}} \right) \times 30.3 \, \text{m/s} + \left( \frac{0.114 \, \text{kg}}{0.368 \, \text{kg}} \right) \times 19.2 \, \text{m/s} \] \[ v_{1f} \approx 0.69 \times 30.3 \, \text{m/s} + 0.31 \times 19.2 \, \text{m/s} \] \[ v_{1f} \approx 20.9 \, \text{m/s} + 5.95 \, \text{m/s} = 26.85 \, \text{m/s} \] ### Final Answer: \[ v_{1f} \approx 26.9 \, \mathrm{m/s} \, \text{(east)} \]