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 find the final velocity of the tennis racket after the elastic collision, we can use the conservation of momentum and the equations for elastic collisions. Let: - \( m_1 = 0.311 \, \text{kg} \) (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) Using the conservation of momentum: \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \] And the formula for the final velocities after an elastic collision: \[ v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2m_2 v_{2i}}{m_1 + m_2} \] Plug in the values: \[ v_{1f} = \frac{(0.311 - 0.0570)(30.3) + 2(0.0570)(19.2)}{0.311 + 0.0570} \] Calculating: 1. \( (0.311 - 0.0570)(30.3) = (0.254)(30.3) \approx 7.6942 \) 2. \( 2(0.0570)(19.2) = 2(1.0944) \approx 2.1888 \) 3. Total momentum for racket: \( 7.6942 + 2.1888 \approx 9.883 \) 4. Total mass: \( 0.311 + 0.0570 = 0.368 \) Now, divide the total momentum by the total mass: \[ v_{1f} \approx \frac{9.883}{0.368} \approx 26.9 \, \text{m/s} \] So, the final velocity of the racket is approximately: \[ v_{1 f} \approx 26.9 \, \mathrm{m/s} \]