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 ball be afterwards? \[ v_{2 f}=[?] \mathrm{m} / \mathrm{s} \] Remember East is positive ( + ): west is negative \( (-) \).

To find the final velocity of the ball after an elastic collision with the racket, you can use the conservation of momentum and the conservation of kinetic energy. For elastic collisions, the following equations apply: 1. **Conservation of Momentum:** \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \] 2. **Conservation of Kinetic Energy:** \[ \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \] Given: - \( m_1 = 0.311 \, \text{kg} \) (mass of racket) - \( v_{1i} = 30.3 \, \text{m/s} \) (initial velocity of racket) - \( m_2 = 0.0570 \, \text{kg} \) (mass of ball) - \( v_{2i} = 19.2 \, \text{m/s} \) (initial velocity of ball) Assuming the final velocity of the racket \( v_{1f} \) is not provided and requires solving: 1. **Momentum equation can be set as follows** (let's denote the final velocity of the ball as \( v_{2f} \)): \[ (0.311)(30.3) + (0.0570)(19.2) = (0.311)v_{1f} + (0.0570)v_{2f} \] 2. **Kinetic energy equation becomes**: \[ \frac{1}{2}(0.311)(30.3^2) + \frac{1}{2}(0.0570)(19.2^2) = \frac{1}{2}(0.311)(v_{1f}^2) + \frac{1}{2}(0.0570)(v_{2f}^2) \] 3. Solving both equations simultaneously, you will arrive at the final velocities. After substituting known values and simplifying, you find that: \[ v_{2f} \approx 32.0 \, \mathrm{m/s} \] Thus the velocity of the ball after the collision is approximately \( v_{2f} \approx 32.0 \, \mathrm{m/s} \).