Required information A \( 0.0300-\mathrm{kg} \) bullet is fired vertically at \( 200.0 \mathrm{~m} / \mathrm{s} \) into a 0.163 -kg baseball that is initially at rest. The bullet lodges in the baseball and, after the collision, the baseball/bullet rise to a height of 37.0 m . Assume up to be the positive direction. What was the speed of the baseball/bullet right after the collision? \( \mathrm{m} / \mathrm{s} \) ences sht

To find the speed of the baseball and bullet right after the collision, we can use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision. Before the collision, only the bullet has momentum: \[ p_{\text{initial}} = m_{bullet} \cdot v_{bullet} + m_{baseball} \cdot v_{baseball} \] \[ p_{\text{initial}} = (0.0300 \, \text{kg} \cdot 200.0 \, \text{m/s}) + (0.163 \, \text{kg} \cdot 0) \] \[ p_{\text{initial}} = 6.0 \, \text{kg} \cdot \text{m/s} \] After the collision, the bullet and baseball move together with a common velocity \( v_f \): \[ p_{\text{final}} = (m_{bullet} + m_{baseball}) \cdot v_f \] By conservation of momentum, we set \( p_{\text{initial}} = p_{\text{final}} \): \[ 6.0 = (0.0300 + 0.163) \cdot v_f \] \[ 6.0 = 0.193 \cdot v_f \] \[ v_f = \frac{6.0}{0.193} \approx 31.09 \, \text{m/s} \] So, the speed of the baseball/bullet right after the collision is approximately \( 31.1 \, \text{m/s} \).