14. A skydiver (54 kg) is coming in for a landing at a brisk speed of 13m/s An unknowing child (22 kg) is watching the landing before getting hit. The skydiver luckily scoops the child with minimal harm and now both are continuing to land What is the final velocity of the skydiver and the child? square

To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision.<br /><br />Let's denote:<br />- \( m_1 \) as the mass of the skydiver (54 kg)<br />- \( v_1 \) as the velocity of the skydiver before the collision (13 m/s)<br />- \( m_2 \) as the mass of the child (22 kg)<br />- \( v_2 \) as the velocity of the child before the collision (0 m/s, since the child is stationary)<br />- \( v_f \) as the final velocity of both the skydiver and the child after the collision<br /><br />The total momentum before the collision is:<br />\[ p_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot v_2 \]<br /><br />Since the child is initially stationary:<br />\[ p_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot 0 = m_1 \cdot v_1 \]<br /><br />The total momentum after the collision is:<br />\[ p_{\text{final}} = (m_1 + m_2) \cdot v_f \]<br /><br />By the conservation of momentum:<br />\[ p_{\text{initial}} = p_{\text{final}} \]<br />\[ m_1 \cdot v_1 = (m_1 + m_2) \cdot v_f \]<br /><br />Now, solve for \( v_f \):<br />\[ v_f = \frac{m_1 \cdot v_1}{m_1 + m_2} \]<br /><br />Substitute the given values:<br />\[ v_f = \frac{54 \, \text{kg} \cdot 13 \, \text{m/s}}{54 \, \text{kg} + 22 \, \text{kg}} \]<br />\[ v_f = \frac{702 \, \text{kg} \cdot \text{m/s}}{76 \, \text{kg}} \]<br />\[ v_f \approx 9.16 \, \text{m/s} \]<br /><br />Therefore, the final velocity of the skydiver and the child is approximately \( 9.16 \, \text{m/s} \).