7. The average space shuttle has a mass of \( 2,051,113 \mathrm{~kg} \). After burning Its fuel for 2 minutes, its mass is \( 880,377 \mathrm{~kg} \). Imagine 2 minutes post-launch, the space shuttle, traveling \( 7850 \mathrm{~m} / \mathrm{s} \), were to hit a 1264 kg plane which has stalled (stationary) over the launch site in an inelastic collision. What is the final velocity of the system?

To solve this inelastic collision problem, we can use the principle of conservation of momentum. The momentum before the collision must equal the momentum after the collision. First, we find the momentum of the space shuttle and the plane before the collision: 1. **Initial momentum of the space shuttle (p_shuttle)**: \( p_{\text{shuttle}} = m_{\text{shuttle}} \times v_{\text{shuttle}} = 880,377 \, \text{kg} \times 7,850 \, \text{m/s} \) 2. **Initial momentum of the plane (p_plane)**: \( p_{\text{plane}} = m_{\text{plane}} \times v_{\text{plane}} = 1,264 \, \text{kg} \times 0 \, \text{m/s} = 0 \) Now, we can calculate the total initial momentum: \( p_{\text{initial}} = p_{\text{shuttle}} + p_{\text{plane}} = 880,377 \times 7,850 + 0 \) Next, after the collision, we have the combined mass of the system moving with a final velocity (v_f). So, the final momentum (p_final) is: \( p_{\text{final}} = (m_{\text{shuttle}} + m_{\text{plane}}) \times v_f = (880,377 + 1,264) \times v_f \) Setting the initial momentum equal to the final momentum gives: \( 880,377 \times 7,850 = (880,377 + 1,264) \times v_f \) Now, we solve for \( v_f \): 1. Calculate \( p_{\text{initial}} \): \[ p_{\text{initial}} = 880,377 \, \text{kg} \times 7,850 \, \text{m/s} \approx 6,903,085,225 \, \text{kg m/s} \] 2. Set up the equation for final momentum: \[ 6,903,085,225 = (881,641) \times v_f \] 3. Solve for \( v_f \): \[ v_f = \frac{6,903,085,225}{881,641} \approx 7,828.94 \, \text{m/s} \] Thus, the final velocity of the system after the collision is approximately \( 7,828.94 \, \text{m/s} \).