9. A 1050 kg car has a velocity of \( 2,65 \mathrm{~m} \cdot \mathrm{~s}^{-1} \) north. The car hits the rear of a stationary truck, and the bumpers lock together. The velocity of the car-truck system immediately after the collision is \( 0,78 \mathrm{~m} \cdot \mathrm{~s}^{-1} \) north. Calculate the mass of the truck? 10. A \( 0,25 \mathrm{~kg} \) volleyball is thrown horizontally at \( 2 \mathrm{~m} \cdot \mathrm{~s}^{-1} \) west. It strikes a \( 0,62 \mathrm{~kg} \) stationary basketball. The volleyball rebounds east at \( 0,79 \mathrm{~m} . \mathrm{s}^{-1} \). Calculate the velocity of the basketball immediately after the collision?

Let's solve both problems step-by-step using the principles of **conservation of momentum**. --- ### **Problem 9** **Given:** - Mass of the car, \( m_{\text{car}} = 1050 \, \text{kg} \) - Initial velocity of the car, \( v_{\text{car}} = 2.65 \, \text{m/s north} \) - Initial velocity of the truck, \( v_{\text{truck initial}} = 0 \, \text{m/s} \) (stationary) - Final velocity of the combined car-truck system, \( v_{\text{final}} = 0.78 \, \text{m/s north} \) - Mass of the truck, \( m_{\text{truck}} = ? \) **Approach:** Apply the **conservation of momentum**: \[ \text{Total initial momentum} = \text{Total final momentum} \] \[ m_{\text{car}} \cdot v_{\text{car}} + m_{\text{truck}} \cdot v_{\text{truck initial}} = (m_{\text{car}} + m_{\text{truck}}) \cdot v_{\text{final}} \] Since the truck is initially stationary: \[ m_{\text{car}} \cdot v_{\text{car}} = (m_{\text{car}} + m_{\text{truck}}) \cdot v_{\text{final}} \] Solving for \( m_{\text{truck}} \): \[ m_{\text{truck}} = \frac{m_{\text{car}} \cdot v_{\text{car}}}{v_{\text{final}}} - m_{\text{car}} \] \[ m_{\text{truck}} = 1050 \cdot \left( \frac{2.65}{0.78} - 1 \right) \] \[ m_{\text{truck}} = 1050 \cdot (3.403 - 1) = 1050 \cdot 2.403 = 2523.15 \, \text{kg} \] **Answer:** The mass of the truck is approximately **2523  kg**. --- ### **Problem 10** **Given:** - Mass of the volleyball, \( m_{\text{volleyball}} = 0.25 \, \text{kg} \) - Initial velocity of the volleyball, \( v_{\text{vol initial}} = 2 \, \text{m/s west} \) - Final velocity of the volleyball, \( v_{\text{vol final}} = 0.79 \, \text{m/s east} \) - Mass of the basketball, \( m_{\text{basketball}} = 0.62 \, \text{kg} \) - Initial velocity of the basketball, \( v_{\text{basket initial}} = 0 \, \text{m/s} \) - Final velocity of the basketball, \( v_{\text{basket final}} = ? \) **Approach:** Choose **east** as the positive direction and **west** as negative. Apply the **conservation of momentum**: \[ m_{\text{volleyball}} \cdot v_{\text{vol initial}} + m_{\text{basketball}} \cdot v_{\text{basket initial}} = m_{\text{volleyball}} \cdot v_{\text{vol final}} + m_{\text{basketball}} \cdot v_{\text{basket final}} \] Substitute the known values: \[ 0.25 \cdot (-2) + 0.62 \cdot 0 = 0.25 \cdot 0.79 + 0.62 \cdot v_{\text{basket final}} \] \[ -0.5 = 0.1975 + 0.62 \cdot v_{\text{basket final}} \] Solve for \( v_{\text{basket final}} \): \[ -0.5 - 0.1975 = 0.62 \cdot v_{\text{basket final}} \] \[ -0.6975 = 0.62 \cdot v_{\text{basket final}} \] \[ v_{\text{basket final}} = \frac{-0.6975}{0.62} \approx -1.124 \, \text{m/s} \] The negative sign indicates the direction is **west**. **Answer:** Immediately after the collision, the basketball moves west at approximately **1.12 m/s**.