aucstion 3 A car with a mass of 650 kg travels at a velocity of \( 60 \mathrm{~km} \mathrm{~h}^{-1} \) and crashes into the back of a truck with a mass of 1000 kg that is travelling in the same direction at \( 30 \mathrm{~km} \cdot \mathrm{~h}^{-1} \) Immedialely after the collision, the truck travels at \( 36 \mathrm{~km} \cdot \mathrm{~h}^{-1} \) in the original direction of motion. (Ignore the effects of friction.) 3.1 Calculate the velocity of the car immediately after the collision 3.2 What was the change in momentum of the car during the collision? 3.3 What should the change in momentum of the truck be? Give a reason for your answer.

Let's analyze the collision between the car and the truck step by step. ### **Given Data:** - **Mass of the car (\( m_{\text{car}} \))**: 650 kg - **Initial velocity of the car (\( u_{\text{car}} \))**: 60 km/h - **Mass of the truck (\( m_{\text{truck}} \))**: 1000 kg - **Initial velocity of the truck (\( u_{\text{truck}} \))**: 30 km/h - **Final velocity of the truck (\( v_{\text{truck}} \))**: 36 km/h **Note:** We'll convert all velocities from km/h to m/s for consistency in units. ### **Conversion from km/h to m/s:** \[ 1 \, \text{km/h} = \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{5}{18} \, \text{m/s} \] \[ \begin{align*} u_{\text{car}} &= 60 \times \frac{5}{18} = 16.6667 \, \text{m/s} \\ u_{\text{truck}} &= 30 \times \frac{5}{18} = 8.3333 \, \text{m/s} \\ v_{\text{truck}} &= 36 \times \frac{5}{18} = 10 \, \text{m/s} \end{align*} \] --- ### **3.1. Calculate the Velocity of the Car Immediately After the Collision (\( v_{\text{car}} \))** **Assumption:** The collision is **elastic**, and **friction is neglected**. **Law of Conservation of Momentum:** \[ \text{Total Momentum Before Collision} = \text{Total Momentum After Collision} \] \[ m_{\text{car}} \times u_{\text{car}} + m_{\text{truck}} \times u_{\text{truck}} = m_{\text{car}} \times v_{\text{car}} + m_{\text{truck}} \times v_{\text{truck}} \] **Plugging in the known values:** \[ 650 \times 16.6667 + 1000 \times 8.3333 = 650 \times v_{\text{car}} + 1000 \times 10 \] \[ 10833.33 + 8333.33 = 650 \times v_{\text{car}} + 10000 \] \[ 19166.66 = 650 \times v_{\text{car}} + 10000 \] \[ 650 \times v_{\text{car}} = 19166.66 - 10000 = 9166.66 \] \[ v_{\text{car}} = \frac{9166.66}{650} \approx 14.111 \, \text{m/s} \] **Converting back to km/h:** \[ v_{\text{car}} \approx 14.111 \times \frac{18}{5} = 50.8 \, \text{km/h} \] **Answer:** **3.1** The velocity of the car immediately after the collision is approximately **50.8 km/h** in the original direction of motion. --- ### **3.2. Change in Momentum of the Car During the Collision (\( \Delta p_{\text{car}} \))** **Momentum (\( p \))** is given by: \[ p = m \times v \] **Change in Momentum (\( \Delta p \))**: \[ \Delta p_{\text{car}} = m_{\text{car}} \times (v_{\text{car}} - u_{\text{car}}) \] \[ \Delta p_{\text{car}} = 650 \times (14.111 - 16.6667) \] \[ \Delta p_{\text{car}} = 650 \times (-2.5557) \approx -1661.25 \, \text{kg·m/s} \] **Interpretation:** A negative sign indicates that the momentum of the car has **decreased** in the direction of motion. **Answer:** **3.2** The change in momentum of the car during the collision is approximately **−1661.25 kg·m/s**. --- ### **3.3. Change in Momentum of the Truck During the Collision and the Reason** **Change in Momentum (\( \Delta p \)) for the Truck:** According to the **Law of Conservation of Momentum**, the total momentum of an isolated system remains constant if no external forces act upon it. Since only the car and the truck are interacting, the momentum lost by the car must be gained by the truck. \[ \Delta p_{\text{truck}} = -\Delta p_{\text{car}} = +1661.25 \, \text{kg·m/s} \] **Reason:** **Newton's Third Law of Motion** states that for every action, there is an equal and opposite reaction. The force exerted by the car on the truck results in an equal and opposite force exerted by the truck on the car. Consequently, the change in momentum of the truck is equal in magnitude but opposite in direction to the change in momentum of the car. **Answer:** **3.3** The change in momentum of the truck is **+1661.25 kg·m/s**. *Reason:* According to Newton's Third Law, the truck experiences an equal and opposite momentum change to that of the car during the collision.