QUESTION 3: Application [10 marks] A 1200 kg car travelling North at \( \mathbf{2 0 . 0} \mathbf{m} \cdot \mathrm{s}^{-1} \) collides with a 1400 kg car travelling South at \( \mathbf{2 2 . 0} \mathrm{m} \cdot \mathrm{s}^{-1} \). The two cars collide and entangle when they move off together. Only \( \mathbf{7 5 \%} \) of the linear momentum is conserved during this collision. Question \( \mathbf{3 . 1} \) [2 marks] Calculate the initial momentum in the system before the collision. Question 3.2 [2 marks] Calculate the final momentum in the system after the collision. Question 3.3 [ 2 marks] Calculate the velocity with which both vehicles move off together. Question 3.4 [2 marks] Calculate the momentum lost due to external forces. Question 3.5 [2 marks] The collision lasts for 0,08 seconds before the cars reach their final velocity. Calculate the net external force that removes momentum from the system.

To solve the problem, we will follow the steps outlined in the questions, calculating the necessary values step by step. ### Known Conditions: - Mass of car 1 (\(m_1\)) = 1200 kg (travelling North) - Velocity of car 1 (\(v_1\)) = 20.0 m/s - Mass of car 2 (\(m_2\)) = 1400 kg (travelling South) - Velocity of car 2 (\(v_2\)) = -22.0 m/s (negative because it's in the opposite direction) - Momentum conservation = 75% ### Question 3.1: Calculate the initial momentum in the system before the collision. The initial momentum (\(p_{initial}\)) of the system can be calculated using the formula: \[ p_{initial} = m_1 \cdot v_1 + m_2 \cdot v_2 \] Substituting the values: \[ p_{initial} = (1200 \, \text{kg} \cdot 20.0 \, \text{m/s}) + (1400 \, \text{kg} \cdot -22.0 \, \text{m/s}) \] Calculating this: \[ p_{initial} = 24000 \, \text{kg m/s} - 30800 \, \text{kg m/s} = -6800 \, \text{kg m/s} \] ### Question 3.2: Calculate the final momentum in the system after the collision. The final momentum (\(p_{final}\)) is 75% of the initial momentum: \[ p_{final} = 0.75 \cdot p_{initial} \] Substituting the value of \(p_{initial}\): \[ p_{final} = 0.75 \cdot (-6800 \, \text{kg m/s}) = -5100 \, \text{kg m/s} \] ### Question 3.3: Calculate the velocity with which both vehicles move off together. After the collision, the two cars move off together, so we can use the combined mass (\(m_{total} = m_1 + m_2\)) and the final momentum to find the final velocity (\(v_f\)): \[ p_{final} = m_{total} \cdot v_f \] \[ m_{total} = 1200 \, \text{kg} + 1400 \, \text{kg} = 2600 \, \text{kg} \] Now substituting the values: \[ -5100 \, \text{kg m/s} = 2600 \, \text{kg} \cdot v_f \] Solving for \(v_f\): \[ v_f = \frac{-5100 \, \text{kg m/s}}{2600 \, \text{kg}} \approx -1.96 \, \text{m/s} \] ### Question 3.4: Calculate the momentum lost due to external forces. The momentum lost (\(p_{lost}\)) can be calculated as: \[ p_{lost} = p_{initial} - p_{final} \] Substituting the values: \[ p_{lost} = -6800 \, \text{kg m/s} - (-5100 \, \text{kg m/s}) = -6800 + 5100 = -1700 \, \text{kg m/s} \] ### Question 3.5: Calculate the net external force that removes momentum from the system. The net external force (\(F_{net}\)) can be calculated using the formula: \[ F_{net} = \frac{p_{lost}}{t} \] where \(t = 0.08 \, \text{s}\). Substituting the values: \[ F_{net} = \frac{-1700 \, \text{kg m/s}}{0.08 \, \text{s}} = -21250 \, \text{N} \] ### Summary of Results: - **Initial Momentum:** \( -6800 \, \text{kg m/s} \) - **Final Momentum:** \( -5100 \, \text{kg m/s} \) - **Final Velocity:** \( -1.96 \, \text{m/s} \) - **Momentum Lost:** \( -1700 \, \text{kg m/s} \) - **Net External Force:** \( -21250 \, \text{N} \) This concludes the calculations for the problem.