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.

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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.
- **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:** $$ -21250 \, \text{N} $$

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