The graph below shows how the momentum of car A changes with time just before and just after a head-on collision with car B - NSC Physical Sciences - Question 4 - 2016 - Paper 1

To find the velocity of car A just before the collision, we use the formula for momentum:

$p = mv$

Given that momentum (p) is 30,000 kg·m/s and the mass (m) of car A is 1,500 kg, we can solve for velocity (v):

$v = \frac{p}{m} = \frac{30000}{1500} = 20 \text{ m/s}$

To find the velocity of car B just after the collision, we apply conservation of momentum. Before the collision, the total momentum is:

$p_{initial} = m_A v_A + m_B v_B = (1500)(20) + (900)(-15) = 30000 - 13500 = 16500 \text{ kg·m/s}$

After the collision, let the velocity of car B be v'_B. The equation becomes:

$p_{final} = m_A v'_A + m_B v'_B$

The momentum of car A drops to 14,000 kg·m/s. Hence, using conservation of momentum:

$16500 = 14000 + 900v'_B$

Solving for v'_B:

ightarrow v'_B = \frac{2500}{900} \approx 2.78 \text{ m/s east}$$

The net average force acting on car A can be calculated using the change in momentum formula:

$F_{net} = \frac{\Delta p}{\Delta t}$

From the graph, the change in momentum (Δp) for car A is:

$\Delta p = p_{final} - p_{initial} = 14000 - 30000 = -16000 \text{ kg·m/s}$

The time interval (Δt) over which this change occurs is 0.1 s (from 20.2 s to 20.3 s). Thus, substituting these values gives:

$F_{net} = \frac{-16000}{0.1} = -160000 \text{ N}$

The magnitude of the net average force is 160,000 N.