A small car of mass 950 kg, travelling to the left at a velocity of 6 m s⁻¹, collided with a stationary truck - NSC Technical Sciences - Question 5 - 2021 - Paper 1

To calculate the net force, we can use the impulse defined earlier, which is also related to the change in momentum:

$F_{net} imes riangle t = riangle p$

The change in momentum ($riangle p$) can be calculated as:

$riangle p = m(v_f - v_i)$

where:

• $m$ is the mass of the car (950 kg),
• $v_f$ is the final velocity (backward at 1.24 m s⁻¹, so it will be -1.24 m s⁻¹),
• $v_i$ is the initial velocity (6 m s⁻¹ to the left, which we consider as negative: -6 m s⁻¹).

Now, substituting:

$riangle p = 950 imes (-1.24 - (-6))$

Calculating:

$riangle p = 950 imes (4.76) = 4522 ext{ kg m s}^{-1}$

Next, substituting $riangle p$ into the impulse equation and using $riangle t = 1.28 ext{ s}$:

$F_{net} imes 1.28 = 4522$

Therefore:

F_{net} = rac{4522}{1.28} \\ F_{net} ext{ } ext{ } ext{ } ext{ } = 3532.81 ext{ N}

Assuming our convention is that left is negative, the force exerted by the truck on the car would be:

$F_{net} ext{ is approximately } 3532.81 ext{ N to the right.}$