During a collision between a bumper car and the barrier, the bumper car and barrier act as a closed system - AQA - GCSE Physics - Question 5 - 2023 - Paper 2

According to Newton's Third Law, for every action, there is an equal and opposite reaction. In the collision, the force exerted by the bumper car on the barrier is equal in magnitude and opposite in direction to the force exerted by the barrier on the bumper car.

To calculate the force, we can use the formula for force derived from momentum change:

$F = \frac{\Delta p}{\Delta t}$

where $\Delta p = 700 \, \text{kg m/s}$ (change in momentum), and $\Delta t = 0.28 \, \text{s}$ (time of collision). Substituting the values:

$F = \frac{700}{0.28} = 2500 \, \text{N}$

Thus, the force on the bumper car during the collision is 2500 N.

A flexible bumper increases the time taken for the collision to occur, which reduces the rate of change of momentum. This decrease in the rate of change of momentum lowers the overall force experienced by the occupants, minimizing the risk of injury as the impact is softened.

Using the kinematic equation:

$v^2 = u^2 + 2as$

where:

• $v = 2.5 \, \text{m/s}$ (final velocity),
• $u$ is the initial velocity (unknown),
• $a = 2.0 \, \text{m/s}^2$ (acceleration),
• $s = 1.5 \, \text{m}$ (distance travelled).

Rearranging the equation gives:

$u^2 = v^2 - 2as$

Substituting in the known values:

$u^2 = (2.5)^2 - 2 \cdot 2.0 \cdot 1.5$ $u^2 = 6.25 - 6.0 = 0.25$

Taking the square root:

$u = 0.5 \, \text{m/s}$

Therefore, the initial constant velocity of the bumper car is 0.5 m/s.