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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

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During a collision between a bumper car and the barrier, the bumper car and barrier act as a closed system. What is meant by a 'closed system'? How does Newton's Th... show full transcript

Worked Solution & Example Answer: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

Step 1

What is meant by a 'closed system'?

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Answer

A 'closed system' refers to a scenario where the total momentum of the objects involved, in this case, the bumper car and the barrier, remains constant. This means that no external forces act on the system, implying that the momentum before and after the collision is the same.

Step 2

How does Newton's Third Law of motion apply to the collision between the bumper car and the barrier?

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Answer

Newton's Third Law states that for every action, there is an equal and opposite reaction. In the collision, the force exerted by the bumper car on the barrier is met with an equal force exerted by the barrier on the bumper car, but in the opposite direction.

Step 3

Calculate the force on the bumper car during the collision.

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Answer

To calculate the force, we use the formula:

F=ΔpΔtF = \frac{\Delta p}{\Delta t}

where Δp is the change in momentum (700 kg m/s) and Δt is the time (0.28 s).

Thus,

F=7000.28=2500NF = \frac{700}{0.28} = 2500 \, N.

Step 4

Explain how the flexible bumper reduces the risk of injury to the people in the bumper car during the collision.

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Answer

The flexible bumper increases the time taken for the collision to occur, which reduces the rate of change of momentum. This decrease in acceleration or deceleration reduces the overall impact force experienced by the occupants, thus lowering the risk of injury.

Step 5

Calculate the initial constant velocity of the bumper car.

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Answer

We can use the kinematic equation:

v2=u2+2asv^2 = u^2 + 2as

Where:

  • v = final velocity = 2.5 m/s
  • a = acceleration = 2.0 m/s²
  • s = distance = 1.5 m

Rearranging the formula to find u (initial velocity):

u2=v22asu^2 = v^2 - 2as u2=(2.52)2(2)(1.5)u^2 = (2.5^2) - 2(2)(1.5) u2=6.256=0.25u^2 = 6.25 - 6 = 0.25 u=0.5m/su = 0.5 \, m/s.

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