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9 (a) A student investigates the relationship between force and acceleration for a trolley on a runaway - Edexcel - GCSE Physics - Question 9 - 2018 - Paper 1

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9 (a) A student investigates the relationship between force and acceleration for a trolley on a runaway. Figure 12 shows some of the apparatus the student uses. (i)... show full transcript

Worked Solution & Example Answer:9 (a) A student investigates the relationship between force and acceleration for a trolley on a runaway - Edexcel - GCSE Physics - Question 9 - 2018 - Paper 1

Step 1

Describe how the student could increase the accelerating force applied to the trolley.

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Answer

The student can increase the accelerating force applied to the trolley by adding additional weights to the trolley or by increasing the angle of the inclined runway. This increase in weight leads to a greater gravitational force acting down the slope, thereby increasing the accelerating force acting on the trolley.

Step 2

Describe how the mass of the moving system can be kept constant.

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Answer

The mass of the moving system can be kept constant by ensuring that the weight attachable to the trolley, such as the weight hanger, remains unchanged throughout the experiment. This can involve consistently using the same weights throughout the entirety of the experiment.

Step 3

Explain how the student could improve the procedure to compensate for the effects of frictional forces acting on the trolley.

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Answer

To improve the procedure for compensating for the effects of friction, the student could raise one end of the runway to allow the trolley to roll freely. Additionally, using a low-friction surface for the runway or adding lubricant to minimize friction could also enhance the accuracy of the measurements. Finally, measuring and subtracting the frictional force from the total force could provide a clearer understanding of the net forces acting on the trolley.

Step 4

Explain how momentum is conserved in the collision.

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Answer

In the collision between objects Q and R, momentum is conserved due to Newton's third law, which states that for every action, there is an equal and opposite reaction. Therefore, the momentum lost by object Q will be equal to the momentum gained by object R. This can be represented mathematically as:

mQvQ+mRvR=mQvQ+mRvRm_Q v_Q + m_R v_R = m_Q v_{Q}' + m_R v_{R}'

where the initial and final momenta of the objects account for their respective masses and velocities before and after the collision.

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