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9 (a) A student investigates the relationship between force and acceleration for a trolley on a runway - 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 runway. 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 runway - 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 more weight to the weight hanger. This increase in weight will apply a greater force due to gravity on the trolley, resulting in increased acceleration according to Newton's second law, where force equals mass times acceleration:

F=mimesaF = m imes a.

Step 2

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

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Answer

To keep the mass of the moving system constant, the student should ensure that the trolley and weight hanger maintain a consistent load throughout the experiment. This can be achieved by keeping the total mass of the trolley and weight hanger unchanged and ensuring that no additional weights are added or removed during 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

The student could improve the procedure by conducting the experiment on a smoother runway or by applying a lubricant to the wheels of the trolley to reduce friction. Alternatively, they could measure the force of friction and subtract it from the total force to accurately determine the net force acting on the trolley. This adjustment would help in accurately measuring acceleration attributable solely to the applied forces.

Step 4

Explain how momentum is conserved in the collision.

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Answer

Momentum is conserved in the collision between objects Q and R due to Newton's third law, which states that for every action, there is an equal and opposite reaction. When Q collides with R, the momentum lost by Q is equal to the momentum gained by R. Therefore, the total momentum before the collision is equal to the total momentum after the collision. In mathematical terms, if the momentum is defined as the product of mass and velocity, then:

extMomentum=mimesv ext{Momentum} = m imes v

The conservation of momentum in this case can be expressed as:

mQvQinitial+mRvRinitial=mQvQfinal+mRvRfinalm_Q v_{Q_{initial}} + m_R v_{R_{initial}} = m_Q v_{Q_{final}} + m_R v_{R_{final}}

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