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Figure 8 shows a model of a system being designed to move concrete building blocks from an upper to a lower level - AQA - A-Level Physics - Question 6 - 2017 - Paper 1

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Question 6

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Figure 8 shows a model of a system being designed to move concrete building blocks from an upper to a lower level. The model consists of two identical trolleys of m... show full transcript

Worked Solution & Example Answer:Figure 8 shows a model of a system being designed to move concrete building blocks from an upper to a lower level - AQA - A-Level Physics - Question 6 - 2017 - Paper 1

Step 1

The tension in the wire when the trolleys are moving is T.

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Answer

To represent the forces acting on trolley A, draw the following arrows:

  • An arrow downwards labeled as weight, pointing to the center of the trolley, representing the gravitational force (Mg).
  • An arrow parallel to the slope of the ramp, labeled as tension, pointing up the ramp (T).
  • An arrow perpendicular to the ramp, showing the normal force (N) acting on the trolley.

Step 2

Show that the acceleration a of trolley B along the ramp is given by a = mg sin 35° / (M + m)

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Answer

By applying Newton’s second law:

  1. For trolley A:
    • The forces along the ramp are: T - mg sin 35° = Ma
  2. For trolley B:
    • The forces along the ramp are: mg sin 35° - T = ma

Adding these two equations gives: (M+m)gsin(35)=(M+m)a(M + m)g \sin(35^\circ) = (M + m)a Thus, simplifying yields: a=mgsin35M+ma = \frac{mg \sin 35^\circ}{M + m}

Step 3

Compare the momentum of loaded trolley A as it moves downwards with the momentum of loaded trolley B.

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Answer

The momentum of an object is given by the product of its mass and velocity. Let:

  • The mass of loaded trolley A be (M + 2m), and its downward velocity be v.
  • The mass of loaded trolley B be M, and its upward velocity also be v.

Therefore, the momentum of trolley A is: pA=(M+2m)vp_A = (M + 2m)v And the momentum of trolley B is: pB=Mvp_B = Mv Here, pAp_A is greater than pBp_B since M+2m>MM + 2m > M.

Step 4

Calculate the time taken for a loaded trolley to travel down the ramp.

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Answer

Given:

  • Distance (s) = 9.0 m
  • Acceleration (a) is calculated as 25% of maximum: a=0.25×gsin35M+ma = 0.25 \times \frac{g \sin 35^\circ}{M + m} Using g9.81m/s2g \approx 9.81 \, m/s^2 and substituting values: a0.25×(9.81×sin35)(95+30)a \approx 0.25 \times \frac{(9.81 \times \sin 35^\circ)}{(95 + 30)} Calculate the final time using: s=ut+12at2s = ut + \frac{1}{2}at^2 Since initial velocity (u) is 0, solve for t.

Step 5

Calculate the number of blocks that can be transferred to the lower level in 30 minutes.

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Answer

If each journey takes T seconds, the number of journeys in 30 minutes (1800 seconds) is: Number of journeys=1800T\text{Number of journeys} = \frac{1800}{T} And if unblocking and reloading takes 12 seconds per operation, the total number of blocks transferred can be calculated by: total blocks=Number of journeys×2 \text{total blocks} = \text{Number of journeys} \times 2

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