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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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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

Arrows should be drawn to represent the following forces on trolley A:

  • Weight (MgMg) acting vertically downwards.
  • Tension (TT) acting upwards along the wire.
  • Normal force perpendicular to the ramp surface.
  • Component of weight acting parallel to the ramp, labeled Mgsin35Mg \sin 35^{\circ}.

Step 2

Assume that no friction acts at the axle of the pulley or at the axles of the trolleys and that air resistance is negligible.

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Answer

To show the acceleration aa of trolley B, we apply Newton's second law:

  1. For trolley A:

    TMgsin35=MaT - Mg \sin 35^{\circ} = Ma

  2. For trolley B:

    T=(M+2m)aT = (M + 2m) a

    Adding both equations:

    Mgsin35=(M+2m)aMg \sin 35^{\circ} = (M + 2m)a

    Substituting gives:

    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 trolley A, pA=MvAp_A = Mv_A, and for trolley B, pB=(M+2m)vBp_B = (M + 2m)v_B. Since the trolleys are connected, both trolleys have the same magnitude of velocity. Therefore, the magnitudes of momentum are proportional to their masses combined with their respective velocities:

The momentum of trolley B is greater due to the additional mass of 2 blocks compared to trolley A.

Step 4

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

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Answer

Given the total mass: M=95extkg,m=30extkgM = 95 ext{ kg}, \, m = 30 ext{ kg}

Total mass of trolley A with 2 blocks: M+2m=95+2(30)=155extkgM + 2m = 95 + 2(30) = 155 ext{ kg}

Initial maximum acceleration without friction: amax=mgsin35M+2m=(30)(9.81)sin35155a_{max} = \frac{mg \sin 35^{\circ}}{M + 2m} = \frac{(30)(9.81) \sin 35^{\circ}}{155}

After accounting for the 25% reduction due to friction, a=0.25amaxa = 0.25 \cdot a_{max} \n Distance = 9.0 m, so using the equation: d=12at2d = \frac{1}{2} a t^2 Solving for time tt, the calculation follows: Use the reduced acceleration value.

Step 5

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

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Answer

In 30 minutes, there are 1800 seconds.

Each journey takes (time to travel and unload) 12 s + tt.

Total number of journeys: number of journeys=1800(12+t)\text{number of journeys} = \frac{1800}{(12 + t)}

Where tt is the calculated time from the previous part.

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