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

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: $(M + m)g \sin(35^\circ) = (M + m)a$ Thus, simplifying yields: $a = \frac{mg \sin 35^\circ}{M + m}$

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: $p_A = (M + 2m)v$ And the momentum of trolley B is: $p_B = Mv$ Here, $p_A$ is greater than $p_B$ since $M + 2m > M$.

Given:

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

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