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

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

1. For trolley A:

   $T - Mg \sin 35^{\circ} = Ma$

2. For trolley B:

   $T = (M + 2m) a$

   Adding both equations:

   $Mg \sin 35^{\circ} = (M + 2m)a$

   Substituting gives:

   $a = \frac{mg \sin 35^{\circ}}{M + m}$

The momentum of trolley A, $p_A = Mv_A$, and for trolley B, $p_B = (M + 2m)v_B$. 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.

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

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

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

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

In 30 minutes, there are 1800 seconds.

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

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

Where $t$ is the calculated time from the previous part.