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

Given that there is no friction, we can assume the net force acting on trolley B is equal to its mass times acceleration:

1. Applying Newton's second law:

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

   and for trolley B:

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

   Combining the two gives:

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

The momentum of a body is given by the product of its mass and velocity:

1. For trolley A loaded with two blocks:
   • Mass = $M + 2m$
   • Momentum $p_A = (M + 2m)v_A$
2. For trolley B loaded with no blocks:
   • Mass = $M + 0m = M$
   • Momentum $p_B = Mv_B$

Both trolleys will have different velocities due to the mass difference, hence:

$p_A > p_B$ as more mass results in greater momentum, assuming velocities are equal.

Using the distance formula:

Given:

• Distance, $d = 9.0 m$
• For maximum acceleration:

$a = \frac{mg \sin 35^{\circ}}{M + 2m}$

After calculating the acceleration: $a ≈ 3.33 m/s^2$

Using the second equation of motion:

$d = ut + \frac{1}{2}at^2$ (initial velocity $u=0$)

This gives:

$9 = \frac{1}{2} \cdot 3.33 \cdot t^2$

Rearranging:

$t^2 = \frac{18}{3.33} \implies t ≈ 2.37 s$.

Total time for one cycle is:

1. Time to move down = 2.37 s
2. Time to reload = 12 s

Total Time per journey = $2.37 + 12 = 14.37 s$

Total number of cycles in 30 minutes:

$\frac{1800 s}{14.37 s} ≈ 125$

Thus, the total number of blocks: $number ≈ 125 imes 2 = 250$