A small stone A of mass 3m is attached to one end of a string - Edexcel - A-Level Maths Mechanics - Question 2 - 2021 - Paper 1

To find (a), we first resolve the forces acting on A:

1. Resolve the weight of A: $3mg \sin(\alpha) = 3mg \cdot \frac{3}{5} = \frac{9}{5} mg$

2. The normal force (R) is: $R = 3mg \cos(\alpha) = 3mg \cdot \frac{4}{5} = \frac{12}{5} mg$

3. The frictional force is given by: $F = \frac{1}{6} R = \frac{1}{6} \cdot \frac{12}{5} mg = \frac{2}{5} mg$

Now, substituting back into the equation:

$\frac{9}{5}mg - \frac{2}{5}mg - T = 3ma$

Substituting ( T = mg ) (for B) gives:

$\frac{7}{5} mg - mg = 3ma =>\frac{2}{5}mg = 3ma$

So,

$a = \frac{2}{15}g$

Dividing both sides by 3 yields:

$a = \frac{1}{10}g$.

The velocity-time graph for stone B, from the instant A is released until just before it reaches the pulley, would be a straight line starting from the origin (0,0). As A accelerates down the plane, B will ascend. Since the motion starts from rest and increases uniformly in speed, the graph will have a linear gradient, representing constant acceleration until the moment just before reaching the pulley.

• The x-axis represents time (t).
• The y-axis represents the velocity of B (v).

As A moves down, B's velocity increases until it reaches a point where it is about to stop when it reaches the pulley.

The fact that the string is not light introduces additional tension onto the forces acting on B. If the mass of the string were to be considered, it would create a different tension in the string compared to if only particles were taken into account, potentially affecting the acceleration determined in part (b). This could lead to an adjustment in the calculations, particularly in how tensions interact and the resulting motion of both A and B.