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

From the equation established earlier, we can substitute the values of friction and normal reaction force:

1. Substitute $F = \frac{1}{6} R$ into the equation of motion:

2. Since $R = 3mg \cos(\alpha)$, we replace it in the equation:

3. Substituting $\sin(\alpha)$ and $\cos(\alpha)$ derived from $\tan(\alpha) = \frac{3}{4}$, we find:

4. Thus,

5. This leads to:

The velocity-time graph for the motion of stone B can be sketched based on the principles of constant acceleration.

• Initially, stone B is at rest. When stone A is released, stone B begins to move downward as A moves down the plane.
• Since the system is set up in such a way that the motion of A influences B, we can describe the velocity of B steadily increasing.
• The graph is a straight line starting from the origin, indicating constant acceleration. The slope of this line represents the acceleration ( ( \frac{1}{10} g )).

It will show time on the x-axis and velocity on the y-axis, with a linear increase until the moment just before B reaches the pulley.

The fact that the string is not light introduces a significant alteration to the system. The tension in the string would vary depending on the weight of the stones and the acceleration of the entire system. Stone B would experience a different tension as compared to A due to its different mass and position in the pulley system. This would affect the derived acceleration formula of A in part (b) as it becomes dependent on these tension values, leading to potential discrepancies in assuming a constant acceleration for both stones.