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

For the motion of A along the plane, we resolve the forces:

• $T = 3mg \sin(\alpha) - F$ and also plug in for (F), thus:

$T = 3mg \sin(\alpha) - \frac{1}{6}R$

Where (R) can be found using the relationship (R = 3mg \cos(\alpha)). Thus substituting this into our previous equation:

$T = 3mg \sin(\alpha) - \frac{1}{6}(3mg \cos(\alpha))$

Using (\tan(\alpha) = \frac{3}{4}), we can derive (\sin(\alpha)) and (\cos(\alpha)):

• (\sin(\alpha) = \frac{3}{5}) and (\cos(\alpha) = \frac{4}{5}) yields:

$T = 3mg \cdot \frac{3}{5} - \frac{1}{6}(3mg \cdot \frac{4}{5})$

From here, solving for (a) will yield:

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

The motion of B can be illustrated in a velocity-time graph beginning from rest. The graph will show a linear increase in the velocity of B as A moves down the plane, up until it reaches a maximum velocity just before reaching the pulley. First, label the time on the x-axis and the velocity on the y-axis. As A starts moving, B starts accelerating:

• The slope of the line indicates constant acceleration equal to (\frac{1}{10}g) in the context of A’s motion influencing B, reflecting a linear shape on the graph.
• Therefore, to the instant right before B reaches the pulley, the graph will flatten out, indicating no further increase in velocity as B cannot gain more speed beyond this point without A being further displaced. Make sure to annotate maximum points depending on how time must be considered until B reaches the pulley.

If the string is not light, there would be additional tension due to the weight of the string itself, influencing the net forces acting on both A and B. This means we cannot simply treat tensions with the assumption that they are constant throughout the string. Thus, we would have different tensions in both segments of the string, impacting the derived acceleration value in part (b) since additional tension from the non-negligible weight of the string needs to be factored into the equations of motion.