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 demonstrate that the acceleration of A is ( \frac{1}{10}g ), we first resolve the forces as previously established.

The normal force R for stone A can be determined by resolving forces perpendicular to the plane:

$R = 3mg \cos(α)$

We know that ( \tan(α) = \frac{3}{4} ) simplifies to:

$\sin(α) = \frac{3}{5}, \quad \cos(α) = \frac{4}{5}$

Substituting these into the expression for R gives:

$R = 3mg \cos(α) = 3mg \times \frac{4}{5} = \frac{12}{5}mg$

Substituting back into the equation of motion gives:

$3mg \cdot \frac{3}{5} - \frac{1}{6} \left( \frac{12}{5}mg \right) - T = 3a$

Solve for a and show that after simplification, ( a = \frac{1}{10}g ).

The velocity-time graph for the motion of B will be a straight line that starts from the origin since at the moment A is released, B begins to move upwards with increasing speed.

The slope of the line represents the acceleration of B. Since B is connected to A, its acceleration will match A's deterministically, which we've shown to be ( \frac{1}{10}g ). Therefore, the increase in velocity over time will be constant.

As A descends, B rises, and thus the graph will not touch the x-axis before reaching the pulley, indicating that B is always in motion as A descends. The area under the graph before the line reaches its peak corresponds to the distance traveled by B before reaching the pulley.

The fact that the string is not light suggests that there might be a negligible tension force affecting the acceleration calculation derived in part (b). If the string had mass, the total force affecting movement would include its weight too, which could alter the acceleration of A and subsequently affect the derived results. Hence, the assumption of a light string simplifies the calculation but may not be entirely accurate in practical scenarios, meaning the computed a of ( \frac{1}{10}g ) may be an underestimate.