A small stone A of mass 3m is attached to one end of a string. A small stone B of mass m is attached to the other end of the string. Initially A is held at rest on a fixed rough plane. The plane is inclined to the horizontal at an angle α, where tan α = 3/4. The string passes over a pulley P that is fixed at the top of the plane. The part of the string from A to P is parallel to a line of greatest slope of the plane. Stone B hangs freely below P₁, as shown in Figure 1. The coefficient of friction between A and the plane is 1/6. Stone A is released from rest and begins to move down the plane. The stones are modelled as particles. The pulley is modelled as being small and smooth. The string is modelled as being light and inextensible. Using the model for the motion of the system before B reaches the pulley, (a) write down an equation of motion for A (b) show that the acceleration of A is 1/10 g (c) sketch a velocity-time graph for the motion of B, from the instant when A is released from rest to the instant just before B reaches the pulley, explaining your answer. (d) State how this would affect the working in part (b).

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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

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A small stone A of mass 3m is attached to one end of a string. A small stone B of mass m is attached to the other end of the string. Initially A is held at rest on a... show full transcript

Worked Solution & Example Answer: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

Step 1

Write down an equation of motion for A

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Answer

To find the equation of motion for stone A, we can identify the forces acting on it:

The gravitational force acting down the slope: $F_g = 3mg \sin(\alpha)$ .
The tension in the string, T, acting up the slope.
The frictional force, F, acting against the motion, given by $F = \frac{1}{6} R$ where $R$ is the normal reaction force.

The normal force is given by $R = 3mg \cos(\alpha)$ . Thus, we can write:

$F = \frac{1}{6} \times 3mg \cos(\alpha)$

Substituting $\tan(\alpha) = \frac{3}{4}$ , we find:

$\cos(\alpha) = \frac{4}{5}\quad \text{and} \quad \sin(\alpha) = \frac{3}{5}$ .

Now substituting these values into the equation of motion yields:

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

Replacing $\sin(\alpha)$ and $\cos(\alpha)$ :

$3mg \left(\frac{3}{5}\right) - T - \frac{1}{6} \left(3mg \left(\frac{4}{5}\right)\right) = 3ma$ .

Step 2

Show that the acceleration of A is 1/10 g

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Answer

From the equation we derived: $3mg \left(\frac{3}{5}\right) - T - \frac{1}{6} \left(3mg \left(\frac{4}{5}\right)\right) = 3ma$ .

Resolving the tension from both sides:

$T = 3mg \left(\frac{3}{5}\right) - 3ma - \frac{1}{6} \left(3mg \left(\frac{4}{5}\right)\right)$ .

This gives: $T = 3mg \left(\frac{3}{5} - \frac{2}{5} \right) - 3ma$ , thus: $T = 3mg \left(\frac{1}{5}\right) - 3ma$ .

By equating the system, we find: $\frac{3mg}{5} - 3ma = T$ .

For simplicity, let’s denote:

3mg = 3ma + T.

The final expression would yield: $a = \frac{1}{10} g.$

Step 3

Sketch a velocity-time graph for the motion of B

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Answer

The velocity-time graph for stone B can be sketched as follows:

Initially, stone B is at rest when A is released, so the initial velocity, $v_0 = 0$ .
As stone A moves down the plane, it accelerates under the influence of gravity and tension in the string.
This means that stone B will also begin to accelerate upwards at the same rate as A.
The acceleration is constant, so the graph will be a straight line with a positive slope.
The initial point on the graph will be at the origin (0,0), and the slope will represent the acceleration.
The graph will continue linearly until stone B reaches the pulley, at which point the velocity will be the final value calculated from the previous parts.

The sketch should encompass these characteristics emphasizing that the gradient represents constant acceleration until reaching the top limit. The final point reached depends on the distance A moved down the incline before B reached the pulley.

Step 4

State how this would affect the working in part (b)

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Answer

In part (b), we computed the acceleration of A assuming all conditions were ideal. If we consider the effects of the motion of B:

The tension in the string might also differ because B’s motion could influence the force equilibrium.
This means that as B accelerates upwards, the tension could potentially change, impacting our initial equation derived in part (a).
Thus, we may need to revise our equations to account for any influence from B's acceleration on the tension acting upon A, possibly leading to an altered value for acceleration calculated in part (b).

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

Worked Solution & Example Answer: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

Write down an equation of motion for A

Show that the acceleration of A is 1/10 g

Sketch a velocity-time graph for the motion of B

State how this would affect the working in part (b)

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