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 P, 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 $ \frac{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

(a) write down an equation of motion for A

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Answer

To derive the equation of motion for stone A, we begin by applying Newton's second law. The forces acting on A include:

The component of weight down the plane: ( W_A = 3mg \sin \alpha )
The tension in the string: ( T )
The frictional force: ( F = \frac{1}{6} R ), where ( R ) is the resultant normal force.

Resolving forces along the plane gives:

$3mg \sin \alpha - T - F = 3ma$

Substituting for ( F ) we get:

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

Next, we must resolve the normal force:

$R = 3mg \cos \alpha$

Thus, the final equation of motion becomes:

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

Step 2

(b) show that the acceleration of A is 1/10 g

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Answer

From the equation derived in part (a), first we resolve the angle α. We know that:

$\tan \alpha = \frac{3}{4}$

Using this, we can find ( \sin \alpha ) and ( \cos \alpha ):

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

Substituting these into the equations, we find:

For the weight component: $3mg \sin \alpha = 3mg \cdot \frac{3}{5} = \frac{9}{5} mg$
For the normal force: $R = 3mg \cos \alpha = 3mg \cdot \frac{4}{5} = \frac{12}{5} mg$
For friction: $F = \frac{1}{6} R = \frac{1}{6} \cdot \frac{12}{5} mg = \frac{2}{5} mg$

Now, substituting these into the equation:

$\frac{9}{5} mg - T - \frac{2}{5} mg = 3ma$ This simplifies to: $\frac{7}{5} mg - T = 3ma$

For equilibrium of the entire system, tension must be equal on both ends of the pulley, and the net motion leads us to solve for acceleration:

Ultimately, solving gives:

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

Step 3

(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.

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Answer

The velocity-time graph for stone B can be sketched based on the acceleration derived in part (b). Initially, when A is released, both stones are at rest, hence the graph starts at the origin (0,0).

As A begins to move down the plane due to gravity, B will start gaining velocity because of the inextensibility of the string. Since the acceleration was found to be ( \frac{1}{10} g ), the graph will be a straight line with a gentle positive slope, indicating uniform acceleration.

The slope of this line represents the acceleration of stone B.

The final point on the graph will represent the velocity of B just before it reaches the pulley. Assuming the distance traveled is moderate, the final velocity will be a value proportional to the acceleration multiplied by the time taken until B reaches the pulley.

Thus the sketch will show a linear increase in velocity over time.

Step 4

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

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Answer

If stone B was not light, the analysis in part (b) would be affected. The tension would vary depending on the mass of B, altering the net forces acting on A. Since we modeled B as a particle with negligible mass, this assumption may lead to incorrect results since a heavier B would create greater tension in the string.

This would change the equations of motion for A and its acceleration would not be ( \frac{1}{10} g ), but would instead depend on the mass of B. Hence, more complex calculations would be required considering the mass of both stones equally.

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

(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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