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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 derive the equation of motion for A, we consider the forces acting on it. The gravitational force acting down the plane is given by:

F_{ ext{gravity}} = 3mg imes rac{3}{5} = rac{9mg}{5}

This is broken down into two components: the tension in the string pulling upwards and the frictional force acting opposite to the direction of motion. The frictional force can be calculated using:

F_{ ext{friction}} = rac{1}{6} R

Where the normal reaction R can be calculated as:

R = 3mg imes rac{4}{5} = rac{12mg}{5}

Substituting for the frictional force:

F_{ ext{friction}} = rac{1}{6} \left( \frac{12mg}{5} \right) = \frac{2mg}{5}

Now, using Newton's second law, the equation of motion for A is given by:

3mgsin(α)FFextfriction=3ma3mg \sin(α) - F - F_{ ext{friction}} = 3ma

This can be expressed as:

3mg×35T2mg5=3a3mg \times \frac{3}{5} - T - \frac{2mg}{5} = 3a

Step 2

show that the acceleration of A is 1/10 g

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Answer

From the previous equation:

3mg×35T2mg5=3a3mg \times \frac{3}{5} - T - \frac{2mg}{5} = 3a

We can express the tension T in terms of the mass B. The equation of motion for B is:

T=mgT = mg

Now substituting into the equation for A:

3mg×35mg2mg5=3a3mg \times \frac{3}{5} - mg - \frac{2mg}{5} = 3a

Combining like terms gives:

9mg5mg2mg5=3a\frac{9mg}{5} - mg - \frac{2mg}{5} = 3a [ = \frac{9mg - 5mg - 2mg}{5} = 3a] [ = \frac{2mg}{5} = 3a] [a = \frac{2g}{15} = \frac{1}{10}g]

Step 3

sketch a velocity-time graph for the motion of B

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Answer

For the velocity-time graph of B:

  1. Initial Stage: When A is released, B will start moving up with an initial velocity of 0.
  2. Acceleration: The acceleration of B will be constant due to the constant force acting from A, giving a linear increase in velocity until the maximum velocity just before reaching the pulley.
  3. Graph Characteristics: The graph will start from the origin (0,0) and increase linearly. The slope of the graph represents the acceleration.

As an illustrative point, labelled axes will show velocity on the y-axis and time on the x-axis. The line will be straight until just before reaching the pulley, indicating a constant acceleration.

Explain that the shape represents constant acceleration from the forces acting on A. The timing and position of the graph intend to reflect that B starts with a very low velocity and then continuously accelerates until it touches the pulley.

Step 4

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

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Answer

In part (b), the conditions underline that the system is not just influenced by mass but also by tension differences and friction forces. The situation described shows that as B approaches the pulley, the tension in the string and friction forces could differ, potentially altering the acceleration derived. This notable difference can impact the equation utilized in part (b) by indicating that as A accelerates or decelerates, it reciprocally influences B, thus modifying the velocity and overall dynamic during the process leading to reaching the pulley.

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