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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 stone A, we can consider the forces acting on it. The forces include:

  1. The gravitational force along the plane:

    F_g = 3mg imes rac{3}{5}

    where the 3 in the numerator represents mass A (3m) and the angle of incline is given by the tangent function.

  2. The frictional force opposing the motion:

    F_f = rac{1}{6} R

    where R is the normal reaction force, defined as:

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

    using the cosine of angle α to resolve it perpendicular to the plane.

The equation of motion can be expressed as:

3mg imes rac{3}{5} - F_f - T = 3ma

Thus, the equation becomes:

3mg imes rac{3}{5} - rac{1}{6} R - T = 3ma.

Step 2

show that the acceleration of A is \frac{1}{10}g

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Answer

First, we know from the previous part the equation:

3mg imes rac{3}{5} - \frac{1}{6} R - T = 3ma

Next, substituting the value of R:

R=3mg×45R = 3mg \times \frac{4}{5}

When substituted, we derive:

3mg×3516(3mg×45)T=3a3mg \times \frac{3}{5} - \frac{1}{6} \left(3mg \times \frac{4}{5}\right) - T = 3a

Resolving the equation further yields:

T=3mg×(3525)T = 3mg \times \left(\frac{3}{5} - \frac{2}{5}\right)

Thus on simplifying:

T=3mg10T = \frac{3mg}{10}

Therefore, the acceleration a can be expressed in terms of g as:

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

Step 3

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 motion of stone B can be visualized on a velocity-time graph of the form:

  1. Initial conditions: When stone A is released, stone B begins with zero velocity (u = 0).
  2. Acceleration: Since the acceleration of A is constant at \frac{1}{10}g, this also affects the velocity of B as the two stones are connected.
  3. Graph representation: The graph starts at the origin (0,0) and rises linearly with a positive slope, representing constant acceleration until B reaches the pulley.
  4. Final velocity: The final velocity right before B reaches the pulley is the last point on the graph, marked based on the time taken to reach this distance.

In conclusion, the velocity-time graph will be a straight line for the motion of B, depicting continuous acceleration.

Step 4

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

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Answer

The working of part (b) would be affected as follows:

  1. Since stone A is accelerating due to gravity, it affects tension experienced by both stones.
  2. The equation derived is dependent on the uniform acceleration of stone A.
  3. If the system is not ideal (e.g., friction was higher or the string was not inextensible), the derived acceleration value would change.
  4. Therefore, any change in the pulling force or resistance would alter the effectiveness of the calculated acceleration of stone A, leading to inaccuracies in applying the formula derived for the motion.

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