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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
Question 2
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
Answer
To find the equation of motion for stone A, we apply Newton's second law. The forces acting on A are the gravitational force component down the slope and the frictional force opposing the motion.
The equation can be expressed as:
where:
- is the tension in the string,
- is the frictional force which can be calculated as , and is the normal reaction force on A.
Step 2
b) show that the acceleration of A is \( \frac{1}{10}g \)
Answer
To demonstrate that the acceleration of A is ( \frac{1}{10}g ), we resolve the forces perpendicular to and parallel to the plane.
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Resolve the weight of A:
- The component of weight parallel to the slope is (3mg \sin(\alpha) = 3mg \cdot \frac{3}{5}) because (\sin(\alpha) = \frac{3}{5}).
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Calculate the normal force and frictional force:
- The normal reaction (R = 3mg \cos(\alpha) = 3mg \cdot \frac{4}{5}).
- Frictional force (F = \frac{1}{6}R = \frac{1}{6} \cdot 3mg \cdot \frac{4}{5} = \frac{2mg}{5}
-
Substitute into the equation of motion:
[3mg \cdot \frac{3}{5} - \frac{2mg}{5} - T = 3ma]
- Solve for acceleration:
- Simplifying gives: [\frac{9mg}{5} - \frac{2mg}{5} - T = 3ma] [\frac{7mg}{5} - T = 3ma]
Considering only the system's motion, we equate: (T = 2mg \quad \text{(based on system equilibrium)})
Therefore: [\frac{7mg}{5} = 3ma + 2mg] Which leads us to: [a = \frac{1}{10}g]
Step 3
c) sketch a velocity-time graph for the motion of B
Answer
The velocity-time graph for B will reflect uniform acceleration starting from rest.
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Since A starts moving down the slope, B will also begin to move upwards due to the tension in the string, gaining speed steadily.
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The graph will be linear: at time (t=0), the velocity will be zero; as time progresses until B reaches the pulley, the velocity will increase uniformly until it reaches its maximum just before hitting the pulley, where it has constant acceleration until terminal speed.
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The slope of the graph represents the acceleration of B, which is constant until the pulley is reached. Thus the graph will start at the origin and show a straight line with a positive slope that levels off just before point P.
Step 4
d) State how this would affect the working in part (b)
Answer
If the string is not light, the tension would vary depending on the mass of B in motion and the acceleration predicted. This means that our calculations assume a tension that would not account for the weight of B. If B has significant mass, the equation of motion derived would need revision, factoring the increased tension impacting the acceleration values for A.