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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
Write down an equation of motion for A
Answer
To analyze the motion of stone A, we need to consider the forces acting on it. The forces can be expressed as:
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The component of gravitational force acting down the slope: F_{gravity} = 3mg imes rac{3}{5} = rac{9mg}{5} Since tanα = 3/4, we can derive the lengths in a right-angle triangle where the hypotenuse represents the length of the slope.
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The frictional force acting against the motion: F_{friction} = rac{1}{6} R
We can substitute R, which will be the normal reaction at the plane: R = 3mg imes rac{4}{5} = rac{12mg}{5} Therefore: F_{friction} = rac{1}{6} imes rac{12mg}{5} = rac{2mg}{5} -
Consequently, the equation of motion for A can be written as: 3mg rac{3}{5} - rac{2mg}{5} - T = 3a
Combine the terms: 3mg imes rac{3}{5} - T - rac{2mg}{5} = 3a Simplifying the equation gives: 3mg imes rac{1}{5} - T = 3a Thus, the equation of motion for A is: 3mg rac{3}{5} - T = 3a
Step 2
Show that the acceleration of A is 1/10 g
Answer
From the derived equation of motion: 3mg rac{3}{5} - T = 3a Next, we resolve the equations for stone B. The weight of stone B is given by: From this equation, we can find T: We also know that A's motion results in B's motion, thus we plug T back into the equation for A: 3mg rac{3}{5} - (mg + ma) = 3a This rearranges to: 3mg rac{3}{5} - mg - ma = 3a Combine like terms: rac{9mg}{5} - mg = 3a + ma Factoring out m gives: m rac{4g}{5} = (3 + 1)a m rac{4g}{5} = 4a Solving for a, we get: a = rac{4g}{20} = rac{1}{10}g
Step 3
Sketch a velocity-time graph for the motion of B
Answer
To sketch the velocity-time graph for stone B:
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Initial Condition: At time t = 0, stone A is released from rest, which means both A and B start from rest. Therefore, the initial velocity of B is zero.
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Acceleration: The acceleration of both stones is constant, calculated as a = rac{1}{10} g, leading to a constant increase in velocity.
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Graph Characteristics:
- The graph starts at the origin (0,0).
- It has a linear positive slope due to the constant acceleration.
- When A reaches the pulley (t = T), B will have reached a maximum velocity given by the formula . This maximum point can be marked on the graph.
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Graph Sketch: The graph will be a straight line increasing linearly from the origin at a constant rate until it reaches the terminal condition just before B gets to the pulley.
Step 4
State how this would affect the working of part (b)
Answer
If the string is not light, the acceleration calculated in part (b) would need to account for the weight of the string itself. A heavier string would add to the tension in the system, which means:
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The acceleration of stone A would be less than rac{1}{10}g, as a portion of the gravitational force would now serve to support the weight of the string.
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For stone B, the increased tension could also lead to a higher effective mass being taken into account, thus altering both the force analysis and the resulting acceleration calculations.
In conclusion, the presence of a significant mass of string reinforces the necessity to reconsider the equations of motion in detail.