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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 derive the equation of motion for A, we can start by analyzing the forces acting on A. The forces include:
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The gravitational force down the plane:
F_g = 3mg imes rac{3}{5} = rac{9mg}{5}
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The frictional force opposing the motion:
F_f = rac{1}{6} R
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The tension in the string:
Applying Newton’s second law along the direction of motion:
3mg rac{3}{5} - F_f - T = 3a
Rearranging, we have:
T = 3mg rac{3}{5} - F_f - 3a
Substituting for the frictional force gives us:
T = 3mg rac{3}{5} - rac{1}{6} R - 3a = 3mg rac{3}{5} - rac{1}{6} (3mg rac{4}{5})
Thus, the equation of motion is:
3mg rac{3}{5} - T - rac{1}{6} imes 3mg rac{4}{5} = 3a
Step 2
show that the acceleration of A is 1/10 g
Answer
To find the acceleration of A, we resolve the forces acting on both stones:
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From the forces acting on A, we can write:
3mg rac{3}{5} - T - rac{2mg}{10} = 3a
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For stone B, the forces acting are:
Substituting the expression for T into the equation for A:
3mg rac{3}{5} - mg - rac{2mg}{10} = 3a
From the above, we can manipulate the equation:
3a = rac{3mg}{5} - mg - rac{2mg}{10}
Combining terms gives:
3a = rac{6mg - 5mg - 2mg}{10}
This simplifies to:
3a = rac{(6-5-2)mg}{10} = rac{-1mg}{10}
Therefore:
a = rac{1}{10}g
Thus, we have shown that the acceleration of A is indeed rac{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
Answer
To sketch the velocity-time graph for stone B:
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Initial Condition: When A is released from rest, B is also at rest; hence, both start off with zero velocity.
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Acceleration: Given that B accelerates as A starts moving down the incline, the velocity of B will increase over time. The acceleration of A is rac{1}{10}g, which translates into a constant acceleration for B, based on their connection via the pulley.
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Graph Characteristics:
- The graph will be a straight line starting from the origin (0,0).
- Since we expect constant acceleration, the slope of this line will be consistent until just before B reaches the pulley.
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Just Before B Reaches the Pulley: As B approaches the pulley, the velocity will increase but the time duration until it reaches the pulley will also define the terminal point on the length of the graph.
In summary, the velocity-time graph will be linear as it depicts a constant acceleration in stone B, leading to a straight line from the origin to the point just before the pulley.
Step 4
State how this would affect the working in part (b)
Answer
This would affect working in part (b) because if stone A is accelerating, then the tension in the string would lead to a difference in the forces acting upon the two stones. Specifically, the acceleration determined for A in part (b) is based on the assumption of unidirectional motion without stalling effects. If B accelerates as A releases, then the value of T in the tension equation would change thereby possibly altering the acceleration calculated. Furthermore, any assumption of constant values could shift due to varying slopes, and this must be incorporated into any calculations, reflecting a need for possible reevaluation of the acceleration results.