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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 derive the equation of motion for stone A, we start by applying Newton's second law. The forces acting on A are:
- The gravitational force component down the slope: F_g = 3mg imes rac{3}{5}, since an heta = rac{3}{4} implies .
- The frictional force acting against the motion: F_f = rac{1}{6} R, where R is the normal reaction.
The normal reaction R can be resolved as: R = 3mg imes rac{4}{5} Substituting R into the frictional force gives: F_f = rac{1}{6} (3mg imes rac{4}{5}) The net force (F_net) acting on A is: F_{ ext{net}} = 3mg imes rac{3}{5} - rac{1}{6}R Thus, the equation of motion for A can be written as: 3mg imes rac{3}{5} - rac{1}{6}(3mg imes rac{4}{5}) - T = 3ma
Step 2
b) show that the acceleration of A is 1/10 g
Answer
From the equation of motion derived in part (a), we can express it as: 3mg imes rac{3}{5} - rac{1}{6}(3mg imes rac{4}{5}) = 3ma + T Now, when B is also considered, the tension T can be expressed as: Substituting this back into the equation leads to: 3mg imes rac{3}{5} - rac{1}{6}(3mg imes rac{4}{5}) - mg = 3ma After simplifying and solving for a, we get: a = rac{1}{10}g
Step 3
c) sketch a velocity-time graph for the motion of B
Answer
To sketch the velocity-time graph for stone B, start by noting that:
- When A is released, B begins to accelerate upwards. Initially, the velocity is zero.
- Stone B’s acceleration is constant at rac{1}{10}g due to the pulley's smooth nature.
- The graph will be a straight line starting from the origin (0,0) with a slope reflecting the constant acceleration.
- At the moment B reaches the pulley, the velocity will just have increased linearly according to the relationship: .
Thus, the graph will extend from the origin upward with a consistent slope until the point where A reaches the pulley.
Step 4
d) State how this would affect the working of part (b)
Answer
If B is not light, the tension would differ due to its mass. In part (b), the calculated acceleration assumes that the mass of B does not significantly affect the overall system. If B has considerable mass, the tension T would not equal mg, and thus the derived acceleration for A would not hold true. Consequently, part (b) would actually produce a different acceleration that accounts for the mass of B.