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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... 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 need to consider the forces acting on mass A. The forces acting along the incline include:
- The component of gravitational force acting down the plane: ( F_{gA} = 3mg \sin(\alpha) )
- The tension in the string: ( T )
- The frictional force: ( F = \frac{1}{6} R ), where ( R ) is the normal reaction force.
The normal force can be resolved as: [ R = 3mg \cos(\alpha) ] Thus, the frictional force becomes: [ F = \frac{1}{6} (3mg \cos(\alpha)) ]
Combining these forces, we have: [ 3mg \sin(\alpha) - T - \frac{1}{6}(3mg \cos(\alpha)) = 3ma ] This is the equation of motion for A.
Step 2
show that the acceleration of A is \frac{1}{10}g
Answer
Using the previous equation: [ 3mg \sin(\alpha) - T - \frac{1}{6}(3mg \cos(\alpha)) = 3ma ] To find the acceleration, we will substitute ( \sin(\alpha) ) and ( \cos(\alpha) ) using the tangent relation: ( \tan(\alpha) = \frac{3}{4} ) implies ( \sin(\alpha) = \frac{3}{5} ) and ( \cos(\alpha) = \frac{4}{5} ).
Substituting these back into our equation: [ 3mg \left( \frac{3}{5} \right) - T - \frac{1}{6} \left( 3mg \cdot \frac{4}{5} \right) = 3ma ] Simplifying: [ \frac{9mg}{5} - T - \frac{2mg}{5} = 3ma ] [ \frac{7mg}{5} - T = 3ma ]
Now, resolving the forces on mass B, we can relate the tensions: [ T = mg + ma ] This results in a full equation to solve for a. Through rearrangement, we demonstrate: [ a = \frac{1}{10}g ]
Step 3
sketch a velocity-time graph for the motion of B
Answer
In the given scenario, when mass A is released from rest, mass B will start moving downwards immediately due to gravitational pull. The initial velocity of B is zero when A is at rest.
The graph can be described as follows:
- Initial state: At time ( t = 0 ), velocity of B is ( v_0 = 0 ).
- Acceleration: Given that A's acceleration will be constant, B will accelerate uniformly.
- Graph characteristics: The slope of the velocity-time graph represents acceleration, which is constant until B reaches the pulley.
As time progresses from 0 to t before reaching the pulley, the graph will be a straight line increasing linearly from the origin. The equation of motion can be employed here to find specific values.
Step 4
State how this would affect the working in part (b)
Answer
The velocity of B increases as it approaches the pulley, which affects the tension in the string. Since mass A accelerates down the plane, the force equations used in part (b) would need to consider dynamic changes in tension. Consequently, the calculation of acceleration may differ slightly, indicating that tension cannot be assumed constant between A and B, which can impact the derived acceleration value in part (b).