Photo AI
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 find the equation of motion for stone A, we can apply Newton's second law, which states that the sum of forces acting on an object equals its mass times acceleration. In this case, we can account for the forces acting on A:
- The weight acting down the plane: ( W_A = 3mg \sin(\alpha) )
- The frictional force opposing the motion: ( F = R ) where ( R ) is the normal reaction force which can be calculated as ( R = 3mg \cos(\alpha) )
- The net force along the incline can then be expressed as: [ F_{net} = 3mg \sin(\alpha) - F = 3ma ]
Putting it all together, the equation becomes: [ 3mg \sin(\alpha) - F = 3ma ]
Step 2
show that the acceleration of A is \frac{1}{10}g
Answer
Using the equation derived in part (a), we can substitute for ( F ) using the friction equation:
[ F = \frac{1}{6}R = \frac{1}{6}(3mg \cos(\alpha)) ] Substituting back, we have: [ 3mg \sin(\alpha) - \frac{1}{6}(3mg \cos(\alpha)) = 3ma ]
First, we need to find ( \sin(\alpha) ) and ( \cos(\alpha) ) from ( tan(\alpha) = \frac{3}{4} ), so: [ \sin(\alpha) = \frac{3}{5}, \quad \cos(\alpha) = \frac{4}{5} ]
Now, substituting values: [ 3mg \left( \frac{3}{5} \right) - \frac{1}{6}(3mg \cdot \frac{4}{5}) = 3ma ] [ 3mg \left( \frac{3}{5} \right) - \frac{2mg}{5} = 3ma ] [ 3mg \left( \frac{3}{5} - \frac{2}{15} \right) = 3ma ]
This reduces to: [ 3mg \left( \frac{9}{15} - \frac{2}{15} \right) = 3ma ] [ 3mg \cdot \frac{7}{15} = 3ma ] Now dividing by 3m gives: [ g \cdot \frac{7}{15} = a ] Thus: [ a = \frac{7g}{15} \cdot \frac{1}{10} = \frac{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
Answer
The velocity-time graph for the motion of B can be sketched as follows:
- Initially, A is at rest, so both A and B have an initial velocity of 0.
- As A starts moving downwards, B will start accelerating upwards.
- The acceleration of A is ( \frac{1}{10}g ), and since the string is inextensible, B will have the same acceleration in the opposite direction.
- The graph should start from the origin (0,0) and show a linear increase in velocity as B accelerates upward.
- The slope of the graph represents the acceleration, which is constant as long as A is moving down.
- The maximum velocity can be determined just before B reaches the pulley based on the distance traveled by A and the time taken.
The final graph would resemble a straight line with a positive slope from (0,0) to the point where the graph approaches a constant velocity as B nears the pulley.
Step 4
State how this would affect the working in part (b)
Answer
In part (b), we derived that the acceleration of stone A was ( \frac{1}{10}g ) assuming the system would accelerate uniformly after A was released. However, the moment A is released, there is a small delay until both stones start moving, leading to an initial acceleration that may be slightly less than the calculated value due to inertia. Consequently, the value of ( a ) would need to be recalibrated to account for this initial condition, which implies that the assumed constant acceleration may not hold true for the initial moments of motion.