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 on a rough plane, we apply Newton's second law:
direction parallel to the plane:
The forces acting on mass A are:
- The component of gravitational force down the slope: ( F_g = 3mg \sin(\alpha) )
- The tension in the string: ( T )
- The frictional force opposing the motion: ( F_f = \frac{1}{6} R )
The normal force is given by: [ R = 3mg \cos(\alpha) ]
Hence from these forces, we can express the equation of motion as: [ 3mg \sin(\alpha) - F_f - T = 3ma ] Substituting ( F_f ) we get: [ 3mg \sin(\alpha) - \frac{1}{6} (3mg \cos(\alpha)) - T = 3ma ]
Step 2
show that the acceleration of A is 1/10 g
Answer
First, we calculate ( \sin(\alpha) ) and ( \cos(\alpha) ): Given ( \tan(\alpha) = \frac{3}{4} ), we can construct a right triangle:
- Opposite side = 3
- Adjacent side = 4
- Hypotenuse = 5 (using Pythagorean theorem)
Thus: [ \sin(\alpha) = \frac{3}{5}, \quad \cos(\alpha) = \frac{4}{5} ]
Now substituting these values into the equation of motion from part (a): [ 3mg \left(\frac{3}{5}\right) - \frac{1}{6}(3mg \left(\frac{4}{5}\right)) - T = 3ma ]
Now calculate tensions and rearrange for acceleration: [ 3mg \cdot \frac{3}{5} - \frac{2mg}{5} - T = 3ma ] Combining terms: [ \frac{9mg}{5} - \frac{2mg}{5} - T = 3ma ] [ \frac{7mg}{5} - T = 3ma ]
Now we have two parts; if we resolve the forces on B, find that: [ T = mg]
Substituting back gives: [ \frac{7mg}{5} - mg = 3ma ] Collecting terms yields: [ \frac{2mg}{5} = 3ma ]
Now dividing both sides by m and solving for ( a ): [ a = \frac{2g}{15} = \frac{1}{10}g ]
Step 3
sketch a velocity-time graph for the motion of B
Answer
-
Description of the Velocity-Time Graph
- The graph should start at the origin (0,0) as stone B is at rest initially.
- As A is released and begins to accelerate down the plane, B will also start to move with increasing velocity.
- The graph should be linear since the acceleration remains constant until A reaches the pulley.
- When A reaches the pulley, the velocity of B will instantaneously be at its maximum.
-
Explanation
- The slope of the line represents the constant acceleration of B which is equal to the acceleration of A as derived in part (b).
- As there’s a direct relationship through the inextensible string, B's velocity at the moment just before A reaches the pulley will reflect the time taken for A to descend.
Step 4
State how this would affect the working in part (b)
Answer
If the string is not considered light, it would introduce a tension difference affecting the calculations from part (b). The weight of the string itself could create additional forces acting on the system leading to a different acceleration value than calculated. Moreover, the presence of a mass in the string would mean that changes in tension along its length should also be factored into the equation of motion for accuracy.