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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

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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

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Answer

To derive the equation of motion for stone A, we analyze the forces acting on it along the inclined plane. The forces acting on A are the gravitational component down the slope and the frictional force opposing the motion.

The gravitational force acting down the slope can be expressed as:

Fgravity=3mgsin(α)F_{gravity} = 3mg \sin(\alpha)

The frictional force opposing the motion is given by:

Ffriction=16RF_{friction} = \frac{1}{6}R

Where ( R = 3mg \cos(\alpha) ) is the normal reaction force. Hence, substituting this in:

Ffriction=16(3mgcos(α))F_{friction} = \frac{1}{6}(3mg \cos(\alpha))

The equation of motion can thus be established as:

3mgsin(α)FfrictionT=3ma3mg \sin(\alpha) - F_{friction} - T = 3ma

Simplifying further will yield the desired equation.

Step 2

Show that the acceleration of A is \( \frac{1}{10}g \)

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Answer

To find the acceleration of stone A, we continue from the equation of motion derived earlier:

3mgsin(α)16(3mgcos(α))T=3ma3mg \sin(\alpha) - \frac{1}{6}(3mg \cos(\alpha)) - T = 3ma

Substituting ( \tan(\alpha) = \frac{3}{4} ) gives us the components for ( \sin(\alpha) ) and ( \cos(\alpha) ):

( \sin(\alpha) = \frac{3}{5} ) and ( \cos(\alpha) = \frac{4}{5} ).

Now substituting these values into the equation yields:

T=3mgsin(α)16(3mgcos(α))3maT = 3mg \sin(\alpha) - \frac{1}{6}(3mg \cos(\alpha)) - 3ma

After simplifying, we find that:

a=110ga = \frac{1}{10}g

Step 3

Sketch a velocity-time graph for the motion of B

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Answer

To create a velocity-time graph for stone B, we note that B will be stationary at first when A is released. As A begins to accelerate downwards due to gravity, B will also start to move upwards with increasing velocity. The graph will start at the origin (0,0) and will show a linear increase in velocity until just before B reaches the pulley, which would be at a consistent acceleration.

The graph can be illustrated as follows:

  • Starting point (0,0): B starts from rest.
  • Increasing slope: As A accelerates, B accelerates upwards, leading to a straight line increasing linearly.
  • Y-intercept represents the velocity of B just before reaching the pulley.

In essence, the graph reflects a constant acceleration pattern as A is released and continues to move until B is about to reach the pulley.

Step 4

State how this would affect the working in part (b)

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Answer

The acceleration of A being calculated as ( \frac{1}{10}g ) implies that the movement of B is directly proportional to the kinematic relationship between A and B. If the string causes an instantaneous change before reaching the pulley, it may suggest a shift in tension values affecting A's motion differently than accounted for, thus requiring a reevaluation of forces at the point of B reaching the pulley.

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