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Two particles, A and B, have masses 2m and m respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2017 - Paper 1

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Two particles, A and B, have masses 2m and m respectively. The particles are attached to the ends of a light inextensible string. Particle A is held at rest on a fix... show full transcript

Worked Solution & Example Answer:Two particles, A and B, have masses 2m and m respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2017 - Paper 1

Step 1

Write down an equation of motion for A.

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Answer

For particle A, the forces acting on it along the horizontal direction are the tension in the string (T) and the frictional force (F) opposing the motion. Therefore, the equation of motion can be expressed as:

TF=2maT - F = 2ma

where F is the friction force given by F = μ × (2m)g (the weight of particle A). Thus, we can rewrite this as:

Tμimes2mg=2maT - μ imes 2mg = 2ma.

This leads to: T=2ma+μimes2mgT = 2ma + μ imes 2mg.

Step 2

Write down an equation of motion for B.

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Answer

For particle B, the forces acting on it include the gravitational force downwards (mg) and the tension in the string (T) acting upwards. The equation of motion for B can thus be expressed as:

mgT=mamg - T = ma.

Step 3

Hence show that, until B hits the floor, the acceleration of A is g/3(1 - 2μ).

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Using the equations derived for A and B, we can substitute for T from the second equation into the first:

Substituting T=mgmaT = mg - ma into the equation for A gives:

mgmaμimes2mg=2mamg - ma - μ imes 2mg = 2ma.

Rearranging gives:

mgμimes2mg=5mamg - μ imes 2mg = 5ma.

Factoring out the m:

g(12μ)=5ag(1 - 2μ) = 5a.

Thus the acceleration a of A can be simplified to:

a = rac{g}{5}(1 - 2μ).

Step 4

Find, in terms of g, h and μ, the speed of A at the instant when B hits the floor.

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Answer

The speed of A at the instant B hits the floor can be derived using the kinematic equation based on its acceleration. The velocity of A, denoted as v, can be calculated using:

v2=u2+2asv^2 = u^2 + 2as,

where u = 0 (initial velocity), a = rac{g}{5}(1 - 2μ), and s = d - h.

Substituting these values into the equation gives:

v^2 = 0 + 2 imes rac{g}{5}(1 - 2μ)(d - h).

Thus, v = rac{g}{5}(1 - 2μ) imes ext{distance}.

Step 5

Describe what would happen if μ = 1/2.

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Answer

If μ = rac{1}{2}$, then the frictional force is reduced. This reduction allows particle A to accelerate more easily. Thus the tension in the string may not be sufficient to balance the gravitational pull on B, leading to a scenario where either particle A does not move or both particles remain in a state of limiting equilibrium. Essentially, the dynamics of the system would change significantly with A being able to slide more freely along the table.

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