Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string. Initially B is held at rest on a rough fixed plane inclined at angle θ to the horizontal, where $ an \theta = \frac{5}{12}$. The part of the string from B to P is parallel to a line of greatest slope of the plane. The string passes over a small smooth pulley, P, fixed at the top of the plane. The particle A hangs freely below P, as shown in Figure 4. The coefficient of friction between B and the plane is $rac{2}{3}$. The particles are released from rest with the string taut and B moves up the plane. (a) Find the magnitude of the acceleration of B immediately after release. (b) Find the speed of B when it has moved 1 m up the plane. (c) When B has moved 1 m up the plane the string breaks. Given that in the subsequent motion B does not reach P,

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Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Question 7

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Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string. Initially B is held at rest on a rough fixed plan... show full transcript

Worked Solution & Example Answer:Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Step 1

Find the magnitude of the acceleration of B immediately after release.

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Answer

First, we need to analyze the forces acting on both particles A and B.

For particle A: The equation of motion can be given as: $7g - T = 7a$ where $T$ is the tension in the string.
For particle B: The forces acting on it (along the incline) are: $T - F - 3g \sin \theta = 3a$ where $F$ is the frictional force and can be expressed as: $F = \mu R = \frac{2}{3} R$ The normal reaction $R$ is: $R = 3g \cos \theta$ Therefore: $F = \frac{2}{3}(3g \cos \theta) = 2g \cos \theta$
Substituting this into our equation for B: From the equation: $T - (2g \cos \theta) - 3g \sin \theta = 3a$ and the $\tan \theta$ value simplifies this calculation further.
Eliminate $T$ between the two equations to find $a$ : Rearranging gives: $7g - 2g \frac{12}{13} - 3g \frac{5}{13} = 7a$ Simplifying results in: $a = \frac{2g}{5}$ Using $g = 9.8$ , the calculated acceleration is approximately: $a \approx 3.92 \, \text{m/s}^2$ .

Step 2

Find the speed of B when it has moved 1 m up the plane.

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Using the equations of motion:

Initial speed $u = 0$ (as it starts from rest)
Distance $s = 1 \, \text{m}$
Acceleration found previously is $a = \frac{2g}{5} = \frac{2 \times 9.8}{5}$ ,

The kinematic equation to use is: $v^2 = u^2 + 2as$ Plugging in the values: $v^2 = 0 + 2 \times \left(\frac{2 \times 9.8}{5}\right) \times 1$ Solving gives: $v \approx 2.8 \, \text{m/s}$ .

Step 3

Find the time between the instants when the string breaks and B comes to instantaneous rest.

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At the point when the string breaks, the equation of motion for B can be written as: $-F + 3g \sin \theta = 3a$ This becomes: $\frac{2}{3}(3g \frac{12}{13}) + 3g \frac{5}{13} = -3a$ Therefore, $2g \frac{12}{13} + 3g \frac{5}{13} = -3a$ Solving this yields: $a = -g$

Now, use $v = u + at$ to find time: When $B$ reaches a speed $v = 0$ initial speed when the string breaks: We have: $2 = 0 - gt$ Solving this gives: $t = \frac{2}{7} \approx 0.286 \, \text{s}$ .

Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Worked Solution & Example Answer:Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Find the magnitude of the acceleration of B immediately after release.

Find the speed of B when it has moved 1 m up the plane.

Find the time between the instants when the string breaks and B comes to instantaneous rest.

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