Two particles P and Q have masses 0.3 kg and m kg respectively. The particles are attached to the ends of a light extensible string. The string passes over a small smooth pulley which is fixed at the top of a fixed rough plane. The plane is inclined to the horizontal at an angle α, where tan α = 3/4. The coefficient of friction between P and the plane is 1/2. The string lies in a vertical plane through a line of greatest slope of the inclined plane. The particle P is held at rest on the inclined plane and the particle Q hangs freely below the pulley with the string taut, as shown in Figure 2. The system is released from rest and Q accelerates vertically downwards at 1.4 m s². Find: (a) the magnitude of the normal reaction of the inclined plane on P, (b) the value of m, (c) when the particles have been moving for 0.5 s, the string breaks. Assuming that P does not reach the pulley, find the further time that elapses until P comes to instantaneous rest.

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Two particles P and Q have masses 0.3 kg and m kg respectively - Edexcel - A-Level Maths Mechanics - Question 6 - 2011 - Paper 1

Question 6

Two particles P and Q have masses 0.3 kg and m kg respectively. The particles are attached to the ends of a light extensible string. The string passes over a small s... show full transcript

Worked Solution & Example Answer:Two particles P and Q have masses 0.3 kg and m kg respectively - Edexcel - A-Level Maths Mechanics - Question 6 - 2011 - Paper 1

Step 1

(a) the magnitude of the normal reaction of the inclined plane on P

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Answer

To find the normal reaction, we need to apply the equations of motion considering the forces acting on particle P. The forces acting on mass P include:

The weight component acting down the incline: $mg \sin(\alpha)$ .
The normal reaction (R) perpendicular to the inclined plane.

The equation can be derived from balancing forces:

$R = mg \cos(\alpha)$

For particle P with mass 0.3 kg,

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

Substituting values:

$R = 0.3 \cdot 9.8 \cdot \cos(\alpha) = 0.3 \cdot 9.8 \cdot \frac{4}{5} = 0.24 \cdot 9.8 = 2.4 \, \text{N}.$

Step 2

(b) the value of m

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Answer

By considering the forces when the system is released, we can apply Newton's second law. The total force acting on the system must equal mass times acceleration:

For particle Q:

$T - mg = -ma \implies T = mg - ma,$ where a = 1.4 m/s².

For particle P, considering the downhill forces:

$mg \sin(\alpha) - T = 0.3 \cdot 1.4$

Substituting T from above into this equation, we eliminate R and T, leading to:

$0.3 g \sin(\alpha) = F = 0.3 \cdot 1.4.$

Solving simultaneously will yield the value of m. After calculations, we can find:

$m = 0.4 \, \text{kg}.$

Step 3

(c) when the particles have been moving for 0.5 s, the string breaks. Assuming that P does not reach the pulley, find the further time that elapses until P comes to instantaneous rest.

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Answer

When the string breaks, the only forces acting on P will be its weight and the friction force. The acceleration of P can be calculated by:

For particle P:

The equation of motion after the string breaks takes into account both forces:

$0.3g - F = 0.3a \. \implies 0.3g - 0.3 \times 9.8 \sin(\alpha) = 0.3a,$

By substituting for a:

The downward force is reduced by friction, leading to a net acceleration:
Assuming g ~ 9.8 m/s².
Then use this to find the velocity after 0.5 s:

$v = 1.4 \times 0.5 = 0.7 \, \text{m/s}.$

After the break, the deceleration will be calculated and the time taken to stop can be determined by finding:

$t = \frac{v}{a} = \frac{0.7}{0.3 \cdot g - 0.3a}.$

After substituting the values, we can find: $t = 0.071 s \text{ or } 0.0714 s.$

Two particles P and Q have masses 0.3 kg and m kg respectively - Edexcel - A-Level Maths Mechanics - Question 6 - 2011 - Paper 1

Worked Solution & Example Answer:Two particles P and Q have masses 0.3 kg and m kg respectively - Edexcel - A-Level Maths Mechanics - Question 6 - 2011 - Paper 1

(a) the magnitude of the normal reaction of the inclined plane on P

(b) the value of m

(c) when the particles have been moving for 0.5 s, the string breaks. Assuming that P does not reach the pulley, find the further time that elapses until P comes to instantaneous rest.

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