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Two particles P and Q have masses 1.5 kg and 3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2016 - Paper 1

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Two particles P and Q have masses 1.5 kg and 3 kg respectively. The particles are attached to the ends of a light inextensible string. Particle P is held at rest on ... show full transcript

Worked Solution & Example Answer:Two particles P and Q have masses 1.5 kg and 3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2016 - Paper 1

Step 1

a) the tension in the string during the motion

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Answer

To find the tension in the string, we start by considering the forces acting on particle P. The weight of particle P is given by:

R=1.5gR = 1.5g

The frictional force, (F_f), acting against the motion of P is:

Ff=μR=15(1.5g)=0.3gF_f = \mu R = \frac{1}{5}(1.5g) = 0.3g

Using Newton's second law for particle P, we can express the forces horizontally:

TFf=1.5aT - F_f = 1.5a

Substituting the frictional force:

T0.3g=1.5aT - 0.3g = 1.5a

For particle Q, which has a mass of 3 kg, we apply Newton's second law vertically:

3gT=3a3g - T = 3a

Now we have two equations:

  1. (T - 0.3g = 1.5a)
  2. (3g - T = 3a)

We can solve these equations simultaneously for T and a.

Step 2

b) the magnitude and direction of the resultant force exerted on the pulley by the string

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Answer

To find the resultant force on the pulley, we need to determine the tension in the string from the previous part. We already established:

R=T2+(3gT)2R = \sqrt{T^2 + \left(3g - T\right)^2}

The angles involved provide:

R=T2+27cos452R = \sqrt{T^2 + 27 \cos{45^\circ}^2}

Substituting the values leads to:

R=T2+18R = \sqrt{T^2 + 18}

Using the found value of T, we can calculate R and determine its direction as being (45^\circ) below the horizontal.

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