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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
Question 8
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
Find the tension in the string during the motion
Answer
To find the tension in the string, we must analyze the forces acting on both particles P and Q.
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For Particle P:
- Weight, ( W_P = m_P g = 1.5g ). The frictional force, ( F_f = \mu W_P = \frac{1}{5}(1.5g) = 0.3g ).
- Using Newton's second law for P, we have: [ T - F_f = m_P a \ T - 0.3g = 1.5a ightarrow T = 1.5a + 0.3g ]
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For Particle Q:
- The weight of Q is ( W_Q = 3g ).
- Applying Newton's second law for Q: [ W_Q - T = m_Q a \ 3g - T = 3a ightarrow T = 3g - 3a ]
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Setting the two equations for T equal gives: [ 1.5a + 0.3g = 3g - 3a \ 4.5a = 2.7g \ a = \frac{2.7g}{4.5} = 0.6g ]
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Plugging a back into T: [ T = 1.5(0.6g) + 0.3g = 0.9g + 0.3g = 1.2g ] Using ( g \approx 9.8 , \text{m/s}^2 ): [ T = 1.2 \times 9.8 \approx 11.76 , \text{N} ightarrow \approx 12N ]
Step 2
Find the magnitude and direction of the resultant force exerted on the pulley by the string
Answer
First, we calculate the tensions acting on both sides of the pulley.
- Tension in the string on side of Particle P is ( T_P = T \approx 12 \text{N} ).
- Tension in the string on side of Particle Q is ( T_Q = T \approx 12 \text{N} ) as they are connected.
To find the resultant force on the pulley, we need to calculate: [ R = \sqrt{T_P^2 + T_Q^2} = \sqrt{T^2 + T^2} = \sqrt{2T^2} = T\sqrt{2} \ R = 12\sqrt{2} \approx 16.97,N ]
Next, to determine the direction:
- The resultant force on the pulley will act at a 45-degree angle from the vertical due to equal tensions on both sides.
Thus, the resultant force exerted on the pulley is approximately ( 16.97N ) at an angle of 45° below the horizontal.