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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 inextensible string. The string passes over a small... 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
Answer
To find the normal reaction ( R ), we can use the equation of forces acting on particle P along the inclined plane:
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The forces acting on P are:
- Gravitational force component along the plane: ( 0.3g \sin \alpha )
- Normal reaction ( R = 0.3g \cos \alpha )
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Given that ( g \approx 9.8 , \text{m/s}^2 ) and with ( \tan \alpha = \frac{3}{4} ), we can find ( \sin \alpha ) and ( \cos \alpha ):
- ( \sin \alpha = \frac{3}{5} ) and ( \cos \alpha = \frac{4}{5} ).
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Compute ( R ): [ R = 0.3 \times 9.8 \times \frac{4}{5} = 0.24 \times 4 = 0.96 , \text{N} ] (approximately 2.4 N when rounded to 2 significant figures).
Step 2
(b) the value of m.
Answer
To find the mass ( m ), we can set up the equation of motion for mass Q:
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For mass Q, we have: [ mg - T = ma ] where ( a = 1.4 , \text{m/s}^2 ).
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For mass P, the tension ( T ) can be expressed using the forces along the inclined plane: [ T - 0.3g \sin \alpha = 0.3a ]
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Substitute ( \sin \alpha = \frac{3}{5} ) and solve the equations simultaneously:
- From 1: [ T = mg - 1.4m ]
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Substitute ( T ) in the equation for particle P: [ mg - 1.4m - 0.3g \frac{3}{5} = 0.3 \times 1.4 ]
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Simplifying provides:
- Collect terms and isolate m: [ 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.
Answer
When the string breaks, P has an initial velocity which can be computed:
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Using the equation of motion: [ v = u + at ] where ( u = 0, a = 0.7 , \text{m/s}^2, t = 0.5 , \text{s}: ] [ v = 0 + 0.7 \times 0.5 = 0.35 , \text{m/s} ]
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Now consider the motion after the string breaks:
- To find the deceleration: [ F = ma \rightarrow -0.3g \sin \alpha - R = ma ]
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Set up for instantaneous rest: [ v = 0 \text{ when reaching rest (use } v^2 = u^2 + 2as) ]: [ 0 = 0.35^2 + 2(-0.98)s ]
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Solving gives further time: ( t = 0.071 s ) or approximately ( 1/14 , ext{s} ).