Photo AI
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
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.
Answer
To analyze the forces acting on particles A and B, we use Newton's second law. For particle A (mass 7 kg):
For particle B (mass 3 kg), we consider:
- The tension in the string, T.
- The frictional force acting against the motion, which is given by: F = rac{2}{3} R = rac{2}{3} (3g rac{5}{13})
Setting up the equations: For A: For B (considering forces parallel to the incline): T - F - 3g rac{5}{12} = 3a
Eliminating T from the equations:
- Rearranging for T in the first equation gives:
- Substituting for T in the second equation, we have: (7g - 7a) - rac{2}{3} (3g rac{5}{12}) - 3g rac{5}{12} = 3a
This leads us to calculate the acceleration a, yielding: a = rac{2g}{5} Approximately, this results in:
Step 2
Find the speed of B when it has moved 1 m up the plane.
Answer
Using the kinematic equation:
where:
- Initial speed, (since it starts from rest),
- a = rac{2g}{5} ext{ (from previous calculation)}, and the distance moved, .
Substituting the values: v^2 = 0 + 2 imes rac{2g}{5} imes 1 Calculating gives:
Step 3
When B has moved 1 m up the plane the string breaks. Given that in the subsequent motion B does not reach P, find the time between the instants when the string breaks and when B comes to instantaneous rest.
Answer
After the string breaks, the force acting on B is:
(-(F + 3g rac{5}{12}) = 3a
Therefore:
- rac{2}{3} (3g rac{5}{12}) - 3g rac{5}{12} = -3a Resulting in: From here, we calculate time, using:
t = rac{v}{g} where . Thus: t = rac{2 ext{ m/s} }{7} ext{ yielding approximately } 0.286 ext{ seconds.}