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A particle A of mass 0.8 kg rests on a horizontal table and is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 8 - 2003 - Paper 1
Question 8
A particle A of mass 0.8 kg rests on a horizontal table and is attached to one end of a light inextensible string. The string passes over a small smooth pulley P fix... show full transcript
Worked Solution & Example Answer:A particle A of mass 0.8 kg rests on a horizontal table and is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 8 - 2003 - Paper 1
Step 1
the tension in the string before B reaches the ground.
Answer
To find the tension in the string before B reaches the ground, we can apply Newton's second law to the two masses A and B. For mass B:
Using the downward acceleration of B as (a), the forces acting on B are:
- Weight of B: (1.2g)
- Tension: (T)
Applying Newton's second law, we get:
[ 1.2g - T = 1.2a ]
For mass A, since there is no friction, the tension will be the only force acting:
[ T = 0.8a ]
We can express 'g' as approximately 9.8 m/s², hence substituting we have:
Substituting for a from A: [ T = 0.8(\frac{1.2g - T}{1.2}) ]
Combining the equations will give us: [ T = 4.7 , N ]
Step 2
the time taken by B to reach the ground.
Answer
Using the distance fallen by B which is 0.6 m:
We can use the equation of motion:
[ s = ut + \frac{1}{2} a t^2 ]
Since B is released from rest, (u = 0), the equation reduces to:
[ 0.6 = \frac{1}{2} a t^2 ]
From the earlier derivation, solving for time, we find: [ a = 5.88 , m/s^2 ]
Substituting it, we have: [ t = \sqrt{\frac{2 * 0.6}{a}} = \sqrt{\frac{1.2}{5.88}} \approx 0.45 , s ]
Step 3
determine the time taken by B to reach the ground.
Answer
Given that the table is now rough and the coefficient of friction between A and the table is (\frac{1}{3}):
Using the refined model:
For mass A, the force of friction acting on it will now affect the motion: [ F_{friction} = \mu R = \frac{1}{3} * 0.8g = \frac{0.8g}{3} ]
Revising the forces acting on A and B: [ T + F_{friction} = 0.8a' ]
For mass B: [ 1.2g - T = 1.2a' ]
We will solve these equations using substitution:
Evaluating for a', we find: [ a' = 0.52 , m/s^2 ]
Again using the distance fallen by B which remains 0.6 m: [ t = \sqrt{\frac{2 * 0.6}{0.52}} \approx 0.49 , s ]