Photo AI
Two particles, A and B, have masses 2m and m respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2017 - Paper 1
Question 8
Two particles, A and B, have masses 2m and m respectively. The particles are attached to the ends of a light inextensible string. Particle A is held at rest on a fix... show full transcript
Worked Solution & Example Answer:Two particles, A and B, have masses 2m and m respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2017 - Paper 1
Step 1
Write down an equation of motion for A.
Answer
For particle A, the forces acting on it are the tension in the string (T) and the frictional force (f). Hence, the equation of motion can be written as:
[ T - f = 2ma ]
Here, ( f = \mu \cdot (2m)g ), the equation can also be expressed as:
[ T - \mu \cdot (2mg) = 2ma ]
This simplifies to:
[ T = 2ma + 2\mu mg ]
Step 2
Step 3
Hence show that, until B hits the floor, the acceleration of A is \( \frac{g}{3}(1-2\mu) \).
Answer
Using the equations from parts (i) and (ii), from particle B's equation, we can substitute T into A's equation:
[ T = mg - ma ].
Substituting into A's equation:
[ mg - ma = 2ma + 2\mu mg ]
Rearranging gives:
[ mg - 2\mu mg = 3ma ]
This yields:
[ a = \frac{g(1-2\mu)}{3} ]
Step 4
Find, in terms of g, h and \( \mu \), the speed of A at the instant when B hits the floor.
Answer
Using the equation of motion for constant acceleration, we have:
[ v^2 = u^2 + 2as ]
Here, ( u = 0 ), ( a = \frac{g(1-2\mu)}{3} ) and ( s = h ).
Thus,
[ v^2 = 0 + 2 \cdot \frac{g(1-2\mu)}{3} \cdot h ]
This simplifies to:
[ v = \sqrt{\frac{2gh(1-2\mu)}{3}} ]
Step 5
Describe what would happen if \( \mu = \frac{1}{2} \).
Answer
If ( \mu = \frac{1}{2} ), the coefficients involved in the calculation would lead to a neutral balance of forces at B. This means particle A would not move due to friction being equal to the tension in the string corresponding to its weight, resulting in A remaining stationary.