Two particles, A and B, have masses 2m and m respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2017 - Paper 1

For particle B, which is influenced by gravity, the equation of motion is:

$mg - T = ma$

where mg is the weight of particle B and T is the tension in the string.

To find the acceleration, we can set the equations for T from both parts. From the equation for A: $T = 2ma + \mu (2m)g$

Substituting T in the equation for B: $mg - (2ma + \mu (2m)g) = ma$

Simplifying this, we have: $mg - \mu (2m)g = 3ma$

Thus, isolating a gives: $a = \frac{mg - \mu (2m)g}{3m} = \frac{g(1 - 2\,\text{µ})}{3}$

Therefore, until B hits the floor, the acceleration of A is: $a = \frac{2g}{3(1 - 2\,\text{µ})}$

Using energy conservation or kinematic equations, we can derive the speed: The potential energy lost by B when it falls a distance h plus the work done against friction equals the kinetic energy gained by A:

$mgh - \mu (2m)g h = \frac{1}{2}(2m)v^2$

Solving for v gives: $v = \sqrt{\frac{2(1 - \text{µ})gh}{1}}$

If ( \mu = \frac{1}{2} ), the frictional force opposing motion would be equal to the weight component acting on A. This means that A would not accelerate; instead, it would remain in (limiting) equilibrium, and B would not move after the initial pull, leading the system to be in a stable position until external forces act on it.