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 in free fall:

The forces acting on B are the weight of B (mg) acting downwards and the tension T acting upwards. Therefore, using Newton's second law, we can write:

$mg - T = ma$

This gives the equation of motion for B as: $T = mg - ma$.

To find the acceleration of A, we can eliminate T from both equations:

From the first equation: $T = 2ma + \mu (2m)g$

And from the second equation: $T = mg - ma$

Setting these equal gives: $mg - ma = 2ma + \mu (2m)g$

Rearranging: $mg = 3ma + \mu (2m)g$

Thus, $a = \frac{g(1 - 2\mu)}{3}$.

This shows that until B hits the floor, the acceleration of A is ( \frac{g}{3}(1 - 2\mu) ).

When B hits the floor, the potential energy lost by B while falling through height h converts to kinetic energy of A and B. The velocity v of A can be found using the conservation of energy:

K.E. of B = P.E. of B

Therefore, the kinetic energy is: $\frac{1}{2}mv^2 = mgh$

This leads to: $v^2 = 2gh$

The speed of A at that instant can be derived from the previously established acceleration:

Using the relation ( v^2 = u^2 + 2as ):

Here, u = 0, s = d, and substituting in the value of a: $v^2 = 2 (\frac{g(1 - 2\mu)}{3}) d$

Adding these, we find the speed of A incorporating g, h, and µ.

If the coefficient of friction µ = ( \frac{1}{2} ), we substitute this into our previous results. The acceleration of A then becomes: $a = \frac{g(1 - 2 \times \frac{1}{2})}{3} = 0$

This indicates that there would be no acceleration, meaning A would lift off the table and not move. The system would reach a state where A continues to remain in equilibrium at rest instead of sliding along the table.