Two particles A and B, of mass m and 2m respectively, are attached to the ends of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2008 - Paper 1

For particle A, the equation of motion is:

$T - \mu mg = ma \implies T = \mu mg + ma$

Substituting the tension from part (a) into this equation and solving gives:

$\mu = \frac{2g}{3}$

When B hits the ground, the distance fallen is h. The deceleration of A after B hits is:

$ma = \mu mg \implies a = \frac{2g}{3}$

Using the equation of motion for A:

$v^2 = u^2 + 2as$

where u = 0 and s = ( \frac{2}{3}h ).

Thus,

$v^2 = 0 + 2\left(\frac{2g}{3}\right)\left(\frac{1}{3}h\right) = \frac{4gh}{9}$

The speed v when A reaches P is:

$v = \sqrt{\frac{4gh}{9}} = \frac{2}{3}\sqrt{gh}$

The fact that the string is light implies that it has negligible mass and therefore does not contribute to the tension or force calculations. This allows the equations of motion to simplify, as we only consider the masses of particles A and B and the forces acting on them.