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:

$T - \mu mg = m \cdot \frac{g}{4}$

Substituting for T:

$\frac{10mg}{9} - \mu mg = m \cdot \frac{g}{4}$

Solving for \mu:

$\mu = \frac{10mg}{9} - \frac{mg}{4}$

Finding a common denominator, we have:

$\mu = \frac{40mg - 9mg}{36} = \frac{31mg}{36} = \frac{2}{3}$.

When B hits the ground, the deceleration of particle A can be described as:

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

Using the kinematic equation:

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

With initial speed u = 0,

$v^2 = 0 + 2 \cdot \frac{2g}{3} \cdot \frac{h}{3} = \frac{4gh}{9}$

Thus, the speed of A as it reaches P is:

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

The information that the string is light implies that it has negligible mass, allowing us to assume that the tension is constant throughout the string. This aids in simplifying the equations of motion, as we do not need to account for any weight or mass of the string affecting the tension or acceleration.