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

From equation (2), we can rewrite it as:

$T - \mu mg = ma$

Substituting for T using equation (1), we have:

$\frac{15mg}{8} - \mu mg = m\cdot\frac{g}{8}$

Dividing through by m gives us:

$\frac{15g}{8} - \mu g = \frac{g}{8}$

Rearranging yields:

$\mu g = \frac{15g}{8} - \frac{g}{8}$

Thus, we arrive at:

$\mu g = \frac{14g}{8}$

Finally, we get:

$\mu = \frac{14}{8} = \frac{7}{4} \implies \, \mu = \frac{2}{3}$.

To find the speed of A, we use the principle of conservation of energy. The potential energy lost by B when it falls a distance h is converted into kinetic energy of A.

Let v be the speed of A as it reaches P:

$mgh = \frac{1}{2} \cdot m v^2$

Solving for v, we get:

The information that the string is light indicates that it has negligible mass. Therefore, we assume there is no loss of energy due to the weight of the string, and that any tension in the string does not contribute to additional forces acting on the particles. This allows us to treat the system as if all forces come solely from the masses of A and B.