Two particles P and Q, of mass 2 kg and 3 kg respectively, are joined by a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 8 - 2008 - Paper 1

Using Newton's second law for the complete system:

$30 - \mu (2g) = 5a$

Where $g = 9 ext{ m/s}^2$ and substituting the calculated acceleration:

$30 - 2\mu(9) = 5 \times \frac{4}{3}$

This simplifies to:

$30 - 18\mu = \frac{20}{3}$

Reorganizing gives:

$18\mu = 30 - \frac{20}{3} = \frac{90 - 20}{3} = \frac{70}{3}$

So,

$\mu = \frac{70}{54} = \frac{35}{27} \approx 0.48$

Applying Newton's second law for particle P:

$T - 2g = 2a$

Substituting $g = 9 ext{ m/s}^2$ and $a = \frac{4}{3} ext{ m/s}^2$ gives:

$T - 18 = 2 \times \frac{4}{3}$

This simplifies to:

$T - 18 = \frac{8}{3} \Rightarrow T = 18 + \frac{8}{3} = \frac{54 + 8}{3} = \frac{62}{3} ext{ N}$

The inextensibility of the string ensures that both particles, P and Q, have the same acceleration. Thus, the acceleration of P is equal to the acceleration of Q when analyzing the motion and applying Newton's laws.

When the force is removed, Q will decelerate due to friction. The deceleration can be calculated as:

$a_{decel} = \mu g = \frac{35}{27} \times 9 = \frac{35}{3}$

Using the equation of motion:

$v = u - at$

Where $u = 6 ext{ m/s}$ (final velocity after 3 s), and setting the final velocity to 0:

$0 = 6 - \frac{35}{3} t$

Solving for $t$ gives:

$t = \frac{6 \times 3}{35} \approx 0.51 ext{ s}$