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

For the system, we apply Newton's Second Law:

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

Where:

• The total mass of the system is (2 + 3 = 5) kg,
• Using (a = \frac{4}{3} \text{ m/s}^2) and substituting it into the equation:

$30 - \mu(2 \times 9.8) = 5 \times \frac{4}{3}$

This leads to:

$30 - 19.6 \mu = \frac{20}{3}$

Rearranging gives:

$19.6 \mu = 30 - \frac{20}{3} = \frac{90 - 20}{3} = \frac{70}{3} \Rightarrow \mu = \frac{70}{3 imes 19.6} ≈ 0.48.$

Applying Newton’s Second Law to particle P:

$T - 2g = 2a$

Using (g \approx 9.8) m/s², we have:

$T - 2(9.8) = 2 \times \frac{4}{3}$

Thus:

$T - 19.6 = \frac{8}{3} \Rightarrow T = 19.6 + \frac{8}{3} ≈ 21.6 \text{ N}.$

In my calculations, the inextensibility of the string implies that both particles P and Q have the same acceleration. Thus, the acceleration of Q determined from its motion also applies to P, ensuring consistency in the derived acceleration and tension equations.

Once the force is removed, only friction acts on Q. The deceleration due to friction is given by:

$F_f = \mu mg$

Substituting the values:

$F_f = 0.48 \times 3 \times 9.8 ≈ 14.1\text{ N}$

Thus, the net force on Q is:

$FNet = -F_f$

Using Newton's Second Law: