(i) Apply Newton's second law to particle Q:

\(0.49g - T = 0.49a\)

For particle P on the inclined plane:

\(T - \frac{40}{160} \times 0.76g = 0.76a\)

Solving these equations simultaneously gives:

\(0.49g - \frac{40}{160} \times 0.76g = (0.49 + 0.76)a\)

Acceleration, \(a = 2.4 \text{ m/s}^2\) and tension, \(T = 3.72 \text{ N}\).

(ii) Using \(v^2 = u^2 + 2as\) with \(u = 0\), \(a = 2.4 \text{ m/s}^2\), and \(s = 0.3 \text{ m}\):

\(v^2 = 2 \times 2.4 \times 0.3\)

Speed, \(v = 1.20 \text{ m/s}\).

(iii) Total distance travelled by P before coming to rest:

Distance while Q is on the ground:

\(\frac{2 \times 2.4 \times 0.3}{2(40 \div 160)g} = 0.288 \text{ m}\)

Total distance travelled, \(0.588 \text{ m}\).

For part (i), apply Newton's second law to both particles. For particle A on the inclined plane:

\(T - 0.26g \sin \alpha = 0.26a\)

For particle B hanging freely:

\(0.52g - T = 0.52a\)

Solving these equations simultaneously:

\(T = 0.26g \sin \alpha + 0.26a\)

\(0.52g - (0.26g \sin \alpha + 0.26a) = 0.52a\)

\(0.52g - 0.26g \sin \alpha = 0.78a\)

\(a = \frac{0.52g - 0.26g \sin \alpha}{0.78}\)

Substitute \(g = 9.8\), \(\sin \alpha = \frac{16}{65}\):

\(a = \frac{0.52 \times 9.8 - 0.26 \times 9.8 \times \frac{16}{65}}{0.78} = 5.85 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 0.26 \times 9.8 \times \frac{16}{65} + 0.26 \times 5.85 = 2.16 \text{ N}\)

For part (ii), use the equation \(v^2 = u^2 + 2as\) for particle B:

\(v^2 = 2 \times 5.85 \times 0.6\)

\(v = \sqrt{2 \times 5.85 \times 0.6} = 2.65 \text{ m/s}\)

For particle A, use \(v^2 = u^2 - 2g \sin \alpha \times s\):

\(0 = 2.65^2 - 2 \times 9.8 \times \frac{16}{65} \times s\)

\(s = \frac{2.65^2}{2 \times 9.8 \times \frac{16}{65}} = 1.425 \text{ m}\)

Distance AP when A comes to rest:

\(2.5 - 0.6 - 1.425 = 0.475 \text{ m}\)

For part (i):

The normal reaction \(R\) on particle A is given by:

\(R = 0.26 \times \left( \frac{12}{13} \right) = 2.4 \text{ N}\)

The frictional force \(F\) is:

\(F = 0.2 \times 2.4 = 0.48 \text{ N}\)

Applying Newton's second law to particle A along the plane:

\(T - 0.26 \times \frac{5}{13} - 0.48 = 0.26a\)

For particle B, applying Newton's second law vertically:

\(0.54g - T = 0.54a\)

Solving these equations simultaneously:

\(T - 1 - 0.48 = 0.26a\)

\(5.4 - T = 0.54a\)

\((5.4 - 1 - 0.48) = (0.54 + 0.26)a\)

Acceleration \(a = 4.9 \text{ m/s}^2\)

Tension \(T = 2.75 \text{ N}\)

For part (ii):

Using the equation of motion:

\(s = \frac{1}{2} a t^2\)

\(s = \frac{1}{2} \times 4.9 \times (0.4)^2\)

\(s = 0.392 \text{ m}\)

(i) The forces acting on P are:

• Weight \(W = 0.13g\) acting vertically downwards.
• Tension \(T\) in the string acting up the plane.
• Frictional force \(F\) acting down the plane.
• Normal reaction \(R\) perpendicular to the plane.

(ii) To show \(T - F > 0.32\):

The component of weight of P down the plane is \(0.13g \sin \theta\), where \(\sin \theta = \frac{16}{65}\).

Thus, \(0.13g \sin \theta = 0.13g \left(\frac{16}{65}\right)\).

The resultant upward force (RUF) is \(T - F - 0.13g \left(\frac{16}{65}\right)\).

For P to move upwards, \(T - F - 0.13g \left(\frac{16}{65}\right) > 0\).

Therefore, \(T - F > 0.13g \left(\frac{16}{65}\right) \approx 0.32\).

(iii) To find the acceleration of P:

The normal reaction \(R = 0.13g \cos \theta\), where \(\cos \theta = \frac{63}{65}\).

Thus, \(R = 0.13g \left(\frac{63}{65}\right) \approx 1.26\).

The frictional force \(F = \mu R = 0.6 \times 1.26 = 0.756\).

Using Newton's second law for P:

\(T - 0.32 = 0.13a\) and \(0.11g - T = 0.11a\).

Solving these equations gives \(a = 0.1 \text{ m/s}^2\).

(a)(i) For particle P, using Newton's second law:

\(2g \sin 30° - F - T = 2a\)

For particle Q:

\(T - 0.25g = 0.25a\)

Solving the system:

\(2g \sin 30° - F - 0.25g = (2 + 0.25)a\)

Friction force:

\(F = 0.3 \times 2g \cos 30° = 3\sqrt{3} = 5.1961...\)

Acceleration from A to B:

\(a = 1.02 \, \text{m/s}^2\)

(a)(ii) Using SUVAT for v^2:

\(v^2 = u^2 + 2as\)

Substitute:

\(v^2 = 0 + 2 \times 1.02 \times 1.2\)

Velocity at C:

\(v = 3.1 \, \text{m/s}\)

(b) Time from A to B:

\(0.8 = \frac{1}{2} \times 1.02 \times t^2\)

Solve for t:

\(t = 1.25 \, \text{s}\)

Time from B to C:

\(1.2 = \frac{1}{2} \times (3.1 + 0) \times \frac{10}{3}\)

Solve for t:

\(t = 0.547 \, \text{s}\)

Total time:

\(1.25 + 0.547 = 1.8 \, \text{s}\)