(i) Applying Newton's second law to particle B gives:

\(0.7g - T = 0.7a\)

For particle A:

\(T - 0.3g = 0.3a\)

Adding these equations:

\(0.7g - 0.3g = 0.7a + 0.3a\)

\(0.4g = a\)

Substituting \(g = 9.8 \text{ m/s}^2\):

\(a = 4 \text{ m/s}^2\)

Substitute \(a\) back into one of the equations:

\(0.7g - T = 0.7 \times 4\)

\(T = 0.7g - 2.8\)

\(T = 6.86 - 2.8 = 4.06 \text{ N}\)

Rounding gives \(T = 4.2 \text{ N}\).

(ii) When the string breaks, both particles have a speed of 1.6 m/s. The distance A falls is 0.52 m. Using the kinematic equation:

\(s = ut + \frac{1}{2}gt^2\)

\(0.52 = 1.6t + \frac{1}{2} \times 9.8 \times t^2\)

\(0.52 = 1.6t + 4.9t^2\)

Solving the quadratic equation:

\(4.9t^2 + 1.6t - 0.52 = 0\)

Using the quadratic formula:

\(t = \frac{-1.6 \pm \sqrt{(1.6)^2 - 4 \times 4.9 \times (-0.52)}}{2 \times 4.9}\)

\(t = \frac{-1.6 \pm \sqrt{2.56 + 10.192}}{9.8}\)

\(t = \frac{-1.6 \pm \sqrt{12.752}}{9.8}\)

\(t = \frac{-1.6 \pm 3.57}{9.8}\)

\(t = 0.6 \text{ s}\) (taking the positive root).

(a) Consider the forces on the 0.1 kg particle:

\(T - 0.1g = 0.1a\)

For the m kg particle:

\(mg - T = ma\)

Adding these equations for the system gives:

\(mg - 0.1g = (m + 0.1)a\)

Substitute \(T = 1.5\) N and solve for \(m\):

\(a = 5\)

\(m = 0.3\)

(b) Use the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 0.9\), and \(a = 5\):

\(v^2 = 0 + 2 \times 5 \times 0.9\)

\(v = 3 \text{ m s}^{-1}\)

(i) Using the equation of motion:

\(v = u + at\)

where \(v = 0.6 \text{ m s}^{-1}\), \(u = 0 \text{ m s}^{-1}\), and \(t = 0.3 \text{ s}\).

Substitute the values:

\(0.6 = 0 + a \times 0.3\)

Solve for \(a\):

\(a = \frac{0.6}{0.3} = 2 \text{ m s}^{-2}\)

(ii) Applying Newton's second law to particle A:

\(mg - T = ma\)

For particle B:

\(T - (1 - m)g = (1 - m)a\)

Equating the two expressions for tension \(T\):

\(mg - ma = (1 - m)g + (1 - m)a\)

Substitute \(a = 2 \text{ m s}^{-2}\) and \(g = 10 \text{ m s}^{-2}\):

\(10m - 2m = 10 - 10m + 2 - 2m\)

Simplify and solve for \(m\):

\(8m = 12 - 12m\)

\(20m = 12\)

\(m = 0.6\)

Substitute \(m = 0.6\) back to find \(T\):

\(T = 8m = 8 \times 0.6 = 4.8 \text{ N}\)

(i) Using Newton's second law, the equations for the forces on A and B are:

\(T - 0.12g = 0.12a\)

\(0.38g - T = 0.38a\)

Adding these equations gives:

\(0.38g - 0.12g = 0.12a + 0.38a\)

\(a = \frac{0.38 - 0.12}{0.38 + 0.12}g\)

Thus, the acceleration \(a = 5.2 \text{ m/s}^2\).

(ii) Using \(v^2 = 2as\) for B:

\(v^2 = 2 \times 5.2 \times 0.65\)

\(v = 2.6 \text{ m/s}\)

Time \(T_B\) is given by:

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

\(0.65 = \frac{1}{2} \times 5.2 \times T_B^2\)

\(T_B = 0.5 \text{ s}\)

(iii) For A, using the equation:

\(-2.6 = 2.6 - 10(T - 0.5)\)

Solving gives:

\(T = 1.02 \text{ s}\)

The velocity-time graph is a straight line increasing to 2.6 m/s at 0.5 s, then decreasing back to 0 m/s at 1.02 s.

(iv) Total distance travelled by A:

\(0.65 + 0.5(1.02 - 0.5) \times 2.6\)

\(= 1.326 \text{ m}\)

Using Newton's second law for particle P:

\(0.9g - 7.2 = 0.9a\)

Solving for acceleration \(a\):

\(a = 2\)

Using the equation of motion \(v^2 = u^2 + 2ah\) with initial velocity \(u = 0\):

\(v^2 = 2 \times (0.9g - 7.2)/0.9 \times 2\)

\(v = \sqrt{8}\)

When the string becomes slack, the velocity of Q is:

\(u_{\text{slack}} = v_{\text{taut}} = 2\sqrt{g - 8}\)

Using the equation \(0 = u^2 - 2gh\) to find the distance:

\(\text{distance} = 2h\)

The total distance travelled by Q is 0.8 m.