(i) Using Newton's Second Law for particle A:

\(1.3g - T = 1.3a\)

For particle B:

\(T - 0.7g = 0.7a\)

Adding these equations gives:

\(1.3g - 0.7g = 1.3a + 0.7a\)

\(0.6g = 2a\)

Thus,

\(a = \frac{0.6g}{2} = 3 ext{ m s}^{-2}\)

Substitute back to find tension:

\(1.3g - T = 1.3 \times 3\)

\(T = 1.3g - 3.9 = 9.1 ext{ N}\)

(ii) The distance while connected is 0.375 m. Using the equation

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

where initial velocity,

\(u = 0,\)

we find

\(v = 1.5 ext{ m s}^{-1}\)

For particle A after the string breaks:

\(1.375 = 1.5t + \frac{1}{2}gt^2\)

Solving gives

\(t = 0.395 ext{ s}\)

For particle B:

\(-1.375 = 1.5t - \frac{1}{2}gt^2\)

Solving gives

\(t = 0.695 ext{ s}\)

\(Difference in times = 0.3 s.\)

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

\(4 - T = 0.4a\)

Apply Newton's second law to particle B:

\(T - 2 = 0.2a\)

For the system:

\(4 - 2 = (0.4 + 0.2)a\)

Solving the system of equations gives:

\(a = \frac{10}{3} \text{ m s}^{-2}, \; T = \frac{8}{3} \text{ N}\)

(ii) Use \(v^2 = u^2 + 2as\) for particle B with \(a = \frac{10}{3}\):

\(v^2 = 0 + 2 \times \frac{10}{3} \times 0.5\)

\(v^2 = \frac{10}{3}\)

After A hits the ground, apply \(v^2 = u^2 + 2as\) to find \(s\):

\(0 = \frac{10}{3} - 2 \times 10 \times s\)

\(s = \frac{1}{6} \text{ m}\)

Maximum height = \(\frac{1}{2} + \frac{1}{2} + \frac{1}{6} = \frac{7}{6} = 1.17 \text{ m}\)

(i) Using Newton's second law for P and Q:

\(T - 0.3g = 0.3a\)

\(0.5g - T = 0.5a\)

Adding these equations gives:

\(0.5g - 0.3g = 0.8a\)

Solving for \(a\):

\(a = \frac{0.2g}{0.8} = 2.5 \text{ m/s}^2\)

Using \(s = ut + \frac{1}{2}at^2\) to find \(h\):

\(h = 0 + \frac{1}{2} \times 2.5 \times 0.6^2 = 0.45 \text{ m}\)

(ii) Velocity of P when Q reaches the floor:

\(v = 0 + 0.6 \times 2.5 = 1.5 \text{ m/s}\)

Using \(v = u + at\) to find time to highest point:

\(0 = 1.5 - 9.8t\)

\(t = \frac{1.5}{9.8} \approx 0.15 \text{ s}\)

Total time:

\(2 \times 0.15 + 0.6 = 0.9 \text{ s}\)

To solve this problem, we use the equations of motion and Newton's second law.

(i) Finding the tension in the string:

First, we find the acceleration of the system. Using the equation of motion:

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

where \(v = 0.6 \text{ m/s}\), \(u = 0 \text{ m/s}\), and \(s = 0.8 \text{ m}\).

Substituting the values, we get:

\(0.6^2 = 0 + 2a \times 0.8\)

\(a = 0.225 \text{ m/s}^2\)

Now, applying Newton's second law to the 0.3 kg particle:

\(T - 0.3g = 0.3a\)

Substituting \(g = 9.8 \text{ m/s}^2\) and \(a = 0.225 \text{ m/s}^2\):

\(T - 0.3 \times 9.8 = 0.3 \times 0.225\)

\(T = 3.07 \text{ N}\)

(ii) Finding the value of \(m\):

Applying Newton's second law to the \(m\) kg particle:

\(mg - T = ma\)

Substituting \(T = 3.07 \text{ N}\) and \(a = 0.225 \text{ m/s}^2\):

\(mg - 3.07 = m \times 0.225\)

Solving for \(m\):

\(m(9.8 - 0.225) = 3.07\)

\(m = 0.314 \text{ kg}\)

Let the tension in the string be denoted by T and the acceleration of the system by a. For particle A (mass 0.8 kg), applying Newton's second law gives:

\(8 - T = 0.8a\)

For particle B (mass 0.2 kg), applying Newton's second law gives:

\(T - 2 = 0.2a\)

Adding these two equations to eliminate T:

\((8 - T) + (T - 2) = 0.8a + 0.2a\)

\(6 = a\)

Thus, the acceleration a is 6 m s^-2.

\(Substitute a = 6 into the equation for particle B:\)

\(T - 2 = 0.2(6)\)

\(T - 2 = 1.2\)

\(T = 3.2\)

Therefore, the tension in the string is 3.2 N.