Since the system is in equilibrium, the tension in the string is equal to the weight of the particle P. The weight of P is given by:

\(T = m_P imes g = 4 imes 10 = 40 \text{ N}\)

Thus, the tension in the string is 40 N.

For block B, the forces acting vertically are its weight, the tension in the string, and the normal reaction from the ground. Since B is in equilibrium, we have:

\(R + T = W_B\)

where \(R\) is the normal reaction, \(T\) is the tension, and \(W_B\) is the weight of block B.

The weight of block B is:

\(W_B = m_B imes g = 5 imes 10 = 50 \text{ N}\)

Substituting the known values:

\(R + 40 = 50\)

Solving for \(R\):

\(R = 50 - 40 = 10 \text{ N}\)

Therefore, the force exerted on B by the ground is 10 N.

(i) (a) To find the acceleration of A, use the equation of motion:

\(v = u + at\)

where \(v = 5 \text{ m s}^{-1}\), \(u = 0\), and \(t = 2 \text{ s}\).

Substitute the values:

\(5 = 0 + 2a\)

\(a = \frac{5}{2} = 2.5 \text{ m s}^{-2}\)

(b) To find the tension in the string, apply Newton's second law to particle A:

\(0.5g - T = 0.5 \times 2.5\)

\(T = 0.5g - 0.5 \times 2.5\)

\(T = 0.5 \times 9.8 - 1.25\)

\(T = 4.9 - 1.25 = 3.75 \text{ N}\)

(ii) To find the value of m, apply Newton's second law to particle B:

\(T - mg = 2.5m\)

Substitute \(T = 3.75\) and solve for \(m\):

\(3.75 - mg = 2.5m\)

\(3.75 = mg + 2.5m\)

\(3.75 = m(g + 2.5)\)

\(m = \frac{3.75}{g + 2.5}\)

\(m = \frac{3.75}{9.8 + 2.5}\)

\(m = \frac{3.75}{12.3} \approx 0.3 \text{ kg}\)

(a) Apply Newton's second law to each particle:

For the 0.8 kg particle:

\(0.8g - T = 0.8a\)

For the 0.2 kg particle:

\(T - 0.2g = 0.2a\)

For the system:

\(0.8g - 0.2g = (0.8 + 0.2)a\)

Solve for \(a\):

\(a = 6 \text{ m s}^{-2}\)

Solve for \(T\):

\(T = 3.2 \text{ N}\)

(b) Use the equation \(v^2 = u^2 + 2as\) to find the velocity \(v\) of the 0.8 kg particle as it reaches the ground:

\(v^2 = 2 \times 6 \times 0.5\)

Find the extra height reached by the 0.2 kg particle using \(v^2\):

\(0 = 6 - 20s\)

\(Greatest height = 0.5 + 0.5 + 0.3 = 1.3 m\)

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

\(0.6g - T = 0.6a\)

Apply Newton's second law to particle Q:

\(T - 0.2g = 0.2a\)

Add the two equations:

\(0.6g - T + T - 0.2g = 0.6a + 0.2a\)

\(0.4g = 0.8a\)

Solve for acceleration \(a\):

\(a = \frac{0.4g}{0.8} = 5 \text{ m/s}^2\)

Substitute \(a = 5\) into the equation for P:

\(0.6g - T = 0.6 \times 5\)

\(T = 0.6g - 3\)

\(T = 3 \text{ N}\)

(ii) Use the equation of motion \(s = ut + \frac{1}{2}at^2\) with \(s = 0.9\), \(u = 0\), \(a = 5\):

\(0.9 = \frac{1}{2} \times 5 \times t^2\)

\(t^2 = \frac{0.9 \times 2}{5}\)

\(t^2 = 0.36\)

\(t = 0.6 \text{ s}\)

(i) Using the equation of motion \(s = ut + \frac{1}{2}at^2\), we have:

\(0.81 = 0 + \frac{1}{2}a \times 0.9^2\)

Solving for \(a\), we get \(a = 2 \text{ m s}^{-2}\).

Using Newton's Second Law for particle A: \(T - mg = ma\) and for particle B: \(kmg - T = kma\).

Solving these equations simultaneously gives:

\(a = \frac{(kg - g)}{k+1}\)

Setting \(a = 2\), we solve for \(k\):

\(\frac{(kg - g)}{k+1} = 2\)

\(k = 1.5\).

The tension \(T = 10m + 2m = 12m \text{ N}\).

(ii) The velocity of A when the string breaks is \(2 \times 0.9 = 1.8 \text{ m s}^{-1}\) upwards.

Using \(v^2 = u^2 + 2gs\) to find \(v\) at the ground:

\(v^2 = 1.8^2 + 2g \times 1.62\)

Speed is \(5.97 \text{ m s}^{-1}\).

Time taken:

\(\frac{(1.8 + 5.97)}{g} = 0.777 \text{ s}\)

(iii) The velocity-time graph is a straight line from (0, 0) to (0.9, 1.8), then from (0.9, 1.8) to approximately (1.7, -6).