Since the system is in equilibrium, the tension \(T\) in the string is equal to the weight of the particle \(P\). Therefore, \(T = 3g\), where \(g\) is the acceleration due to gravity. Assuming \(g = 10 \text{ m/s}^2\), we have:

\(T = 3 \times 10 = 30 \text{ N}\).

For the normal component of the contact force on \(B\), resolve the forces perpendicular to the plane. The normal force \(R\) is given by:

\(R = (4g - T) \cos \alpha\).

Substitute \(T = 30 \text{ N}\) and \(\cos \alpha = 0.8\):

\(R = (4 \times 10 - 30) \times 0.8 = 10 \times 0.8 = 8 \text{ N}\).

(i) The area under the velocity-time graph represents the distance traveled. The graph shows a triangle with base 0.5 s and height 2 m/s. The area is given by:

\(h = \frac{1}{2} \times 0.5 \times 2 = 0.5\)

(ii) The acceleration \(a\) can be found using the gradient of the velocity-time graph:

\(a = \frac{2}{0.5} = 4 \text{ m/s}^2\)

Using Newton's second law for both particles:

For P: \(T - mg = ma\)

For Q: \((1-m)g - T = (1-m)a\)

Solving these equations simultaneously gives:

\(a = \frac{(1-2m)}{(1+m)}g\)

Substituting \(a = 4\) and \(g = 10\):

\(4 = \frac{(1-2m)}{(1+m)} \times 10\)

Solving for \(m\):

\(m = 0.3\)

Substitute \(m = 0.3\) back to find tension \(T\):

\(T = 0.3 \times 10 + 4 \times 0.3 = 4.2 \text{ N}\)

(iii) The string is slack while Q is at rest on the ground. The graph shows that P moves from 2 m/s to -2 m/s, taking 0.5 s to reach 0 m/s. The total time is:

\(T = 0.5 + 0.4 = 0.9 \text{ s}\)

(i) From the velocity-time graph, the acceleration can be found using the gradient. The acceleration is 4 m/s². Using Newton's second law for both particles, we have:

\(T - mg = 4m\)

\((1 - m)g - T = 4(1 - m)\)

Solving these equations gives:

\(4 = (1 - m - m)g\)

Thus, P has mass 0.3 kg and Q has mass 0.7 kg.

(ii) Using the area under the velocity-time graph or the formula \(h = \frac{1}{2} a t^2\), we find \(h = 2\) m.

(iii) The distance travelled upwards by P is given by the area under the graph:

\(\text{Distance} = \frac{1}{2} \times 1.4 \times 4 = 2.8 \text{ m}\)

Adding the initial height, the greatest height is 4.8 m.

(i) Using Newton's second law for both particles and eliminating tension, we have:

\((M + m)a = (M - m)g\)

where \(M = 0.75\) kg, \(m = 0.25\) kg, and \(g = 9.8\) m/s².

Solving gives:

\(a = \frac{(0.75 - 0.25) \times 9.8}{0.75 + 0.25} = 5\) m/s².

Using the equation \(s = ut + \frac{1}{2}at^2\) with \(u = 0\), \(t = 0.6\) s:

\(s = 0 + \frac{1}{2} \times 5 \times (0.6)^2 = 0.9\) m.

(ii) From the velocity-time graph, the area under the graph from 0 to 0.6 s is:

\(\frac{1}{2} \times 0.6 \times V = 0.9\)

Solving gives:

\(V = 3\) m/s.

Using the equation \(0 = V - g(T - 0.6)\):

\(0 = 3 - 9.8(T - 0.6)\)

Solving gives:

\(T = 0.9\) s.

(iii) The distance travelled upwards is given by:

\(s_{\text{up}} = \frac{1}{2} \times 0.9 \times 3 = 1.35\) m.

The distance travelled downwards is given by:

\(s_{\text{down}} = 0 + \frac{1}{2} \times 9.8 \times (1.6 - 0.9)^2 = 2.45\) m.

Thus, the height \(h\) is:

\(h = s_{\text{down}} - s_{\text{up}} = 2.45 - 1.35 = 1.1\) m.

Let the acceleration of the system be \(a\). Using Newton's second law for both particles:

For particle A: \(T - 0.2g = 0.2a\)

For particle B: \(0.6g - T = 0.6a\)

Adding these equations gives:

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

\(0.4g = 0.8a\)

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

When B reaches the floor, using \(v^2 = u^2 + 2as\) with \(u = 0\), \(a = 5\), and \(s = 1.6\):

\(v^2 = 2 \times 5 \times 1.6\)

\(v^2 = 16\)

\(v = 4 \, \text{m/s}\)

For A to reach 3 m above the floor, the total distance moved by A is \(3 - h\). Using energy conservation:

\(Potential energy gain = Kinetic energy loss\)

\(0.2g(3 - h) = \frac{1}{2} \times 0.2 \times 4^2\)

\(0.2g(3 - h) = 1.6\)

\(2g(3 - h) = 16\)

\(3 - h = 0.8\)

\(h = 3 - 0.8 = 0.6\)