For particle Q, using the equation of motion for free fall:

\(3.2 = \frac{1}{2} g t^2\)

Solving for t, we get:

\(t = \sqrt{\frac{2 \times 3.2}{g}} = 0.8 ext{ s}\)

For particle P, using the equation of motion along the ramp:

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

Where a is the component of gravitational acceleration along the ramp:

\(a = g \sin 30° = 0.5g\)

Substituting the values:

\(6.4 = u(0.8) + \frac{1}{2} (0.5g)(0.8)^2\)

Solving for u:

\(u = 6 ext{ m s}^{-1}\)

To find the speed of P at the bottom, use:

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

Substitute the known values:

\(v^2 = 6^2 + 2 \times 0.5g \times 6.4\)

Solving for v:

\(v = 10 ext{ m s}^{-1}\)

The acceleration of the particle down the plane is given by:

\(a = g \sin 30^{\circ}\)

where \(g\) is the acceleration due to gravity \((9.8 \text{ m/s}^2)\).

Thus, \(a = 9.8 \times 0.5 = 4.9 \text{ m/s}^2\).

For part (i):

Using the equation \(v^2 = u^2 + 2as\), where \(u = 0\), \(s = 0.9 \text{ m}\), and \(a = 4.9 \text{ m/s}^2\):

\(v^2 = 2 \times 4.9 \times 0.9\)

\(v^2 = 8.82\)

\(v = \sqrt{8.82} \approx 2.97 \text{ m/s}\)

Rounding to the nearest whole number, the speed is 3 m/s.

For part (ii):

Using the equation \(v = u + at\), where \(u = 0\), \(a = 4.9 \text{ m/s}^2\), and \(t = 0.8 \text{ s}\):

\(v = 0 + 4.9 \times 0.8\)

\(v = 3.92 \text{ m/s}\)

Rounding to the nearest whole number, the speed is 4 m/s.

(i) Using the equation of motion:

\(v = u + at\)

where \(v = 3.5\) m/s, \(u = 0\), and \(t = 2\) s, we find:

\(3.5 = 0 + 2a\)

Solving for \(a\), we get \(a = 1.75\) m/s².

The acceleration \(a = g \sin \alpha\), where \(g = 9.8\) m/s². Thus:

\(1.75 = 9.8 \sin \alpha\)

Solving for \(\sin \alpha\), we get \(\alpha = 10.1^{\circ}\) or \(0.176\) radians.

(ii) For particle P, the distance \(s_P\) is given by:

\(s_P = \frac{1}{2} a (2^2) + a(2)t + \frac{1}{2} at^2\)

For particle Q, the distance \(s_Q\) is:

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

The separation is given by:

\(s_P - s_Q = 4.9\)

Substituting the expressions for \(s_P\) and \(s_Q\):

\(\frac{1}{2} \times 1.75 \times 4 + 1.75 \times 2 \times t + \frac{1}{2} \times 1.75 \times t^2 - \frac{1}{2} \times 1.75 \times t^2 = 4.9\)

Simplifying, we get:

\(2 \times 1.75 + 2 \times 1.75t = 4.9\)

Solving for \(t\), we find \(t = 0.4\) s.

To find the acceleration of the particle, we use the equation of motion:

\(v = u + at\)

where \(v = 4.5 \text{ m s}^{-1}\), \(u = 1.5 \text{ m s}^{-1}\), and \(t = 1.2 \text{ s}\).

Substituting the values, we have:

\(4.5 = 1.5 + 1.2a\)

Solving for \(a\), we get:

\(a = \frac{4.5 - 1.5}{1.2} = 2.5 \text{ m s}^{-2}\)

To find the angle \(\alpha\), we use the equation:

\(mg \sin \alpha = ma\)

\(g \sin \alpha = a\)

\(\sin \alpha = \frac{a}{g} = \frac{2.5}{9.8}\)

\(\alpha = \sin^{-1}\left(\frac{2.5}{9.8}\right) \approx 14.5^\circ\)

(a) Using the conservation of momentum, the initial momentum of the hammerhead is \(1.2v\). The final momentum of the combined hammerhead and nail is \((1.2 + 0.004) \times 40\). Setting these equal gives:

\(1.2v = (1.2 + 0.004) \times 40\)

\(1.2v = 48.16\)

Solving for \(v\), we get:

\(v = \frac{48.16}{1.2} = \frac{602}{15}\)

(b) Using the equation of motion \(v^2 = u^2 + 2as\), where \(v = 0\), \(u = 40\), and \(s = 0.04\), we have:

\(0 = 40^2 + 2 \times 0.04 \times a\)

\(0 = 1600 + 0.08a\)

\(a = -20000 \text{ m s}^{-2}\)

Using Newton's Second Law vertically, \(-R + (1.2 + 0.004)g = (1.2 + 0.004)a\):

\(-R + 12.04 = 1.204 \times (-20000)\)

\(-R + 12.04 = -24080\)

\(R = 24092.04\)

Rounding to the nearest whole number, \(R = 24,100 \text{ N}\).