(a) Using the equation of motion:

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

where \(u = 25 \text{ m s}^{-1}, a = -g = -9.8 \text{ m s}^{-2}, s = 20 \text{ m}\).

\(v^2 = 25^2 + 2(-9.8)(20)\)

\(v^2 = 625 - 392\)

\(v^2 = 233\)

\(v = \sqrt{233} \approx 15 \text{ m s}^{-1}\)

(b) Using conservation of momentum:

Taking upwards as positive:

\(0.5v_A + 0.3(-32.5) = 0.5(-15) + 0\)

\(0.5v_A - 9.75 = -7.5\)

\(0.5v_A = 2.25\)

\(v_A = 4.5 \text{ m s}^{-1}\) downwards

(c) For particle A:

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

\(20 = 4.5t_A + \frac{1}{2}(9.8)t_A^2\)

Solving gives \(t_A = 1.6 \text{ s}\)

For particle B:

\(20 = \frac{1}{2}(9.8)t_B^2\)

Solving gives \(t_B = 2 \text{ s}\)

Time interval = \(t_B - t_A = 2 - 1.6 = 0.4 \text{ s}\)

(a) Using conservation of momentum for the collision between P and Q:

The initial momentum is given by:

\(0.3 \times 4 + 0 = 1.2 \text{ kg m s}^{-1}\)

The final momentum is:

\(0.3v + 0.2 \times 3\)

Equating initial and final momentum:

\(1.2 = 0.3v + 0.6\)

Solving for \(v\):

\(0.3v = 0.6\)

\(v = 2 \text{ m s}^{-1}\)

(b) Using conservation of momentum for the collision between Q and R:

The initial momentum is given by:

\(0.2 \times 3 + 0 = 0.6 \text{ kg m s}^{-1}\)

The final momentum is:

\((0.2 + m) \times 2\)

Equating initial and final momentum:

\(0.6 = (0.2 + m) \times 2\)

Solving for \(m\):

\(0.6 = 0.4 + 2m\)

\(0.2 = 2m\)

\(m = 0.1 \text{ kg}\)

(a) Using the conservation of momentum, the initial momentum of the system is given by:

\(5 \times 8.5 + 3 \times 0 = 5 \times 0.25v + 3v\)

Solving for \(v\), we have:

\(5 \times 8.5 = 5 \times 0.25v + 3v\)

\(42.5 = 1.25v + 3v\)

\(42.5 = 4.25v\)

\(v = 10 \text{ m/s}\)

Thus, the speed of B after the collision is 10 m/s.

(b) The initial kinetic energy (KE) of the system is:

\(\text{KE before} = \frac{1}{2} \times 5 \times 8.5^2 = 180.625 \text{ J}\)

The kinetic energy after the collision is:

\(\text{KE after} = \frac{1}{2} \times 5 \times 2.5^2 + \frac{1}{2} \times 3 \times 10^2\)

\(= 15.625 + 150 = 165.625 \text{ J}\)

The loss of kinetic energy is:

\(\text{KE loss} = 180.625 - 165.625 = 15 \text{ J}\)

(a) Using conservation of momentum for the collision between A and B:

\(0.4 \times 3 + 0.2 \times 2 = 0.4 \times 2.5 + 0.2v\)

Solving for \(v\):

\(1.2 + 0.4 = 1.0 + 0.2v\)

\(1.6 = 1.0 + 0.2v\)

\(0.2v = 0.6\)

\(v = 3\) m/s

(b) For B when it hits the barrier:

Using the equation of motion \(v^2 = u^2 + 2as\):

\(v^2 = 3^2 + 2 \times 5 \times 1.6\)

\(v = 5\) m/s

Speed after hitting the barrier is reduced by 90%:

\(0.1 \times 5 = 0.5\) m/s

For the second collision, using the equation of motion \(v = u + at\) for both particles:

For A: \(v_A = 2.5 + 5 \times 0.44 = 4.7\) m/s

For B: \(v_B = -0.5 + 5 \times 0.04 = -0.3\) m/s

Using conservation of momentum for the second collision:

\(0.4 \times 4.7 + 0.2 \times (-0.3) = 0.6 \times v_{comb}\)

\(1.88 - 0.06 = 0.6 \times v_{comb}\)

\(v_{comb} = \frac{1.82}{0.6} = 3.03\) m/s

(a) Using the equation of motion: \(0.6^2 = 0 + 2a \times 0.45\), we find \(a = 0.4\).

The normal reaction \(R = 0.1g \cos \alpha = 0.1g \times \frac{24}{25} = 0.1g \times \cos 16.3^\circ\).

Using Newton's second law: \(0.1g \frac{7}{25} - F = 0.1 \times 0.4\), where \(F = \mu R\).

Substituting \(F = \mu \times 0.1g \times \frac{24}{25}\), we solve for \(\mu\) and find \(\mu = 0.25\).

(b) Using conservation of momentum: \(0.1 \times 0.6 = 0.5v\), we find \(v = 0.12\).

For B, \(0.5g \frac{7}{25} - 0.275 \times 0.5g \times \frac{24}{25} = 0.5a\), leading to \(a = 0.16\).

Using \(s = ut + \frac{1}{2}at^2\), for A: \(s_A = 0 + \frac{1}{2} \times 0.4 \times t^2\), and for B: \(s_B = 0.12t + \frac{1}{2} \times 0.16 \times t^2\).

Setting \(s_A = s_B\) and solving for \(t\), we find \(t = 1\) s.