(a) Apply the conservation of momentum. The initial momentum of the system is given by the momentum of particle A since B is at rest:

\(6 \times 2.5 = 2.5v + 5v\)

Solving for \(v\):

\(15 = 7.5v\)

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

(b) Calculate the kinetic energy before and after the collision. The kinetic energy before the collision is:

\(\text{KE(before)} = \frac{1}{2} \times 2.5 \times 6^2 = 45 \text{ J}\)

The kinetic energy after the collision is:

\(\text{KE(after)} = \frac{1}{2} \times 7.5 \times 2^2 = 15 \text{ J}\)

The loss of kinetic energy is:

\(\text{Loss of KE} = 45 - 15 = 30 \text{ J}\)

By the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

Before the collision, the momentum is given by:

\(m \times 2 + 0 \times 0.2 = 2m\)

After the collision, the momentum is given by:

\(m \times (-0.5) + 0.2 \times 1 = -0.5m + 0.2\)

Setting the total momentum before and after the collision equal gives:

\(2m = -0.5m + 0.2\)

Solving for \(m\):

\(2m + 0.5m = 0.2\)

\(2.5m = 0.2\)

\(m = \frac{0.2}{2.5} = 0.08\)

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

\(4 \times 10 + [0] = 4 \times 0.5v + 2v\)

Solving for \(v\):

\(40 = 2v + 2v\)

\(40 = 4v\)

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

(b) For the collision between B and C, using conservation of momentum:

\(2 \times 10 + [0] = 2v + 3v\)

Solving for \(v\):

\(20 = 5v\)

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

Since \(v_A > v_B\), another collision occurs.

(c) For the coalescence of A and B, using conservation of momentum:

\(4 \times 5 + 2 \times 4 = 4v + 2v\)

Solving for \(v\):

\(34 = 6v\)

\(v = \frac{14}{3} \text{ m/s}\)

Initial kinetic energy:

\(KE_{\text{initial}} = \frac{1}{2} \times 4 \times 10^2 = 200 \text{ J}\)

Final kinetic energy:

\(KE_{\text{final}} = \frac{1}{2} \times 6 \times \left( \frac{14}{3} \right)^2 + \frac{1}{2} \times 1 \times 12^2\)

\(KE_{\text{final}} = \frac{412}{3} \text{ J}\)

Loss of kinetic energy:

\(200 - \frac{412}{3} = \frac{188}{3} \text{ J}\)

(a) To find the speed of Q after the collision, we use the conservation of momentum. Before the collision, the velocity of P is 5 m s^-1 (calculated using energy conservation on the incline: \(v^2 = 0 + 2 \times 2.5 \times a\) where \(a = g \sin 30^{\circ} = 5\)).

Using conservation of momentum:

\(0.3 \times 5 + 0 = 0.3 \times 2 + 0.2 \times w\)

Solving for \(w\):

\(1.5 = 0.6 + 0.2w\)

\(0.2w = 0.9\)

\(w = 4.5\) m s^-1

(b) For the rough plane, after the collision, P and Q coalesce and move with speed 1.2 m s^-1. Using conservation of momentum:

\(0.3 \times z + 0 = 0.5 \times 1.2\)

Solving for \(z\):

\(0.3z = 0.6\)

\(z = 2\) m s^-1

To find the coefficient of friction \(\mu\), consider the work-energy principle:

Friction force on P after reaching the horizontal plane \(F = \mu \times 0.3g\).

Using energy conservation:

\(\mu \times 0.3g \times 1.5 = \frac{1}{2} \times 0.3 \times 5^2 - \frac{1}{2} \times 0.3 \times 2^2\)

Solving for \(\mu\):

\(\mu \times 0.3g \times 1.5 = 3.75 - 0.6\)

\(\mu \times 0.3g \times 1.5 = 3.15\)

\(\mu = \frac{3.15}{0.3 \times 9.8 \times 1.5}\)

\(\mu = 0.7\)

(a) Apply Newton's second law to the trailer:

\(T - 200 = 700 \times (-12)\)

\(T = 700 \times 12 + 200 = 8200 \text{ N}\)

The magnitude of the force in the tow-bar is 8200 N.

(b) Apply Newton's second law to the car:

\(T - 600 - F = 1600 \times (-12)\)

Using the value of \(T\) from part (a):

\(8200 - 600 - F = 1600 \times (-12)\)

\(F = 26800 \text{ N}\)

The braking force is 26800 N.

(c) Use the equation of motion:

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

where \(u = 22 \text{ m s}^{-1}, a = -12 \text{ m s}^{-2}, s = 17.5 \text{ m}\).

\(v^2 = 22^2 + 2 \times (-12) \times 17.5\)

\(v^2 = 484 - 420\)

\(v^2 = 64\)

\(v = 8 \text{ m s}^{-1}\)

The car hits the van at a speed of 8 m s^-1.

(d) Apply the conservation of momentum:

\((2300 \times 8) + (m \times 0) = (2300 \times 2) + (m \times 5)\)

\(18400 = 4600 + 5m\)

\(5m = 13800\)

\(m = 2760 \text{ kg}\)

The mass of the van is 2760 kg.