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

\(Initial momentum = Final momentum\)

\(0.1 \times 5 + 0 = 0.1 \times (-1) + 0.2 \times v\)

Solving for \(v\):

\(0.5 = -0.1 + 0.2v\)

\(0.6 = 0.2v\)

\(v = 3 \text{ m s}^{-1}\)

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

\(Initial momentum = Final momentum\)

\(0.2 \times 3 + 0 = 0.2 \times u + 0.5V\)

Given \(u \geq -1\) to prevent a subsequent collision between P and Q, solve for the greatest \(V\):

\(0.6 = 0.2(-1) + 0.5V\)

\(0.6 = -0.2 + 0.5V\)

\(0.8 = 0.5V\)

\(V = 1.6\)

(a) To find the speed of P immediately before the collision, use the equation of motion:

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

where \(u = 20\) m/s, \(a = -0.4 \times g = -4\) m/s², and \(s = 12.5\) m.

Substitute the values:

\(v^2 = 20^2 + 2(-4)(12.5)\)

\(v^2 = 400 - 100 = 300\)

\(v = \sqrt{300} = 10\sqrt{3} \text{ m/s}\)

(b) Use conservation of momentum to find the speed of R after the collision:

\(6 \times 10\sqrt{3} = (6 + 2)v'\)

\(v' = \frac{60\sqrt{3}}{8} = 7.5\sqrt{3} \text{ m/s}\)

Initial kinetic energy:

\(KE_i = \frac{1}{2} \times 6 \times (10\sqrt{3})^2 = 900 \text{ J}\)

Final kinetic energy:

\(KE_f = \frac{1}{2} \times 8 \times (7.5\sqrt{3})^2 = 675 \text{ J}\)

Loss of kinetic energy:

\(\text{Loss} = 900 - 675 = 225 \text{ J}\)

(c) To find the distance travelled by R before coming to rest, use:

\(0 = (7.5\sqrt{3})^2 + 2(-4)s\)

\(0 = 168.75 - 8s\)

\(s = \frac{168.75}{8} = 21.1 \text{ m}\)

Using the conservation of momentum, the total initial momentum is equal to the total final momentum.

\(Initial momentum of P = 0.2 imes 0.5 = 0.1 ext{ kg m s}^{-1}\)

\(Initial momentum of Q = 0.3 imes (-1) = -0.3 ext{ kg m s}^{-1}\)

\(Total initial momentum = 0.1 - 0.3 = -0.2 ext{ kg m s}^{-1}\)

Let the speed of P after the collision be v. Since Q comes to rest, its final momentum is 0.

\(Total final momentum = 0.2 imes v + 0 = 0.2v\)

By conservation of momentum:

\(0.2v = -0.2\)

Solving for v gives:

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

Since speed is the magnitude of velocity, the speed of P after the collision is 1 m s^-1.

Using conservation of momentum, we have:

\(6 \times 4 = 3m + 4 \times 1.5\) or \(6 \times 4 = 3m - 4 \times 1.5\)

Solving these equations gives \(m = 6\) and \(m = 10\).

Calculate the initial kinetic energy of sphere A:

\(\text{KE}_{A, \text{initial}} = \frac{1}{2} \times 4 \times 6^2 = 72 \text{ J}\)

Calculate the kinetic energy of sphere A after the collision:

\(\text{KE}_{A, \text{after}} = \frac{1}{2} \times 4 \times 1.5^2 = 4.5 \text{ J}\)

Calculate the kinetic energy of sphere B after the collision for each mass:

For \(m = 6\):

\(\text{KE}_{B, \text{after}} = \frac{1}{2} \times 6 \times 3^2 = 27 \text{ J}\)

For \(m = 10\):

\(\text{KE}_{B, \text{after}} = \frac{1}{2} \times 10 \times 3^2 = 45 \text{ J}\)

Calculate the loss of kinetic energy:

For \(m = 6\):

\(\text{KE}_{\text{loss}} = [\frac{1}{2} \times 4 \times 6^2 - \frac{1}{2} \times 4 \times 1.5^2 - \frac{1}{2} \times 6 \times 3^2] = 40.5 \text{ J}\)

For \(m = 10\):

\(\text{KE}_{\text{loss}} = [\frac{1}{2} \times 4 \times 6^2 - \frac{1}{2} \times 4 \times 1.5^2 - \frac{1}{2} \times 10 \times 3^2] = 22.5 \text{ J}\)

(a) The momentum of a particle is given by the product of its mass and velocity. For particle P, the momentum is:

\(\text{Momentum of } P = 0.2 \times 2 = 0.4 \text{ kg m s}^{-1}\)

(b) Using the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Initially, only P is moving, so the initial momentum is:

\(0.4 = 0.2 \times 0.3 + 0.5v\)

Solving for \(v\):

\(0.4 = 0.06 + 0.5v\)

\(0.34 = 0.5v\)

\(v = \frac{0.34}{0.5} = 0.68 \text{ m s}^{-1}\)