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

\(0.1 imes 4 + 0 imes 0 = 0.4v + 0.1v\)

Solving for \(v\), we have:

\(0.4v + 0.1v = 0.4 \times v + 0.1 \times v = 0.4\)

Thus, \(0.5v = 0.4\)

\(v = \frac{0.4}{0.5} = 0.8 \text{ m/s}\)

Alternatively, considering the possibility of opposite direction:

\(0.1 imes 4 + 0 = 0.4v + 0.1(-v)\)

\(0.4 = 0.4v - 0.1v\)

\(0.4 = 0.3v\)

\(v = \frac{0.4}{0.3} = \frac{4}{3} \approx 1.33 \text{ m/s}\)

Therefore, the two possible speeds of P after the collision are 0.8 m/s and 1.33 m/s.

(a) By the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the particles come to rest after the collision, the total momentum after the collision is 0.

Initial momentum: \(3.2v + 2.4(-6) = 0\)

Solving for \(v\):

\(3.2v - 14.4 = 0\)

\(3.2v = 14.4\)

\(v = \frac{14.4}{3.2} = 4.5\)

(b) The initial kinetic energy of the system is the sum of the kinetic energies of both particles:

\(KE_{initial} = \frac{1}{2} \times 3.2 \times (4.5)^2 + \frac{1}{2} \times 2.4 \times 6^2\)

\(= \frac{1}{2} \times 3.2 \times 20.25 + \frac{1}{2} \times 2.4 \times 36\)

\(= 32.4 + 43.2 = 75.6\)

The final kinetic energy is 0 since both particles come to rest.

Thus, the loss of kinetic energy is 75.6 J.

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

\(m \times 5 + 0 = m \times 2 + 0.3v\)

Solving for \(v\):

\(5m = 2m + 0.3v\)

\(3m = 0.3v\)

\(v = \frac{3m}{0.3} = 10m \text{ m s}^{-1}\)

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

\(0.3 \times 10m + 0 = 0 + 0.6 \times 1.5\)

Solving for \(m\):

\(3m = 0.9\)

\(m = 0.3\)

(a) Use conservation of energy for particle P from A to B:

\(\frac{1}{2} \times 0.5 \times v^2 = 0.5 \times 9.8 \times 1.8\)

Solving gives \(v = 6\) m/s.

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

\(0.5 \times 6 + 0 = 0.5 \times 4 + 0.1 \times w\)

Solving gives \(w = 10\) m/s.

(b) Use conservation of momentum for the collision at C:

\(0.1 \times 10 = (0.1 + 0.4) \times z \Rightarrow z = 2\)

Use conservation of energy for the motion from C to F:

\(\frac{1}{2} \times (0.1 + 0.4) \times (2)^2 = (0.1 + 0.4) \times 9.8 \times h \Rightarrow h = 0.2\)

Using trigonometry, \(\theta = \sin^{-1} \left( \frac{0.2}{0.4} \right) \Rightarrow \theta = 30^\circ\).

(c) Q takes 0.7 s to travel from B to C.

\(0.4 = \frac{(\text{their } 2) \times 0.8}{2} \Rightarrow t = 0.4\)

Distance moved by P is \((0.7 + 0.8) \times 4 = 6\) m.

Distance from B is \(\frac{6}{2}\) m.

(a) For the 3 kg particle in limiting equilibrium at \(B\), resolve forces parallel and perpendicular to the plane. The normal reaction \(R = 3g \cos \alpha\). The frictional force \(F = 3g \sin \alpha\). Given \(\mu = \frac{F}{R}\), we have:

\(\mu = \frac{3g \sin \alpha}{3g \cos \alpha} = \tan \alpha = 0.75\)

(b) For the 1.5 kg particle, use energy conservation. The potential energy loss \(= 1.5gx \sin \alpha\) is converted to kinetic energy \(\frac{1}{2} \times 1.5 \times v_1^2\). Thus:

\(v_1^2 = 2gx \sin \alpha\)

After collision, use conservation of momentum:

\(1.5v_1 = 4.5v_2\)

\(v_2 = \frac{1}{3}v_1\)

For \(BC\), \(s = vt\) gives:

\(4 = 2v_2\)

\(v_2 = 2\)

Substitute back to find \(x\):

\(x = 3\)

(c) Calculate the total energy loss. Initial potential energy loss for \(AC\):

\(\text{PE loss} = (1.5 + 3)g \times 4 \sin \alpha = 135 \text{ J}\)

Kinetic energy gain for \(AC\):

\(\text{KE gain} = \frac{1}{2} \times 4.5 \times 2^2 = 9 \text{ J}\)

Total energy loss:

\(135 - 9 = 126 \text{ J}\)