(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.

(a) Using conservation of momentum:

The initial momentum of the object is given by:

\(120 \times 8\)

The final momentum of the combined system (object + post) is:

\((120 + 40) \times v\)

Equating the initial and final momentum:

\(120 \times 8 = 160v\)

Solving for \(v\):

\(v = \frac{960}{160} = 6 \text{ m/s}\)

(b) Using Newton's second law and constant acceleration equations:

The net force on the system is:

\(1600 - 4800 = 160a\)

Solving for \(a\):

\(a = -20 \text{ m/s}^2\)

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

\(0 = 6^2 + 2 \times (-20) \times s\)

Solving for \(s\):

\(0 = 36 - 40s\)

\(40s = 36\)

\(s = 0.9 \text{ m}\)

Alternative method using work-energy principle:

Initial kinetic energy (KE) of the system:

\(\frac{1}{2} \times 160 \times 6^2\)

Work done against the resistive force:

\(4800s\)

Equating initial KE to work done:

\(\frac{1}{2} \times 160 \times 6^2 = 4800s\)

Solving for \(s\):

\(s = 0.9 \text{ m}\)

(a) For particle Q, apply Newton's second law along the plane:

\(-2mg \sin α - F = 2ma\)

Where \(F = 0.6 \times 2mg \cos α\).

Substitute \(\sin α = 0.8\) and \(\cos α = 0.6\):

\(-2m(9.8)(0.8) - 0.6 \times 2m(9.8)(0.6) = 2ma\)

\(-15.68m - 7.056m = 2ma\)

\(-22.736m = 2ma\)

\(a = -11.368 \approx -11.6 \text{ m/s}^2\)

(b) For particle P, apply Newton's second law:

\(mg \sin α - 0.6mg \cos α = ma\)

\(9.8 \times 0.8 - 0.6 \times 9.8 \times 0.6 = a\)

\(7.84 - 3.528 = a\)

\(a = 4.312 \text{ m/s}^2\)

Using constant acceleration equations, find the time \(T_1\) for Q to come to rest:

\(10 - 11.6T_1 = 0\)

\(T_1 = \frac{25}{29} \approx 0.862 \text{ s}\)

Calculate the distance traveled by P and Q:

\(s_P = \frac{1}{2} \times 4.4 \times (0.862)^2 = 1.635 \text{ m}\)

\(s_Q = 10 \times 0.862 - \frac{1}{2} \times 11.6 \times (0.862)^2 = 4.31 \text{ m}\)

Find the extra distance needed for collision:

\(d = 6.4 - (s_P + s_Q) = 0.455 \text{ m}\)

Find the time \(T_2\) for P to travel this extra distance:

\(d = 4.4T_2\)

\(T_2 = \frac{0.455}{4.4} \approx 0.103 \text{ s}\)

Total time before collision:

\(T = T_1 + T_2 = 0.862 + 0.103 = 0.965 \text{ s}\)

(c) Using conservation of momentum for the collision:

\(m \times 4.4 \times 0.965 + 2m \times 0 = (m + 2m)v\)

\(4.4m \times 0.965 = 3mv\)

\(v = \frac{4.4 \times 0.965}{3} \approx 1.79 \text{ m/s}\)

(a) Using 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:

\(km \times 6 - m \times 2\)

After the collision, the momentum is:

\((km + m) \times 4\)

Equating the two expressions:

\(km \times 6 - m \times 2 = (km + m) \times 4\)

\(6km - 2m = 4km + 4m\)

\(2km = 6m\)

\(k = 3\)

(b) The initial kinetic energy (KE) is given by:

\(\frac{1}{2} \times km \times 6^2 + \frac{1}{2} \times m \times (-2)^2\)

\(= 18km + 2m\)

The kinetic energy after the collision is:

\(\frac{1}{2} \times (km + m) \times 4^2\)

\(= 8(km + m)\)

\(= 8km + 8m\)

The loss of kinetic energy is:

\((18km + 2m) - (8km + 8m)\)

\(= 10km - 6m\)

Substituting \(k = 3\):

\(= 30m - 6m\)

\(= 24m\)

First, calculate the total momentum before the collision:

\(0.4 \times 2.5 - 0.5 \times 1.5\)

\(= 1.0 - 0.75 = 0.25 \text{ kg m s}^{-1}\)

Let the speed of P after the collision be \(v\) and the speed of Q be \(2v\). Using conservation of momentum:

\(0.4 \times 2.5 - 0.5 \times 1.5 = 0.4v + 0.5 \times 2v\)

\(0.25 = 0.4v + v\)

\(0.25 = 1.4v\)

\(v = \frac{0.25}{1.4} \approx 0.179 \text{ m s}^{-1}\)

Alternatively, consider the opposite direction for P:

\(0.4 \times 2.5 - 0.5 \times 1.5 = -0.4v + 0.5 \times 2v\)

\(0.25 = -0.4v + v\)

\(0.25 = 0.6v\)

\(v = \frac{0.25}{0.6} \approx 0.417 \text{ m s}^{-1}\)

Thus, the two possible speeds of P after the collision are 0.179 m s^-1 or 0.417 m s^-1.

Let the mass of particle A be m and the mass of particle B be 2m. The displacement equations for particles A and B are:

\(s_A = 30t - 5t^2\)

\(s_B = 15 - 5t^2\)

At collision, \(s_A + s_B = 15\), leading to \(15 = 30t\), so \(t = 0.5\).

At \(t = 0.5\), the velocities are:

\(v_A = 30 - 10 \times 0.5 = 25\) m/s

\(v_B = -10 \times 0.5 = -5\) m/s

Using conservation of momentum:

\(25(2m) - 5(m) = (3m)v\)

\(v = 15\) m/s

For the second case, let the mass of particle A be 2m and the mass of particle B be m:

\(25(m) - 5(2m) = (3m)v\)

\(v = 5\) m/s

The equations of motion for particle C are:

\(s = -13.75 + 15t - 5t^2\)

\(s = -13.75 + 5t - 5t^2\)

Solving for \(t\) when \(s = 0\):

\(t_{C_1} = 3.74\), \(t_{C_2} = 2.23\)

The difference in time is:

\(T = t_{C_1} - t_{C_2} = 1.50\) seconds

(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 \(v\):

\(0.4 = 0.06 + 0.5v\)

\(0.34 = 0.5v\)

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

(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.

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:

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

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

Initial momentum: \(0.3 \times 2 + 0.4 \times 0 = 0.6\).

Final momentum: \(0.3 \times 0.6 + 0.4 \times v\).

Equating initial and final momentum:

\(0.6 = 0.18 + 0.4v\).

Solving for \(v\):

\(0.4v = 0.42\).

\(v = 1.05\) m/s.

(b) Using conservation of momentum for the collision between B and C:

Initial momentum: \(0.4 \times 1.05 + m \times 0 = 0.42\).

Final momentum: \((0.4 + m) \times 0.5\).

Equating initial and final momentum:

\(0.42 = (0.4 + m) \times 0.5\).

\(0.42 = 0.2 + 0.5m\).

\(0.5m = 0.22\).

\(m = \frac{11}{25}\).

(c) To find the time from when B and C collide until A collides with the combined particle:

Distance traveled by B and C together: \(0.5t\).

Distance traveled by A: \(0.6t\).

Since the total distance is 2.1 m, we have:

\(0.6t - 0.9 = 0.5t\).

\(0.1t = 0.9\).

\(t = 9\) s.

Using conservation of momentum:

\(6 \times 5 + 2 \times (-3) = 6v_A + 2v_B\)

\(30 - 6 = 6v_A + 2v_B\)

\(24 = 6v_A + 2v_B\)

Given that the difference in speeds is 2 m s^-1, we have:

\(v_B = v_A + 2\)

Substitute into the momentum equation:

\(24 = 6v_A + 2(v_A + 2)\)

\(24 = 6v_A + 2v_A + 4\)

\(24 = 8v_A + 4\)

\(20 = 8v_A\)

\(v_A = 2.5 \text{ m s}^{-1}\)

Then:

\(v_B = 2.5 + 2 = 4.5 \text{ m s}^{-1}\)

Calculate initial kinetic energy (KE):

\(\text{Initial KE} = \frac{1}{2} \times 6 \times 5^2 + \frac{1}{2} \times 2 \times (-3)^2\)

\(= 75 + 9 = 84 \text{ J}\)

Calculate final kinetic energy (KE):

\(\text{Final KE} = \frac{1}{2} \times 6 \times (2.5)^2 + \frac{1}{2} \times 2 \times (4.5)^2\)

\(= 18.75 + 20.25 = 39 \text{ J}\)

Loss of KE:

\(\text{Loss of KE} = 84 - 39 = 45 \text{ J}\)