(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