(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 \(s\):

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