(a) Using the conservation of momentum, the initial momentum of the hammerhead is \(1.2v\). The final momentum of the combined hammerhead and nail is \((1.2 + 0.004) \times 40\). Setting these equal gives:

\(1.2v = (1.2 + 0.004) \times 40\)

\(1.2v = 48.16\)

Solving for \(v\), we get:

\(v = \frac{48.16}{1.2} = \frac{602}{15}\)

(b) Using the equation of motion \(v^2 = u^2 + 2as\), where \(v = 0\), \(u = 40\), and \(s = 0.04\), we have:

\(0 = 40^2 + 2 \times 0.04 \times a\)

\(0 = 1600 + 0.08a\)

\(a = -20000 \text{ m s}^{-2}\)

Using Newton's Second Law vertically, \(-R + (1.2 + 0.004)g = (1.2 + 0.004)a\):

\(-R + 12.04 = 1.204 \times (-20000)\)

\(-R + 12.04 = -24080\)

\(R = 24092.04\)

Rounding to the nearest whole number, \(R = 24,100 \text{ N}\).