(a) Is the momentum of the pumpkin conserved?
No. The pumpkin alone is not an isolated system — the bullet exerts a large horizontal impulse during penetration. Its momentum changes from 0 to about 0.50 kg·m/s. However, the total momentum of bullet + pumpkin is conserved (smooth plane, neglect flying debris).
(b) Estimate the pumpkin speed after the bullet leaves
Using conservation of momentum with \(m_b = 0.005\,\text{kg}\), \(m_p = 4\,\text{kg}\), \(u_b = 300\,\text{m s}^{-1}\), \(v_b = 200\,\text{m s}^{-1}\), \(u_p = 0\):
\[v_p = \frac{m_b(u_b - v_b)}{m_p} = \frac{0.005 \times 100}{4} = 0.125\,\text{m s}^{-1}\]
Check:
\(p_\text{before} = 0.005 \times 300 = 1.50\,\text{kg·m/s}\)
\(p_\text{after} = 0.005 \times 200 + 4 \times 0.125 = 1.50\,\text{kg·m/s}\) ✓