Q.A bullet of mass $0.01\ \text{kg}$, moving horizontally at $300\ \text{m/s}$, embeds itself in a stationary wooden block of mass $2\ \text{kg}$ suspended by strings as a ballistic pendulum. Find the common speed of the bullet-and-block just after impact, and the height through which the block subsequently rises (take $g = 9.8\ \text{m/s}^2$).

Given: bullet mass $m=0.01\ \text{kg}$, bullet speed $u=300\ \text{m/s}$; block mass $M=2\ \text{kg}$, initially at rest; perfectly inelastic (bullet embeds in block); $g=9.8\ \text{m/s}^2$.

Common speed just after impact. By momentum conservation (a very short collision, so gravity's effect during the impact itself is negligible): $mu = (m+M)v' \quad\Longrightarrow\quad v' = \frac{mu}{m+M} = \frac{0.01\times300}{2.01} = \frac{3}{2.01} \approx 1.49\ \text{m/s}$ …

Use momentum conservation for the (perfectly inelastic) impact itself to find the common post-collision speed, then use energy conservation for the subseq …

• Applying energy conservation to the collision itself (kinetic energy is not conserved in a perfectly inelastic collision) instead of correctly using momentum conservation for that stage. …