Ballistic pendulum (velocity)

Problem

Ballistic pendulum. A $20\;\text{g}$ bullet with unknown velocity $v$ embeds itself in a $5\;\text{kg}$ wooden block hanging from a string. The block swings up by $20\;\text{cm}$. Calculate the initial velocity of the bullet by combining conservation of $\vv{p}$ at the collision with conservation of energy during the rise.

Solution

Data: $m = 20\;\text{g} = 0{,}020\;\text{kg}$, $M = 5\;\text{kg}$, $h = 20\;\text{cm} = 0{,}20\;\text{m}$, $g = 9{,}81\;\text{m/s}^2$.

The problem has two phases to be treated separately.

Phase 1 — totally inelastic collision (bullet embedding itself): momentum is conserved, energy is not. Let $V$ be the common velocity of block plus bullet right after the collision. $m v = (m + M)\,V \quad\Rightarrow\quad v = \frac{m+M}{m}\,V$

Phase 2 — rise: after the collision only gravity acts, so mechanical energy is conserved. All the kinetic energy becomes gravitational potential energy. $\tfrac{1}{2}(m+M)\,V^2 = (m+M)\,g\,h \quad\Rightarrow\quad V = \sqrt{2 g h}$ $V = \sqrt{2 \cdot 9{,}81 \cdot 0{,}20} = \sqrt{3{,}924} \approx 1{,}98\;\mathrm{m/s}$

Substituting $V$ into the phase-1 relation: $v = \frac{0{,}020 + 5}{0{,}020}\cdot 1{,}98 = 251 \cdot 1{,}98 \approx 497\;\mathrm{m/s}$ Result: $\ev{v \approx 497\;\mathrm{m/s}}$

Links

Topics: Momentum and collisions Concepts: Inelastic collision · Conservation of momentum · Kinetic energy · Gravitational potential energy Skills: Conservation of momentum Objects: Ballistic pendulum · Bullet