Problem — Ballistic pendulum

Ballistic pendulum. A $20\;\mathrm{g}$ bullet at $400\;\mathrm{m/s}$ embeds itself in a $2\;\mathrm{kg}$ block hanging from a string. Combining conservation of momentum in the collision and conservation of mechanical energy afterwards, calculate the height to which the block rises.

Solution

Data: bullet $m = 20\;\mathrm{g} = 0{,}02\;\mathrm{kg}$, $v_0 = 400\;\mathrm{m/s}$; block $M = 2\;\mathrm{kg}$.

The problem has two stages requiring different principles.

Stage 1 — inelastic collision (conservation of momentum). The collision is extremely brief: the bullet and block stick together. Kinetic energy is not conserved (part becomes heat and deformation), but momentum is:

$m v_0 = (m + M)\,v_f \quad\Rightarrow\quad v_f = \frac{m v_0}{m + M} = \frac{0{,}02\cdot 400}{0{,}02 + 2} = \frac{8}{2{,}02} \approx 3{,}96\;\mathrm{m/s}$

$\ev{v_f \approx 3{,}96\;\mathrm{m/s}}$

Stage 2 — rise (conservation of mechanical energy). After the collision the block+bullet swings upward: the kinetic energy just gained becomes potential energy:

$\tfrac{1}{2}(m+M)v_f^2 = (m+M)gh \quad\Rightarrow\quad h = \frac{v_f^2}{2g} = \frac{(3{,}96)^2}{2\cdot 9{,}81} = \frac{15{,}7}{19{,}62} \approx 0{,}80\;\mathrm{m}$

$\ev{h \approx 0{,}80\;\mathrm{m}}$

Physical reading. The classic mistake is to apply conservation of energy to the collision as well: wrong, because the inelastic collision dissipates kinetic energy (here over $99\%$: from $\tfrac12 m v_0^2 = 1600\;\mathrm{J}$ to $\tfrac12(m+M)v_f^2 \approx 15{,}8\;\mathrm{J}$). Momentum, however, is conserved in the collision because no impulsive external horizontal forces act. Measuring $h$ lets you work back to $v_0$: this is how the ballistic pendulum is used to measure projectile speeds.

Collegamenti

Argomenti: Work and energy Concetti: Momentum · Conservation of mechanical energy Competenze: Conservation of momentum Oggetti: Ballistic pendulum