Inelastic Collisions in One Dimension — Question 3

Question 3

A ballistic pendulum requires two different balances

A $0.020\,\mathrm{kg}$ pellet embeds in a stationary $0.980\,\mathrm{kg}$ bob suspended by a light cord. The combined bob rises by $0.20\,\mathrm m$. The short collision occurs at the bottom, and air drag is negligible. Use $g=9.8\,\mathrm{m/s^2}$.

1. Find the pellet speed immediately before impact.

2. Find the kinetic energy lost during embedding and explain why one energy equation from launch to peak would fail.

Question 3 - Solution

A ballistic pendulum requires two different balances

(a) Derivation and calculation

Use mechanical energy conservation during the swing, then momentum conservation during capture: $$\tfrac12(m+M)V^2=(m+M)gh\Rightarrow V=\sqrt{2gh}=\sqrt{3.92}=1.980\ \mathrm{m/s}.$$ $$mu=(m+M)V\Rightarrow\boxed{u=\frac{1.000}{0.020}\sqrt{3.92}=99.0\ \mathrm{m/s}}.$$

(b) Result and interpretation

Immediately before and after capture, $$K_i=\tfrac12(0.020)u^2=98.0\ \mathrm J,\qquad K_+=\tfrac12(1.000)(3.92)=1.96\ \mathrm J.$$ $$\boxed{K_{\mathrm{lost,impact}}=96.04\ \mathrm J}.$$ Impact is inelastic; the subsequent ideal swing converts the retained kinetic energy into gravitational potential energy.

Physical check

The impact retains only the fraction $m/(m+M)=0.020$ of the initial kinetic energy.