Bowling ball versus pin

Problem

The elastic collision. A bowling ball of mass $M$ elastically strikes, in a straight line, a pin of mass $m \ll M$ initially at rest. (a) After the collision, how does the ball move? (b) How does the pin move? (c) What if the roles were reversed (light ball against a very heavy pin)? Answer qualitatively, reasoning about the limits of the 1D elastic collision formulas.

Solution

1D elastic collision formulas, with the target at rest ($v_2 = 0$) and the incoming ball at velocity $V$: $\begin{aligned} v_1' &= \frac{M-m}{M+m}\,V \\ v_2' &= \frac{2M}{M+m}\,V \end{aligned}$

(a) Ball ($M$ large, $m \ll M$). In the limit $m/M \to 0$: $v_1' = \frac{M-m}{M+m}\,V \;\longrightarrow\; V$ The ball carries on at almost the same speed (it slows down very little).

(b) Pin. Again with $m \ll M$: $v_2' = \frac{2M}{M+m}\,V \;\longrightarrow\; 2V$ The pin shoots forward at about twice the ball’s speed.

(c) Roles reversed: light ball $m$ against a very heavy pin $M$, now the ball is the incoming object. With $m \ll M$: $v_\text{pallino}' = \frac{m-M}{m+M}\,v \;\longrightarrow\; -v, \qquad v_\text{birillo}' = \frac{2m}{m+M}\,v \;\longrightarrow\; 0$ The ball bounces back at almost the same speed; the heavy pin remains practically still.

Final result: $\ev{\text{(a) ball} \approx V;\quad \text{(b) pin} \approx 2V;\quad \text{(c) ball} \to -v,\ \text{pin} \to 0}$

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Topics: Quantità di moto e urti Concepts: Urto elastico · Quantità di moto Skills: Impostazione simbolica