Elastic tennis

Problem

Elastic tennis. Two students imagine a game of “elastic tennis”: both the ball and the racket are perfectly elastic (coefficient of restitution $e = 1$) and frictionless. One of them claims that, with a sharp enough stroke, the ball can fail to bounce off the racket and stay stuck to it. The other firmly disagrees. (a) Who is right? (b) Show that, in a one-dimensional elastic collision between a ball of mass $m$ and a racket of mass $M$, the relative velocity always reverses: $v_\text{rel,dopo} = -v_\text{rel,prima}$. (c) Conclude: with $e = 1$ it is impossible for the ball to “follow” the racket.

Problem adapted from (Povey 2015, §4.2).

Solution

(a) The one who disagrees is right: the ball always bounces off. We prove it from the conservation laws.

(b) Let $v_1, v_2$ be the velocities before the collision (ball $m$ and racket $M$) and $v_1', v_2'$ those after. Conservation of momentum and kinetic energy hold: $\begin{aligned} m v_1 + M v_2 &= m v_1' + M v_2' \\ \tfrac{1}{2}m v_1^2 + \tfrac{1}{2}M v_2^2 &= \tfrac{1}{2}m v_1'^2 + \tfrac{1}{2}M v_2'^2 \end{aligned}$ I rewrite grouping by mass: $\begin{aligned} m(v_1 - v_1') &= M(v_2' - v_2) \\ m(v_1^2 - v_1'^2) &= M(v_2'^2 - v_2^2) \end{aligned}$ I divide the second by the first (factoring the differences of squares): $\frac{m(v_1-v_1')(v_1+v_1')}{m(v_1-v_1')} = \frac{M(v_2'-v_2)(v_2'+v_2)}{M(v_2'-v_2)}$ $v_1 + v_1' = v_2' + v_2 \;\Longrightarrow\; v_1 - v_2 = -(v_1' - v_2')$ That is, the relative velocity reverses: $v_\text{rel,dopo} = -v_\text{rel,prima}$

(c) “Staying stuck” would mean $v_1' = v_2'$, i.e. $v_\text{rel,dopo} = 0$. But from (b), with $e = 1$, $v_\text{rel,dopo} = -v_\text{rel,prima} \neq 0$ (the ball arrives with a non-zero relative velocity). So it is impossible: the ball always separates.

Final result: $\ev{v_\text{rel,dopo} = -v_\text{rel,prima}\ \Rightarrow\ \text{the ball always bounces off}}$

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