Three bodies in cascade

Problem

Challenge: three bodies in cascade. A ball $A$ (mass $m$, velocity $+v$) collides in a straight line with a ball $B$ (mass $2m$, at rest), which in turn collides with a ball $C$ (mass $m$, at rest). Both collisions are elastic and happen in cascade. Calculate the final velocities of the three balls.

Solution

For a 1D elastic collision with a stationary target (mass $m_1$ hitting $m_2$ at rest), conservation of $\vv{p}$ and kinetic energy give: $v_1' = \frac{m_1 - m_2}{m_1 + m_2}\,v_1, \qquad v_2' = \frac{2 m_1}{m_1 + m_2}\,v_1$

Collision 1 — $A$ ($m$, $+v$) against $B$ ($2m$, at rest): $m_1 = m$, $m_2 = 2m$. $v_A = \frac{m - 2m}{m + 2m}\,v = -\frac{1}{3}\,v$ $v_B' = \frac{2m}{m + 2m}\,v = \frac{2}{3}\,v$ $A$ bounces back; $B$ moves forward at $\tfrac{2}{3}v$. $A$ no longer touches anything, so $v_A = -v/3$ is final.

Collision 2 — $B$ ($2m$, now at $\tfrac{2}{3}v$) against $C$ ($m$, at rest): $m_1 = 2m$, $m_2 = m$, initial velocity $\tfrac{2}{3}v$. $v_B = \frac{2m - m}{2m + m}\cdot \frac{2}{3}v = \frac{1}{3}\cdot \frac{2}{3}v = \frac{2}{9}\,v$ $v_C = \frac{2\cdot 2m}{2m + m}\cdot \frac{2}{3}v = \frac{4}{3}\cdot \frac{2}{3}v = \frac{8}{9}\,v$ $B$ continues slower than $C$, so it never catches up: the collisions are finished. Result: $\ev{v_A = -\frac{v}{3},\quad v_B = \frac{2v}{9},\quad v_C = \frac{8v}{9}}$

Links

Topics: Quantità di moto e urti Concepts: Urto elastico · Conservazione della quantità di moto · Energia cinetica Skills: Conservazione della quantità di moto