Elastic collision between different masses

Problem

A $2\;\text{kg}$ ball travelling at $5\;\text{m/s}$ collides elastically with a stationary $3\;\text{kg}$ ball. Calculate the velocities after the collision and comment on which one bounces back and which one moves forward.

Solution

Elastic collision in one dimension: both momentum and kinetic energy are conserved. Data: $m_1=2\;\text{kg}$, $v_1=5\;\text{m/s}$, $m_2=3\;\text{kg}$, $v_2=0$. Setting up the before/after tables:

body	mass	$v$ before	$v$ after
ball 1	$2\;\text{kg}$	$5\;\text{m/s}$	$v_1'$
ball 2	$3\;\text{kg}$	$0$	$v_2'$

For an elastic collision with a stationary target, the solving formulas hold: $v_1' = \frac{m_1 - m_2}{m_1 + m_2}\,v_1, \qquad v_2' = \frac{2\,m_1}{m_1 + m_2}\,v_1$ Substituting the values: $v_1' = \frac{2 - 3}{2 + 3}\cdot 5 = \frac{-1}{5}\cdot 5 = -1\ \mathrm{m/s}, \qquad v_2' = \frac{2\cdot 2}{5}\cdot 5 = 4\ \mathrm{m/s}$ Momentum check: $2\cdot(-1)+3\cdot 4 = -2+12 = 10 = 2\cdot 5$. ✓ Kinetic energy check: $\tfrac12\cdot 2\cdot(-1)^2 + \tfrac12\cdot 3\cdot 4^2 = 1+24 = 25 = \tfrac12\cdot 2\cdot 5^2$. ✓ The light ball (2 kg) bounces backwards (negative velocity) because it hits a heavier target; the heavy ball (3 kg) moves forward. Note (typo): the original text and figure give $v_1' = -0{,}2\ \mathrm{m/s}$, but the $\vv{p}$ and $E_\text{kin}$ balances give $v_1' = -1\ \mathrm{m/s}$: the correct value is the latter. $\ev{v_1' = -1\ \mathrm{m/s}\ (\text{bounces back}),\quad v_2' = 4{,}0\ \mathrm{m/s}}$

Links

Topics: Momentum and collisions Concepts: Elastic collision · Kinetic energy Skills: Two-table method Methods: Parallel two-table method