Two identical balls collide head-on. The initial velocity of one is $0.75 \mathrm{~m} / \mathrm{s}$ - EAST, while that of the other is $0.43 \mathrm{~m} / \mathrm{s}$ -WEST. If the collision is perfectly elastic, what is the final velocity of each ball?
Because the collision is perfectly elastic, both momentum and KE are conserved. Since the collision is head-on, all motion takes place along a straight line. Take east as positive and call the mass of each ball $m$. Momentum is conserved in a collision, so we can write
Momentum before $=$ Momentum after
$$
m(0.75 \mathrm{~m} / \mathrm{s})+m(-0.43 \mathrm{~m} / \mathrm{s})=m v_{1}+m v_{2}
$$
where $v_{1}$ and $v_{2}$ are the final values. This equation simplifies to
$$
0.32 \mathrm{~m} / \mathrm{s}=v_{1}+v_{2}
$$
Because the collision is assumed to be perfectly elastic, KE is also conserved. Thus,
$$
\begin{array}{c}
\text { KE before }=\text { KE after } \\
\frac{1}{2} m(0.75 \mathrm{~m} / \mathrm{s})^{2}+\frac{1}{2} m(-0.43 \mathrm{~m} / \mathrm{s})^{2}=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2}
\end{array}
$$
This equation can be simplified to
$$
0.747=v_{1}^{2}+v_{2}^{2}
$$
We can solve for $u_{2}$ in Eq. (1) to get $v_{2}=0.32-v_{1}$ and substitute that into Eq. (2). This yields
$$
0.747=\left(0.32-v_{1}\right)^{2}+v_{1}^{2}
$$
$2 v_{1}^{2}-0.64 v_{1}-0.645=0$
Using the quadratic formula,
$$
v_{1}=\frac{0.64 \pm \sqrt{(0.64)^{2}+5.16}}{4}=0.16 \pm 0.59 \mathrm{~m} / \mathrm{s}
$$
from which $v_{1}=0.75 \mathrm{~m} / \mathrm{s}$ or $-0.43 \mathrm{~m} / \mathrm{s}$. Substitution back into $\mathrm{Eq} .$
(1) gives $v_{2}=-0.43 \mathrm{~m} / \mathrm{s}$ or $0.75 \mathrm{~m} / \mathrm{s}$. Two choices for answers are
available:
$\left(v_{1}=0.75 \mathrm{~m} / \mathrm{s}, v_{2}=-0.43 \mathrm{~m} / \mathrm{s}\right) \quad$ and $\quad\left(v_{1}=-0.43 \mathrm{~m} / \mathrm{s}, v_{2}=0.75 \mathrm{~m} / \mathrm{s} \mathrm{s}\right.$
We must discard the first choice because it implies that the balls continue on unchanged; that is to say, no collision occurred. The correct answer is therefore $v_{1}=-0.43 \mathrm{~m} / \mathrm{s}$ and $v_{2}=0.75 \mathrm{~m} / \mathrm{s}$, which tells us that in a perfectly elastic, head-on collision between equal masses, the two bodies simply exchange velocities. Hence, $\overrightarrow{\mathbf{v}}_{1}=0.43 \mathrm{~m} / \mathrm{s}-$ WEST and $\overrightarrow{\mathbf{v}}_{2}=0.75 \mathrm{~m} / \mathrm{s}$ - EAST.
Alternative Method
If we recall that $e=1$ for a perfectly elastic head-on collision, then
$$
e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad 1=\frac{v_{2}-v_{1}}{(0.75 \mathrm{~m} / \mathrm{s})-(20.43 \mathrm{~m} / \mathrm{s})}
$$
which gives
$$
v_{2}-v_{1}=1.18 \mathrm{~m} / \mathrm{s}
$$
Equations (1) and (3) determine $v_{1}$ and $v_{2}$ uniquely.