A $1.0$ -kg ball moving at $12 \mathrm{~m} / \mathrm{s}$ collides head-on with a $2.0-\mathrm{kg}$ ball moving in the opposite direction at $24 \mathrm{~m} / \mathrm{s}$. Determine the motion of each after impact if $(a) e=2 / 3,(b)$ the balls stick together, and
(c) the collision is perfectly elastic.
In all three cases the collision occurs along a straight line, and momentum is conserved. Hence,
$$
\begin{aligned}
\text { Momentum before } &=\text { Momentum after } \\
(1.0 \mathrm{~kg})(12 \mathrm{~m} / \mathrm{s})+(2.0 \mathrm{~kg})(-24 \mathrm{~m} / \mathrm{s}) &=(1.0 \mathrm{~kg}) v_{1}+(2.0 \mathrm{~kg}) v_{2}
\end{aligned}
$$
which becomes
$$
-36 \mathrm{~m} / \mathrm{s}=v_{1}+2 v_{2}
$$
(a) When $e=2 / 3$,
$$
e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad \frac{2}{3}=\frac{v_{2}-v_{1}}{(12 \mathrm{~m} / \mathrm{s})-(-24 \mathrm{~m} / \mathrm{s})}
$$
from which $24 \mathrm{~m} / \mathrm{s}=v_{2}-v_{1}$. Combining this with the momentum equation found above gives $v_{2}=-4.0 \mathrm{~m} / \mathrm{s}$ and $v_{1}=-28 \mathrm{~m} / \mathrm{s}$
(b) In this case $v_{1}=v_{2}=v$, and so the momentum equation becomes
$$
3 v=-36 \mathrm{~m} / \mathrm{s} \quad \text { or } \quad v=-12 \mathrm{~m} / \mathrm{s}
$$
(c) Here $e=1$, and
$$
e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad 1=\frac{v_{2}-v_{1}}{(12 \mathrm{~m} / \mathrm{s})-(-24 \mathrm{~m} / \mathrm{s})}
$$
from which $v_{2}-v_{1}=36 \mathrm{~m} / \mathrm{s}$. Adding this to the momentum equation yields $v_{2}=0$. Using this value for $v_{2}$ then leads to $v_{1}=-36 \mathrm{~m} / \mathrm{s}$.