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.
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