Solved Problem on Collisions
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Two bodies, A and B, identical and of the same mass, are on a horizontal surface. Initially, body A has a speed v0 = 5 m/s, and body B is at rest. Body A collides with body B in a head-on elastic collision. Show that, in these conditions, after the collision, the speeds of bodies will be exchanged.
Problem data:
- Mass of ball A: mA = m;
- Mass of ball B: mB = m;
- Initial speed of ball A: v0A = 5 m/s;
- Initial speed of ball B: v0B = 0 m/s;
- Coefficient of restitution (elastic collision): e = 1.
Solution
The momentum of a body is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{p=mv}
\end{gather}
\]
From the Principle of Conservation of Momentum, we have that the initial momentum is equal to the
final momentum
\[
\begin{gather}
p_{i}=p_{f}\\[5pt]
p_{\small A i}+p_{\small B i}=p_{\small A f}+p_{\small B f}\\[5pt]
mv_{0 \small A}+mv_{0 \small B}=mv_{\small A}+mv_{\small B}
\end{gather}
\]
factoring the mass m on both sides
\[
\begin{gather}
\cancel{m}\left(v_{0\small A}+v_{0\small B}\right)=\cancel{m}\left(v_{\small A}+v_{\small B}\right)
\end{gather}
\]
substituting the problem data
\[
\begin{gather}
v_{0\small A}+v_{0\small B}=v_{\small A}+v_{\small B}\\[5pt]
5\;\mathrm{\frac{m}{s}}+0=v_{\small A}+v_{\small B}\\[5pt]
v_{\small A}+v_{\small B}=5\;\mathrm{\frac{m}{s}} \tag{I}
\end{gather}
\]
The coefficient of restitution is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{e=-{\left[\frac{v_{\small B}-v_{\small A}}{v_{0\small B}-v_{0\small A}}\right]}}
\end{gather}
\]
\[
\begin{gather}
1=-{\left[\frac{v_{\small B}-v_{\small A}}{0-5\;\mathrm{\frac{m}{s}}}\right]}\\[5pt]
-1=\frac{v_{\small B}-v_{\small A}}{-5\;\mathrm{\frac{m}{s}}}\\[5pt]
(-1)\times\left(-5\;\mathrm{\frac{m}{s}}\right)=v_{\small B}-v_{\small A}\\[5pt]
v_{\small B}-v_{\small A}=5\;\mathrm{\frac{m}{s}} \tag{II}
\end{gather}
\]
Equations (I) and (II) can be written as a system of two equations with two unknowns (vA
and vB), and adding the two equations
\[
\begin{gather}
\frac{
\left\{
\begin{matrix}
\phantom{\text{--}}v_{\small A}+v_{\small B}=5\;\mathrm{\frac{m}{s}}\\
-v_{\small A}+v_{\small B}=5\;\mathrm{\frac{m}{s}}
\end{matrix}
\right.
}
{0+2v_{\small B}=10\;\mathrm{\frac{m}{s}}}\\[5pt]
v_{\small B}=\frac{10\;\mathrm{\frac{m}{s}}}{2}\\[5pt]
v_{\small B}=5\;\mathrm{m/s}
\end{gather}
\]
substituting the value of vB found in the first equation
\[
\begin{gather}
v_{\small A}+5\;\mathrm{\frac{m}{s}}=5\;\mathrm{\frac{m}{s}}\\[5pt]
v_{\small A}=5\;\mathrm{\frac{m}{s}}-5\;\mathrm{\frac{m}{s}}\\[5pt]
v_{\small A}=0
\end{gather}
\]
How we wanted
vA = 0
and
vB = 5 m/s.
The bodies exchanged velocities after the collision.
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .