Solved Problem on Collisions
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Two elastic balls, with masses m1 and m2 and speeds v1 and v2 respectively, collide head-on, their speeds are in the direction of the line joining their centers. Determine the speeds of the balls after the collision in the following cases:
a) The speed of the second ball before the impact is equal to zero;
b) The masses of the balls are equal.
Problem data:
- Mass of ball 1: m1;
- Mass of ball 2: m2;
- Initial speed of ball 1: v1;
- Initial speed of ball 2: v2.
We choose a reference frame pointing to the right, with the speed of ball 1 in the same direction as the reference frama, v1 > 0, and the speed of ball2 in the opposite direction, v2 < 0 (Figure 1).
Solution
As the collision is elastic, the momentum and kinetic energy of the system are conserved. We write the equations for balls 1 and 2 in the initial and final situations.
The momentum is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\vec Q=m\vec v}
\end{gather}
\]
The kinetic energy is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{K=\frac{mv^{2}}{2}}
\end{gather}
\]
- Before the collision
\[
\begin{gather}
Q_{1i}=m_{1}v_{1} \tag{I}
\end{gather}
\]
\[
\begin{gather}
Q_{2i}=m_{2}v_{2} \tag{II}
\end{gather}
\]
\[
\begin{gather}
K_{1i}=\frac{m_{1}v_{1}^{2}}{2} \tag{III}
\end{gather}
\]
\[
\begin{gather}
K_{2i}=\frac{m_{2}v_{2}^{2}}{2} \tag{IV}
\end{gather}
\]
- After the collision
\[
\begin{gather}
Q_{1f}=m_{1}v_{1f} \tag{V}
\end{gather}
\]
\[
\begin{gather}
Q_{2f}=m_{2}v_{2f} \tag{VI}
\end{gather}
\]
\[
\begin{gather}
K_{1f}=\frac{m_{1}v_{1f}^{2}}{2} \tag{VII}
\end{gather}
\]
\[
\begin{gather}
K_{2f}=\frac{m_{2}v_{2f}^{2}}{2} \tag{VIII}
\end{gather}
\]
Applying the Law of Conservation of linear Momentum, applying equations (I), (II), (V) and (VI)
\[
\begin{gather}
Q_{i}=Q_{f}\\[5pt]
m_{1}v_{1}-m_{2}v_{2}=m_{1}v_{1f}+m_{2}v_{2f} \tag{IX}
\end{gather}
\]
Applying the Law of Conservation of Energy, applying equations (III), (IV), (VII) and (VIII)
\[
\begin{gather}
K_{i}=K_{f}\\[5pt]
\frac{m_{1}v_{1}^{2}}{\cancel{2}}+\frac{m_{2}v_{2}^{2}}{\cancel{2}}=\frac{m_{1}v_{1f}^{2}}{\cancel{2}}+\frac{m_{2}v_{2f}^{2}}{\cancel{2}}\\[5pt]
m_{1}v_{1}^{2}+m_{2}v_{2}^{2}=m_{1}v_{1f}^{2}+m_{2}v_{2f}^{2} \tag{X}
\end{gather}
\]
Equations (IX) and (X) can be written as a system of two equations with two unknowns
(v1f and v2f)
\[
\begin{gather}
\left\{
\begin{matrix}
\;m_{1}v_{1}-m_{2}v_{2}=m_{1}v_{1f}+m_{2}v_{2f}\\
\;m_{1}v_{1}^{2}+m_{2}v_{2}^{2}=m_{1}v_{1f}^{2}+m_{2}v_{2f}^{2}
\end{matrix} \tag{XI}
\right.
\end{gather}
\]
a) Letting v2 = 0 in system (XI)
\[
\begin{gather}
\left\{
\begin{matrix}
\;m_{1}v_{1}-m_{2}.0=m_{1}v_{1f}+m_{2}v_{2f}\\
\;m_{1}v_{1}^{2}+m_{2}.0^{2}=m_{1}v_{1f}^{2}+m_{2}v_{2f}^{2}
\end{matrix}
\right.
\\[5pt]
\left\{
\begin{matrix}
\;m_{1}v_{1}=m_{1}v_{1f}+m_{2}v_{2f}\\
\;m_{1}v_{1}^{2}=m_{1}v_{1f}^{2}+m_{2}v_{2f}^{2}
\end{matrix}
\right.
\end{gather}
\]
solving the first equation for v2f
\[
\begin{gather}
m_{2}v_{2f}=m_{1}v_{1}-m_{1}v_{1f}\\[5pt]
v_{2f}=\frac{1}{m_{2}}(m_{1}v_{1}-m_{1}v_{1f}) \tag{XII}
\end{gather}
\]
and substituting in the second equation
\[
\begin{gather}
m_{1}v_{1}^{2}=m_{1}v_{1f}^{2}+m_{2}\left[\frac{1}{m_{2}}(m_{1}v_{1}-m_{1}v_{1f})\right]^{2}\\[5pt]
m_{1}v_{1}^{2}=m_{1}v_{1f}^{2}+\cancel{m_{2}}\frac{1}{m_{2}^{\cancel{2}}}(m_{1}v_{1}-m_{1}v_{1f})^{2}
\end{gather}
\]
the term in parentheses on the right-hand side of the equality is a special binomial of the type
\( (a-b)^{2}=a^{2}-2ab+b^{2} \)
\[
\begin{gather}
m_{1}v_{1}^{2}=m_{1}v_{1f}^{2}+\frac{1}{m_{2}}\left(m_{1}^{2}v_{1}^{2}-2m_{1}^{2}v_{1}v_{1f}+m_{1}^{2}v_{1f}^{2}\right) \tag{XIII}
\end{gather}
\]
multiplying equation (XIII) by m2
\[
\begin{gather}
\qquad \qquad \qquad m_{1}v_{1}^{2}=m_{1}v_{1f}^{2}+\frac{1}{m_{2}}\left(m_{1}^{2}v_{1}^{2}-2m_{1}^{2}v_{1}v_{1f}+m_{1}^{2}v_{1f}^{2}\right)\qquad (\;\times m_{2}\;)\\[5pt]
\; m_{1}m_{2}v_{1}^{2}=m_{1}m_{2}v_{1f}^{2}+m_{1}^{2}v_{1}^{2}-2m_{1}^{2}v_{1}v_{1f}+m_{1}^{2}v_{1f}^{2}\\[5pt]
\; \cancel{m_{1}}m_{2}v_{1}^{2}=\cancel{m_{1}}\left( m_{2}v_{1f}^{2}+m_{1}v_{1}^{2}-2m_{1}v_{1}v_{1f}+m_{1}v_{1f}^{2}\right)\\[5pt]
m_{2}v_{1}^{2}=m_{2}v_{1f}^{2}+m_{1}v_{1}^{2}-2m_{1}v_{1}v_{1f}+m_{1}v_{1f}^{2}\\[5pt]
\left(m_{2}+m_{1}\right)v_{1f}^{2}-2m_{1}v_{1}v_{1f}+m_{1}v_{1}^{2}-m_{2}v_{1}^{2}=0
\end{gather}
\]
This is a Quadratic Equation of type
\( ax^{2}+bx+c=0 \)
where the unknown is the value v1f.
Solution ofde
\( \underbrace{\left(m_{2}+m_{1}\right)}_{a}\underbrace{v_{1f}^{2}}_{x^{2}}-\underbrace{2m_{1}v_{1}}_{b}\underbrace{v_{1f}}_{x}+\underbrace{m_{1}v_{1}^{2}-m_{2}v_{1}^{2}}_{c}=0 \)
\[ \underbrace{\left(m_{2}+m_{1}\right)}_{a}\underbrace{v_{1f}^{2}}_{x^{2}}-\underbrace{2m_{1}v_{1}}_{b}\underbrace{v_{1f}}_{x}+\underbrace{m_{1}v_{1}^{2}-m_{2}v_{1}^{2}}_{c}=0 \]
\[
\begin{align}
& \Delta=b^{2}-4ac=\left(-2m_{1}v_{1}\right)^{2}-4\left(m_{2}+m_{1}\right)\left(m_{1}v_{1}^{2}-m_{2}v_{1}^{2}\right)\\[5pt]
& \Delta=4m_{1}^{2}v_{1}^{2}-4\left(m_{1}m_{2}v_{1}^{2}-m_{2}^{2}v_{1}^{2}+m_{1}^{2}v_{1}^{2}-m_{1}m_{2}v_{1}^{2}\right)\\[5pt]
& \Delta=4m_{1}^{2}v_{1}^{2}-4\left(-m_{2}^{2}v_{1}^{2}+m_{1}^{2}v_{1}^{2}\right)\\[5pt]
& \Delta=4m_{1}^{2}v_{1}^{2}+4m_{2}^{2}v_{1}^{2}-4m_{1}^{2}v_{1}^{2}\\[5pt]
& \Delta =4m_{2}^{2}v_{1}^{2}\\[10pt]
& v_{1f}=\dfrac{-b\pm\sqrt{\Delta \;}}{2a}=\dfrac{-\left(-2m_{1}v_{1}\right)\pm\sqrt{4m_{2}^{2}v_{1}^{2}\;}}{2(m_{1}+m_{2})}\\[5pt]
& v_{1f}=\dfrac{2m_{1}v_{1}\pm2m_{2}v_{1}}{2(m_{1}+m_{2})}
\end{align}
\]
\[
\begin{gather}
v_{1f}=\frac{(m_{1}+m_{2})}{(m_{1}+m_{2})}v_{1}\\[5pt]
v_{1f}=v_{1} \tag{XIV}
\end{gather}
\]
\[
\text{or}
\]
\[
\begin{gather}
v_{1f}=\frac{(m_{1}-m_{2})}{(m_{1}+m_{2})}v_{1}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_{1f}=\frac{(m_{1}-m_{2})}{(m_{1}+m_{2})}v_{1}}
\end{gather}
\]
substituting this solution into equation (XII)
\[
\begin{gather}
v_{2f}=\frac{1}{m_{2}}\left[m_{1}v_{1}-m_{1}\frac{\left(m_{1}-m_{2}\right)}{(m_{1}+m_{2})}v_{1}\right]\\[5pt]
v_{2f}=\frac{1}{m_{2}}\left[\frac{m_{1}v_{1}(m_{1}+m_{2})-m_{1}\left(m_{1}-m_{2}\right)v_{1}}{(m_{1}+m_{2})}\right]\\[5pt]
v_{2f}=\frac{1}{m_{2}}\left[\frac{m_{1}^{2}v_{1}+m_{1}m_{2}v_{1}-m_{1}^{2}v_{1}+m_{1}m_{2}v_{1}}{(m_{1}+m_{2})}\right]\\[5pt]
v_{2f}=\frac{1}{\cancel{m_{2}}}\left[\frac{2m_{1}\cancel{m_{2}}v_{1}}{(m_{1}+m_{2})}\right]
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_{2f}=\frac{2m_{1}v_{1}}{(m_{1}+m_{2})}}
\end{gather}
\]
If m1 > m2, the final speed of ball 1 will be positive, v1f > 0, since m1 − m2 > 0, the final speed of ball 2 will also be positive, v2f > 0. This means that after the collision, ball 1 continues to move in the same direction of the trajectory with a lower speed, and ball 2, which was at rest, begins to move in the same direction (Figure 2-A).
If m1 < m2, the final speed of ball 1 will be negative, v1f < 0, since m1 − m2 < 0, the final speed of ball 2 will be positive, v2f > 0. This means that ball 1 moveing in the same direction of the trajectory, reverses its movement after the impact and returns in the oposite direction the trajectory, ball 2 begins to move in the same direction of the trajectory (Figure 2-B).
Substituting solution (XIV) into equation (XII)
\[
\begin{gather}
v_{2f}=\frac{1}{m_{2}}(m_{1}v_{1}-m_{1}v_{1})\\[5pt]
v_{2f}=0 \tag{XV}
\end{gather}
\]
Note: Solutions (XIV) and (XV) were not used, as they are physically impossible.
Ball 1 would continue with its speed in the same direction of the trajectory and ball 2 would
continue at rest, there would be no collision, ball 1 would be a “ghost ball” passing through
ball 2 (Figure 3).
b) Lettng m1 = m2 = m in system (XI)
\[
\begin{gather}
\left\{
\begin{matrix}
\;\cancel{m}v_{1}-\cancel{m}v_{2}=\cancel{m}v_{1f}+\cancel{m}v_{2f}\\
\;\cancel{m}v_{1}^{2}+\cancel{m}v_{2}^{2}=\cancel{m}v_{1f}^{2}+\cancel{m}v_{2f}^{2}
\end{matrix}
\right.
\\[5pt]
\left\{
\begin{matrix}
\;v_{1}-v_{2}=v_{1f}+v_{2f}\\
\;v_{1}^{2}+v_{2}^{2}=v_{1f}^{2}+v_{2f}^{2}
\end{matrix}
\right.
\end{gather}
\]
solving the first equation for v2f
\[
\begin{gather}
v_{2f}=v_{1}-v_{2}-v_{1f} \tag{XVI}
\end{gather}
\]
and substituting in the second equation
\[
\begin{gather}
v_{1}^{2}+v_{2}^{2}=v_{1f}^{2}+(v_{1}-v_{2}-v_{1f})^{2}
\end{gather}
\]
the term in parentheses on the right-hand side of the equality is of type
\( (a-b-c)^{2}=a^{2}+b^{2}+c^{2}-2ab-2ac+2bc \)
\( (a-b-c)^{2}=a^{2}+b^{2}+c^{2}-2ab-2ac+2bc \)
applying to the equation above
\[
\begin{gather}
v_{1}^{2}+v_{2}^{2}=v_{1f}^{2}+v_{1}^{2}-2v_{1}v_{2}-2v_{1}v_{1f}+v_{2}^{2}+2v_{2}v_{1f}+v_{1f}^{2}
\end{gather}
\]
Note: We can directly multiply the two terms
\[
\begin{array}{l}
\begin{array}{l}
v_{1}-v_{2}-v_{1f}\\
v_{1}-v_{2}-v_{1f}
\end{array}\\
\hline
\begin{alignat}{4}
& v_{1}^{2} & - & v_{1}v_{2} & - & v_{1}v_{1f}\\
& & - & v_{1}v_{2} & & & + & v_{2}^{2} & + & v_{2}v_{1f}\\
& & & & - & v_{1}v_{1f} & & & + & v_{2}v_{1f} & + & v_{1f}^{2}\\
\hline
& v_{1}^{2} & - & 2 v_{1}v_{2} & - & 2 v_{1}v_{1f} & + & v_{2}^{2} & + & 2 v_{2}v_{1f} & + & v_{1f}^{2}
\end{alignat}\\
\end{array}
\]
\[
\begin{gather}
\cancel{v_{1}^{2}}+\cancel{v_{2}^{2}}=\cancel{v_{1}^{2}}-2v_{1}v_{2}-2v_{1}v_{1f}+\cancel{v_{2}^{2}}+2v_{2}v_{1f}+v_{1f}^{2}\\[5pt]
0=v_{1f}^{\;2}-2v_{1}v_{2}-2v_{1}v_{1f}+2v_{2}v_{1f}+v_{1f}^{2}\\[5pt]
2v_{1f}^{;2}+\left(2v_{2}-2v_{1}\right)v_{1f}-2v_{1}v_{2}=0
\end{gather}
\]
This is a Quadratuc Equation of type
\( ax^{2}+bx+c=0 \)
where the unknown is the value v1f.
Solution of
\( \underbrace{2}_{a}\underbrace{v_{1f}^{2}}_{x^{2}}+\underbrace{\left(2v_{2}-2v_{1}\right)}_{b}\underbrace{v_{1f}}_{x}-\underbrace{2v_{1}v_{2}}_{c}=0 \)
\( \underbrace{2}_{a}\underbrace{v_{1f}^{2}}_{x^{2}}+\underbrace{\left(2v_{2}-2v_{1}\right)}_{b}\underbrace{v_{1f}}_{x}-\underbrace{2v_{1}v_{2}}_{c}=0 \)
\[
\begin{align}
& \Delta=b^{2}-4ac=\left(2v_{2}-2v_{1}\right)^{2}-4.2\left(-2v_{1}v_{2}\;\right)\\[5pt]
& \Delta=4v_{2}^{2}-8v_{1}v_{2}+4v_{1}^{2}-8\left(-2v_{1}v_{2}\right)\\[5pt]
& \Delta =4v_{1}^{2}-8v_{1}v_{2}+4v_{2}^{2}+16v_{1}v_{2}\\[5pt]
& \Delta=4v_{1}^{2}+8v_{1}v_{2}+4v_{2}^{2}\\[5pt]
& \Delta=4\left(v_{1}^{2}+2v_{1}v_{2}+v_{2}^{2}\right)
\end{align}
\]
the term in parentheses is a special binomia of the type
\( (a^{2}+2ab+b^{2})=(a+b)^{2} \)
\[
\begin{align}
& \Delta =4\left(v_{1}+v_{2}\right)^{2} \\[10pt]
& v_{1f}=\frac{-b\pm \sqrt{\Delta}}{2a}=\frac{-\left(2v_{2}-2v_{1}\right)\pm \sqrt{4\left(v_{1}+v_{2}\right)^{2}}}{2\times 2}\\[5pt]
& v_{1f}=\frac{\cancel{2}\left(v_{1}-v_{2}\right)\pm \cancel{2}\left(v_{1}+v_{2}\right)}{\cancel{2}\times 2}\\[5pt]
& v_{1f}=\frac{v_{1}-v_{2}+v_{1}+v_{2}}{2}\qquad \text{ou}\qquad v_{1f}=\frac{v_{1}-v_{2}-v_{1}-v_{2}}{2}\\[5pt]
& v_{1f}=\frac{2}{2}v_{\;1}\qquad \text{ou}\qquad v_{1f}=-\frac{2}{2}v_{\;2}
\end{align}
\]
\[
\begin{gather}
v_{1f}=v_{1} \tag{XVII}
\end{gather}
\]
\[
\text{or}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_{1f}=-v_{2}}
\end{gather}
\]
substituting this solution into equation (XVI)
\[
\begin{gather}
v_{2f}=v_{1}-v_{2}-(-v_{2})\\[5pt]
v_{2f}=v_{1}-v_{2}+v_{2}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_{2f}=v_{1}}
\end{gather}
\]
In this case, the balls change their speed, ball 1 returns in the oposite direction of the trajectory
with the speed equal to the speed of ball 2, while ball 2 returns in the same direction of the
trajectory with the speed equal to the speed of ball 1 (Figure 4).
Substituting the solution (XVII) into equation (XVI)
\[
\begin{gather}
v_{2f}=v_{1}-v_{2}-v_{1}\\[5pt]
v_{2f}=-v_{2} \tag{XVIII}
\end{gather}
\]
Note: Solutions (XVII) and (XVIII) were not used, as they are physically impossible.
Ball 1 would continue with its speed in the same direction of the trajectory and ball 2 would
continue with its speed in the oposite direction of the trajectory, there would be no collision,
there would be two “ghost balls” passing through each other (Figure 5).
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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 .