Solved Problem on Collisions
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Calculate the energy loss in an inelastic head-on collision between two spheres of masses m1 and m2 that move in the same direction with speeds v1 and v2.


Problem data:
  • Mass of sphere 1:    m1;
  • Mass of sphere 2:    m2;
  • Speed of sphere 1:    v1;
  • Speed of sphere 2:    v2.
Problem diagram:

Figure 1

Solution

For the shock to occur, we assume v1 > v2, as the collision is inelastic, the two spheres stick together after the collision, the momentum is conserved, and the kinetic energy of the system, after the collision is less than the kinetic energy before the collision.
The momentum is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {Q=mv} \end{gather} \]
The kinetic energy is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {K=\frac{mv^{2}}{2}} \end{gather} \]
Writing the equations for spheres 1 and 2 in the situations before and after the collision

Before the collision:
\[ \begin{gather} Q_{1}=m_{1}v_{1} \tag{I} \end{gather} \]
\[ \begin{gather} Q_{2}=m_{2}v_{2} \tag{II} \end{gather} \]
\[ \begin{gather} K_{1}^{i}=\frac{m_{1}v_{1}^{2}}{2} \tag{III} \end{gather} \]
\[ \begin{gather} K_{2}^{i}=\frac{m_{2}v_{2}^{2}}{2} \tag{IV} \end{gather} \]
After the collision:
\[ \begin{gather} Q=m_{1}v_{1}+m_{2}v_{2} \tag{V} \end{gather} \]
\[ \begin{gather} K^{f}=\frac{m_{1}v^{2}}{2}+\frac{m_{2}v^{2}}{2} \tag{VI} \end{gather} \]
where \( K_{1}^{i} \) and \( K_{2}^{i} \) are the initial kinetic energies of spheres 1 and 2, \( K^{f} \) is the final kinetic energy of the set, Q, and v are the momentum and velocity of the system after the collision
The dissipated energy ΔE will be the difference between the final energy and the initial energy of the spheres
\[ \begin{gather} \Delta E=K^{f}-\left(K_{1}^{i}+K_{2}^{i}\right) \end{gather} \]
substituting equations (VI), (III), and (IV) into this equation
\[ \begin{gather} \Delta E=\frac{m_{1}v^{2}}{2}+\frac{m_{2}v^{2}}{2}-\left(\frac{m_{1}v_{1}^{2}}{2}+\frac{m_{2}v_{2}^{2}}{2}\right)\\[5pt] \Delta E=\frac{v^{2}}{2}\left(m_{1}+m_{2}\right)-\left(\frac{m_{1}v_{1}^{2}}{2}+\frac{m_{2}v_{2}^{2}}{2}\right) \tag{VII} \end{gather} \]
To obtain v, we use the Law of Conservation of Momentum using equations (I) and (II) before the collision and equation (VI) after the collision
\[ \begin{gather} Q^{i}=Q^{f}\\[5pt] m_{1}v_{1}+m_{2}v_{2}=m_{1}v+m_{2}v\\[5pt] m_{1}v_{1}+m_{2}v_{2}=v(m_{1}+m_{2})\\[5pt] v=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}} \tag{VIII} \end{gather} \]
substituting equation (VIII) into equation (VII)
\[ \begin{gather} \Delta E=\frac{1}{2}\left(\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}\right)^{2}\left(m_{1}+m_{2}\right)-\frac{m_{1}v_{1}^{2}}{2}-\frac{m_{2}v_{2}^{2}}{2}\\[5pt] \Delta E=\frac{1}{2}\frac{\left(m_{1}v_{1}+m_{2}v_{2}\right)^{2}}{\left(m_{1}+m_{2}\right)^{\cancel{2}}}\cancel{\left(m_{1}+m_{2}\right)}-\frac{m_{1}v_{1}^{2}}{2}-\frac{m_{2}v_{2}^{2}}{2}\\[5pt] \Delta E=\frac{\left(m_{1}v_{1}+m_{2}v_{2}\right)^{2}}{2\left(m_{1}+m_{2}\right)}-\frac{m_{1}v_{1}^{2}}{2}-\frac{m_{2}v_{2}^{2}}{2} \end{gather} \]
In the first term on the right-hand side of the equation, the denominator is a Special Binomial Product   \( \left(a+b\right)^{2}=a^{2}+2 ab+b^{2} \)
\[ \left(a+b\right)^{2}=a^{2}+2 ab+b^{2} \]

writing the three terms on the right-hand side over the denominator 2(m1+m2), and expanding the Special Binomial Product
\[ \begin{gather} \Delta E=\frac{m_{1}^{2}v_{1}^{2}+2m_{1}v_{1}m_{2}v_{2}+m_{2}^{2}v_{2}^{2}-m_{1}v_{1}^{2}\left(m_{1}+m_{2}\right)-m_{2}v_{2}^{2}\left(m_{1}+m_{2}\right)}{2\left(m_{1}+m_{2}\right)}\\[5pt] \Delta E=\frac{m_{1}^{2}v_{1}^{2}+2m_{1}v_{1}m_{2}v_{2}+m_{2}^{2}v_{2}^{2}-m_{1}^{2}v_{1}^{2}-m_{1}m_{2}v_{1}^{2}-m_{1}m_{2}v_{2}^{2}-m_{2}^{2}v_{2}^{2}}{2\left(m_{1}+m_{2}\right)}\\[5pt] \Delta E=\frac{2m_{1}v_{1}m_{2}v_{2}-m_{1}m_{2}v_{1}^{2}-m_{1}m_{2}v_{2}^{2}}{2\left(m_{1}+m_{2}\right)}\\[5pt] \Delta E=\frac{-m_{1}m_{2}\left(v_{1}^{2}-2v_{1}v_{2}+v_{2}^{2}\right)}{2\left(m_{1}+m_{2}\right)} \end{gather} \]
The term in parentheses in the numerator is a Special Binomial Product   \( \left(a-b\right)^{2}=a^{2}-2 ab+b^{2} \)
\[ \left(a-b\right)^{2}=a^{2}-2 ab+b^{2} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\Delta E=\frac{-m_{1}m_{2}\left(v_{1}-v_{2}\right)^{2}}{2\left(m_{1}+m_{2}\right)}} \end{gather} \]
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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 .