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Proof 31: Elastic-collision speed “conservation”

We call a collision elastic if the total kinetic energy is conserved (same before as after the collision) and there are no other energies involved. Let’s imagine the most general case of two objects – we’ll call them $\text A$ and $\text B$ – colliding elastically.

The momentum conservation law with momentum being $p=mv$:

$$\begin{align}
\sum p_\text{before}&=\sum p_\text{after} \quad\Leftrightarrow\quad\\
p_\text{A,before}+p_\text{B,before}&=p_\text{A,after}+p_\text{B,after} \quad\Leftrightarrow\quad\\
m_\text{A}v_\text{A,before}+m_\text{B}v_\text{B,before}&=m_\text{A}v_\text{A,after}+m_\text{B}v_\text{B,after} \quad\Leftrightarrow\quad\\
m_\text{A}v_\text{A,before}-m_\text{A}v_\text{A,after}&=m_\text{B}v_\text{B,after}-m_\text{B}v_\text{B,before} \quad\Leftrightarrow\quad\\
m_\text{A}\left(v_\text{A,before}-v_\text{A,after}\right) &=m_\text{B}\left(v_\text{B,after}-v_\text{B,before}\right) \quad\Leftrightarrow\quad\\
m_\text{A}&=m_\text{B}\frac{v_\text{B,after}-v_\text{B,before}}{v_\text{A,before}-v_\text{A,after}} \quad\Leftrightarrow\quad\\
\end{align}$$

We save this result for later. Now, on to the kinetic-energy conservation requirement with kinetic energy being $K=\frac 12 mv^2$:1

$$\begin{align}
\sum K_\text{before}&=\sum K_\text{after} \quad\Leftrightarrow\quad\\
K_\text{A,before}+K_\text{B,before}&=K_\text{A,after}+K_\text{B,after} \quad\Leftrightarrow\quad\\
\frac 12 m_\text{A}v_\text{A,before}^2+\frac 12 m_\text{B}v_\text{B,before}^2&=\frac 12 m_\text{A}v_\text{A,after}^2+\frac 12 m_\text{B}v_\text{B,after}^2 \quad\Leftrightarrow\quad\\
m_\text{A}v_\text{A,before}^2+m_\text{B}v_\text{B,before}^2&=m_\text{A}v_\text{A,after}^2+m_\text{B}v_\text{B,after}^2 \quad\Leftrightarrow\quad\\
m_\text{A}v_\text{A,before}^2-m_\text{A}v_\text{A,after}^2&=m_\text{B}v_\text{B,after}^2-m_Bv_\text{B,before}^2 \quad\Leftrightarrow\quad\\
m_\text{A}\left(v_\text{A,before}^2-v_\text{A,after}^2\right) &=m_\text{B}\left(v_\text{B,after}^2-v_\text{B,before}^2 \right)\quad\Leftrightarrow\quad\\
\end{align}$$

Insert the result from the momentum conservation:

$$\begin{align}
\underbrace{\cancel{m_\text{B}}\frac{v_\text{B,after}-v_\text{B,before}}{v_\text{A,before}-v_\text{A,after}}}_{m_{\text A}}\left(v_\text{A,before}^2-v_\text{A,after}^2\right) &=\cancel{m_\text{B}}\left(v_\text{B,after}^2-v_\text{B,before}^2 \right) \quad\Leftrightarrow\quad\\

\frac{v_\text{B,after}-v_\text{B,before}}{v_\text{A,before}-v_\text{A,after}}\left(v_\text{A,before}^2-v_\text{A,after}^2\right) &=v_\text{B,after}^2-v_\text{B,before}^2 \quad\Leftrightarrow\quad\\
\end{align}$$

Use the formula for a sum times a difference:

$$\begin{align}
\frac{v_\text{B,after}-v_\text{B,before}}{\cancel{v_\text{A,before}-v_\text{A,after}}}\underbrace{(v_\text{A,before}+v_\text{A,after})\cancel{(v_\text{A,before}-v_\text{A,after})}}_{v_\text{A,before}^2-v_\text{A,after}^2} &=\underbrace{(v_\text{B,after}+v_\text{B,before})(v_\text{B,after}-v_\text{B,before})}_{v_\text{B,after}^2-v_\text{B,before}^2} \quad\Leftrightarrow\quad\\

\cancel{(v_\text{B,after}-v_\text{B,before})}(v_\text{A,before}+v_\text{A,after})&=(v_\text{B,after}+v_\text{B,before})\cancel{(v_\text{B,after}-v_\text{B,before})} \quad\Leftrightarrow\quad\\

v_\text{A,before}+v_\text{A,after}&=v_\text{B,after}+v_\text{B,before} \quad\Leftrightarrow\quad\\

v_\text{A,before}-v_\text{B,before}&=v_\text{B,after}-v_\text{A,after} \quad\Leftrightarrow\quad\\

v_\text{A,before}-v_\text{B,before}&=-(v_\text{A,after}-v_\text{B,after})\quad\Leftrightarrow\quad\\

\Delta v_\text{before}&=-\Delta v_\text{after} \qquad\blacksquare
\end{align}$$

The difference in speed between the two colliding objects $\Delta v$2 is here shown to remain constant before and after the collision, although flipped (changing sign).

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