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[SOLVED] Elastic Collision

dwsmith

Well-known member
Feb 1, 2012
1,673
We have 2 masses: one with mass \(M\) with velocity \(V_0\) and the other with mass \(m\) and velocity \(0\).
\begin{align}
MV_0 &= MV_0' + mv'\\
M(V_0 - V_0') &= mv'\qquad (*)\\
MV_0^2 &= MV_0^{'2} + mv^{'2}\\
M(V_0 - V_0')(V_0 + V_0') &= mv^{'2}\qquad (**)
\end{align}
So let's take \(\frac{(**)}{(*)}\Rightarrow V_0 + V_0' = v'\)

How do I write \(v'\) and \(V_0'\) in terms of their masses and \(V_0\)?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
We have 2 masses: one with mass \(M\) with velocity \(V_0\) and the other with mass \(m\) and velocity \(0\).
\begin{align}
MV_0 &= MV_0' + mv'\\
M(V_0 - V_0') &= mv'\qquad (*)\\
MV_0^2 &= MV_0^{'2} + mv^{'2}\\
M(V_0 - V_0')(V_0 + V_0') &= mv^{'2}\qquad (**)
\end{align}
So let's take \(\frac{(**)}{(*)}\Rightarrow V_0 + V_0' = v'\)

How do I write \(v'\) and \(V_0'\) in terms of their masses and \(V_0\)?
So you could do
\begin{align*}
v'&= \frac{M(V_0-V_0')}{m} \\
V_0+V_0'&= \frac{M(V_0-V_0')}{m}
\end{align*}
Solve for $V_0'$ ...