MHB Elastic Collision of 2 Masses: Calculating \(v'\) and \(V_0'\)

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The discussion focuses on calculating the velocities \(v'\) and \(V_0'\) after an elastic collision between two masses, \(M\) and \(m\). The conservation of momentum and kinetic energy equations are presented, leading to the relationships \(MV_0 = MV_0' + mv'\) and \(MV_0^2 = MV_0^{'2} + mv^{'2}\). By manipulating these equations, it is derived that \(V_0 + V_0' = v'\). The participants suggest expressing \(v'\) and \(V_0'\) in terms of the masses and initial velocity \(V_0\). The discussion emphasizes the mathematical approach to solving for the post-collision velocities.
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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 \(V_0\)?
 
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dwsmith said:
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 $V_0'$ ...
 
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