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

AI Thread Summary
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.
Dustinsfl
Messages
2,217
Reaction score
5
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 \(V_0\)?
 
Mathematics news on Phys.org
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 $V_0'$ ...
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top