Let's consider a collision in a lab frame
in which one particle is coming in
and the other particle is at rest.
So here's particle 2.
V2 initial is 0.
And we could say that m2 is twice m1.
And this is our lab frame.
And now let's consider the same collision
in the center of mass frame.
And in that reference frame, we have
particle 1 moving with velocity V1 prime initial
and particle 2 moving with velocity V2 initial prime.
Now, the way we're going to analyze this problem
is that we know the speeds in the center of mass frame
relative to the lab frame.
So we actually know these initial speeds.
And we'll write that result down in a moment.
Because we know the initial speeds,
we also know that V1 final in the center of mass frame
is just minus the initial velocity.
That's the beauty of the center of mass frame.
The velocities just change direction.
They don't change magnitude.
Because we know this we can get V1 final.
And then it's just a simple exercise
to go back to the lab frame to calculate
the quantity, the velocity of the object 1 in the lab frame.
So that will be our sequence of ideas.
And the key fact that we know is that V1 prime
is equal initially to the reduced mass divided
by m1 times V1, 2, where V1, 2 initial
is just equal to V1 initial minus 0.
So that's V1 initial.
And the ratio, mu over m1, it's very simple to calculate that.
That's just m2 over m1 plus m2, or our case, that's 2/3.
And so from our result, we now have very simply
that V1 final prime is minus V1 initial prime.
So it's minus 2/3 V1 initial.
And that's a very straightforward calculation.
We can do exactly the same thing with V2 final prime.
V2 final prime is minus V2 initial prime.
And V2 initial prime is equal to minus mu over m2 times V1, 2
initial.
This ratio instead of being 2/3, it's a simple exercise.
It will be m1 over m1 plus m2 is 1/3 V1 initial.
And so we have solved for the final velocities in the center
of mass reference frame.
Now let's just double check our results.
We have V2 final is minus V2 initial.
So that's minus.
But there's another minus sign here.
So we have 2 minus signs.
So that's a plus.
And that's why we have a plus 1/3 V1 initial.
And finally, if we want to ask the question, what
are the velocities in the lab frame,
it's now a very simple exercise to do reference frame change.
We do need to know what V center of mass is.
That's V1 V initial over m1 plus m2, because remember
that V2 initial is 0.
So that's another factor, V1 initial.
And now for conclusion, we have that the final velocity
in the lab frame is equal to the velocity
in the center of mass frame plus V center of mass.
And we just collect our results, minus 2/3 V1
initial plus 1/3 V1 initial.
So that's minus 1/3 V1 initial.
The outgoing velocity of particle 1 in the lab frame
and the outgoing velocity of particle 2 in the lab frame,
again, we have 1/3 V1 initial plus another 1/3 V1 initial
is equal to 2/3 V1 initial.
So we were able to solve this problem by switching reference
frame using our basic fact in the center of mass frame
that the speeds remain the same but the direction changes
and able to solve this problem without any
of the traditional ways of applying the energy momentum
relationship and kinetic energy or having quadratic equations.