# How is v1i-v2i=-(v1f-v2f) Derived?

• Alameen Damer
Alameen Damer
Hello, I was looking at some questions for head on perfect elastic collisions, and I ran by this equation:

v1i-v2i=-(v1f-v2f)

I want to know how that equation is derived.

Note: v1i is the initial velocity of the first mass
v2i is the initial velocity of the second mass
v1f is the final velocity of the first mass
v2f is the final velocity of the second mass

Homework Helper
Does not make much sense to me. You sure about the signs ? And are the masses perhaps equal ?

Homework Helper
In any collision, whether elastic or not, momentum is conserved: m1v1i+ m2vi= m1v1f+ m2v2f.

If m1= m2 those reduce to v1i+ v2i= v1f+ v2f. By subtracting v1f and v2i from both sides you can get v1i- v2f= v1f- v2i= -(v2i- v1f).

But you cannot get v1i-v2i=-(v1f-v2f). That simply isn't true, even in an elastic collision.

Staff Emeritus
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But you cannot get v1i-v2i=-(v1f-v2f). That simply isn't true, even in an elastic collision.
It is true for an elastic collision in one dimension. The easiest way is to derive it in the CoM frame and then translate the result to an arbitrary frame:
In the CoM frame, the kinematics give:
##v_1 = -u_1##
##v_2 = -u_2##
where I have exchanged the i/f subscripts for using v and u instead (less clogged notation). Obviously, this means that ##v_1 - v_2 = -(u_1 - u_2)##. Translate to an arbitrary frame to yield the result.

Homework Helper
You can get it in the lab frame if you combine the two conservation laws (momentum and KE). Just group the terms by masses in each equation.
The relationship tells you that for ellastic collision the relative velocity is reversed (but keeps same magnitude). An obvious statement for collision with a wall but true "in general" (I mean for any masses).

Look up wiki for "elastic collisions" for details about obtaining the relationship.

Staff Emeritus
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You can get it in the lab frame if you combine the two conservation laws (momentum and KE).

You do not need to do this. It holds in the CoM frame and therefore by transforming it to the lab frame, in the lab frame (velocities transform as ##v \to v + v_0##, where ##v_0## is the transformation velocity - the ##v_0## cancels between the velocities since it is a difference on both sides). Of course you can do it in the lab frame, but it gets more complicated than simply following the very simple outline I gave above (which of course is based on the conservation laws).

Alameen Damer
Thank you very much for the explanation, however is there a simpler form of derivation that can be explicitly be explained with conservation of momentum, without using CoM or arbitrary frames as my textbook has not covered that? This is 12 physics by the way.

Staff Emeritus
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Thank you very much for the explanation, however is there a simpler form of derivation that can be explicitly be explained with conservation of momentum, without using CoM or arbitrary frames as my textbook has not covered that? This is 12 physics by the way.
You cannot do it only with conservation of momentum, you need conservation of energy as well. You can do the derivation in an arbitrary frame, but the algebra is much simpler in the CoM frame. This is why physicists like to work in frames where they can easily compute invariant properties.

Alameen Damer
My question is stemming from a textbook question stating:
"A neutron in a nuclear reactor moves at 1.0 x 10^6 m/s and makes a head on elastic collision with a carbon atom initially at rest. The carbon atom is 12 times the mass of the neutron. Determine the fraction of the neutron's kinetic energy transferred to the carbon atom"
Can this be solved without the above mentioned formula?

Staff Emeritus
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My question is stemming from a textbook question stating:
"A neutron in a nuclear reactor moves at 1.0 x 10^6 m/s and makes a head on elastic collision with a carbon atom initially at rest. The carbon atom is 12 times the mass of the neutron. Determine the fraction of the neutron's kinetic energy transferred to the carbon atom"
Can this be solved without the above mentioned formula?
Yes, you can go on to just apply energy and momentum conservation as it stands. Please keep in mind that specific problem questions should be asked in the homework forums and that you need to provide your attempted solution when posting there.

• Alameen Damer
Alameen Damer
Oh okay i see. Thanks again

Mentor
My question is stemming from a textbook question stating:
"A neutron in a nuclear reactor moves at 1.0 x 10^6 m/s and makes a head on elastic collision with a carbon atom initially at rest. The carbon atom is 12 times the mass of the neutron. Determine the fraction of the neutron's kinetic energy transferred to the carbon atom"
Can this be solved without the above mentioned formula?

If you want to do it the hard way, you have two unknowns (the speeds of the two particles after the collision) and two equations in those unknowns (momentum before equals momentum after, kinetic energy before equals kinetic energy after) so a bit of algebra will see you home.

I called that "the hard way", but when you're learning it's a good idea to work through a few problems this way. Once you have a feel for how the two conservation laws work together, you'll be better able to appreciate the power of the CoM approach and more comfortable using it.

• Alameen Damer
Homework Helper
You do not need to do this. It holds in the CoM frame and therefore by transforming it to the lab frame, in the lab frame (velocities transform as ##v \to v + v_0##, where ##v_0## is the transformation velocity - the ##v_0## cancels between the velocities since it is a difference on both sides). Of course you can do it in the lab frame, but it gets more complicated than simply following the very simple outline I gave above (which of course is based on the conservation laws).
I did not say that you need to.
Just that it may be more "comfortable" for someone not so familiar with going between reference frames.
And the algebra is pretty simple anyway.