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

• Alameen Damer
In summary, the conversation discusses the equation v1i-v2i=-(v1f-v2f) for head on perfect elastic collisions and how it is derived. It is explained that this equation is only true for elastic collisions in one dimension and can be derived in the center of mass frame. The conservation of momentum and energy are also mentioned as important factors in understanding this equation. A specific textbook problem is also discussed, and it is determined that the problem can be solved without the mentioned equation, but it is helpful to understand the two conservation laws.
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

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

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.

HallsofIvy said:
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.

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.

nasu said:
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).

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.

Alameen Damer said:
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.

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?

Alameen Damer said:
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
Oh okay i see. Thanks again

Alameen Damer said:
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
Orodruin said:
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.

## 1. How is the equation v1i-v2i=-(v1f-v2f) derived?

The equation v1i-v2i=-(v1f-v2f) is derived from the principle of conservation of momentum, which states that the total momentum of a closed system remains constant. In this equation, v1i and v2i represent the initial velocities of two interacting objects, while v1f and v2f represent their final velocities. By applying the conservation of momentum principle, we can equate the total initial momentum of the system to the total final momentum, resulting in the equation v1i-v2i=-(v1f-v2f).

## 2. Why is the equation v1i-v2i=-(v1f-v2f) important?

The equation v1i-v2i=-(v1f-v2f) is important because it allows us to calculate the final velocities of two interacting objects based on their initial velocities and masses. This is useful in many scientific fields, such as physics, engineering, and astronomy, as it helps us understand and predict the behavior of objects in motion.

## 3. Can the equation v1i-v2i=-(v1f-v2f) be applied to all types of collisions?

Yes, the equation v1i-v2i=-(v1f-v2f) can be applied to all types of collisions, including elastic and inelastic collisions. In elastic collisions, both momentum and kinetic energy are conserved, while in inelastic collisions, only momentum is conserved. This equation takes into account both cases by considering the change in velocity of each object, regardless of whether kinetic energy is conserved or not.

## 4. What are the limitations of the equation v1i-v2i=-(v1f-v2f)?

The equation v1i-v2i=-(v1f-v2f) assumes that there are no external forces acting on the system, such as friction or air resistance. In real-world situations, these external forces may affect the final velocities of the objects, making the equation less accurate. Additionally, this equation only applies to two objects interacting with each other, and cannot be used for systems with more than two objects.

## 5. How can the equation v1i-v2i=-(v1f-v2f) be used in practical applications?

The equation v1i-v2i=-(v1f-v2f) can be used in a variety of practical applications, such as analyzing car crashes, designing sports equipment, and understanding the motion of celestial bodies. It can also be used to calculate the recoil of firearms and the motion of particles in particle accelerators. Overall, this equation is a valuable tool in understanding and predicting the behavior of objects in motion.

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