## Homework Statement

Two identical molecules, A and B, move with the same horizontal velocities but opposite vertical velocities. Which of the following is always NOT true after they collide:

A - the sum of kinetic energy before collision is less than the sum after
B - the sum of kinetic energy before collision is greater than the sum after
C - molecule A will have greater momentum after the collision than molecule B
D - molecule A will have greater vertical velocity than molecule B

## Homework Equations

conservation of momentum: mavai+mbvbi=mavaf+mbvbf

## The Attempt at a Solution

The answer in the back of the book says "The question does not specify any specific velocities, so assign any number that you like. If both molecules initially have horizontal components of +5m/s and vertical components of +3m/s and -3m/s, then possible values after collision could be horizontal components of +8m/s and +2m/s and vertical components of +10m/s and -10m/s. This would conserve momentum, but not the kinetic energies of the molecules. Kinetic energy does not have to be conserved because you are not told that this is an elastic collision. While the magnitude of the vertical components must remain the same for it to add up to 0, the magnitude of the horizontal components can vary as long as their sum adds up to +10. This implies that either molecule A or B may have a greater horizontal component, and therefore greater overall speed and momentum, so (C) is incorrect. (D) is wrong, because it would violate the law of conservation of momentum."

I don't understand how this was done at all, I've gone over it multiple times and I'm boggled. Can someone break this answer down for me?

How were the values of +2, +8, +10, & -10 acquired exactly?

I get that momentum must be conserved in elastic and inelastic collisions, whereas kinetic energy is only conserved in completely elastic collisions. So the equation that applies here is:

mavai+mbvbi=mavaf+mbvbf

Since the 2 molecules are identical we can drop all the m's so:

vai+vbi = vaf+vbf ...

from here on I'm stumped :( any clarification would be appreciated!

## The Attempt at a Solution

Related Introductory Physics Homework Help News on Phys.org
This may not be something you were taught, however, you can always view a collision from the center of mass frame. In the center of mass frame, the particles come in, collide, and then leave with the same momentum (just rotated). In this problem A and B have the same momentum. When they collide, they will each leave with their own momentum (which happens to be the same in this case).

Hi reshmaji
The numerical values given are just examples, don't think about it, in the end you only have 2 components:
Vah, Vav (velocity of A horizontal and vertical)
and for B, they are given by Vbh=Vah and Vbv=-Vav (before the collision)

Since the collision is not necessarily elastic, there is nothing you can do about assertions (A) and (B), they are not necessarily true, but it could happen, so they are not 'always NOT true'

So you can just focus on (C) and (D) which talks about momentum.
(C) talks about the momentum of a being different than b, this is possible because only the total momentum is conserved as a vector quantity (which means the all the components of the sum vector are each conserved too)
So it is possible that the momentum of a is less or more than the momentum of b as long as they compensate each other and the sum is the same

now for (D) we do have a problem: it is impossible for a to have a greater velocity than b in the vertical direction because of the symetry before the collision: the vertical component of the total momentum was 0, it must therefore still be 0. both velocities of a and b can change all you want, they will have to be the same (and opposite of course) (since they are alike and have the same mass)
otherwise, the system as a whole would suddenly gain momentum in the vertical direction, which is impossible.

Hope this helps.
cheers...

That totally cleared it up, thank you!!