# Elastic and inelastic collisions

1. Sep 17, 2011

### Bernoulli19

Hey everyone,
Reviewing material for a technical interview and I have stumbled into some problems that I sadly have issues with. Old review for me :D. Any help would be appreciated.

I can follow up through the conservation of energy equation. However at the in an elastic collision line, I do not know why those are being subtracted. That would not seem to follow the conservation of momentum since the masses are not the same.

assuming i can get that part, the Vo = v - V is just substituted into the simplified conservation of momentum equation? Is nothing done with the conservation of energy equation?

2. Sep 17, 2011

### PeterO

That velocity arrangement is actually true. It is saying that the difference in velocity before collision is the reverse of the difference in velocity after. If you visualise the collision from the point of view of the centre of mass - you would see both masses approach the C of M at a certain speed, then bounce off at the same speed [but opposite direction].

I will demonstrate:

The way I work these problems is as follows.

If you calculate as if for an inelastic collision [totally - the masses stick together] in the first instance, you can note how much the velocity of each body changes for that to happen.
When the collision is actually elastic, the velocities change by that much again.

eg if M = 10, m = 1 and the initial vel of M was 11 m/s

During collision, at the point when the masses are traveling at the same velocity, that common velocity will be 10m/s [conservation of momentum will get you that - I chose figures wisely to avoid decimals]

If the collision was inelastic, you have finished

If the collision is elastic we look at the changes.
VM reduced by 1 - from 11 to 10.
Vm increased by 10 - from 0 to 10

They will change by that much again!!

So afterwards:
VM = 9 [reduced by another 1]
and
Vm = 20 [increased by another 10]

before collision, M was traveling 11 m/s faster than m
after collision, m is traveling 11 m/s faster than M

NOTE: The Centre of Mass of this system has been travelling at a constant 10 m/s.

Before, M was doing 11 m/s, so approaching the CofM at 1 m/s, afterwards it was moving away at 1 m/s [or traveling at 9 m/s relative to a stationary observer]
Before, m was approaching at 10 m/s [some would say the CofM was approaching it at 10 m/s], afterwards it was moving away from the CofM at 10 m/s - [or travelling at 20 m/s relative to a stationary observer]

3. Sep 17, 2011

### Bernoulli19

Thank you for the quick reply. I am going on a quick trip in the morning but i read this a few times and will work on it at the airport and let you know wednesdayish. Thanks!

4. Sep 18, 2011

### Jokerhelper

Another way to look at this is to see that those equations can be derived from the coefficient of restitution formula:
$$e = \frac{v^{'}_{2} - v^{'}_{1}}{v_{1} - v_{2}}$$
Where e = 1 in an elastic collision, and e = 0 in a perfectly inelastic collision. All other inelastic collisions range in between.