# Converting Elastic Collision to Inelastic

Hi,

If I have two 1 kg masses that are going to hit each other. Mass A is traveling at 10 ms and Mass B is at rest but I wish to cause for an Inelastic Collision. How do I achieve this?

As best I can make of it, I would require a buffer between mass A & B... and I would expect this buffer could be a small explosion that would be less or equal momentum to that of mass A before it hits B.

Am I correct in saying this?

Thanks!

tiny-tim
Homework Helper
Welcome to PF!

Hi Billy! Welcome to PF!
If I have two 1 kg masses that are going to hit each other. Mass A is traveling at 10 ms and Mass B is at rest but I wish to cause for an Inelastic Collision. How do I achieve this?

Glue.

(a purely inelastic collision is where the two objects have the same final velocity)

Glue is good - for a lot of things!

But let say I am not able to use glue only an explosion.

So based on the definition of an inelastic collision, in creating a small explosion that allows the two masses to meet and now travel at the same velocity as opposed to bouncing off each other would be an inelastic collision?

So the explosion pushes off mass A & B in opposite directions, slowing mass A to the same speed as mass B.

I believe the end result would be the same momentum of an inelastic collision plus the force of the explosion? (or is the force of the explosion simply nullified?)

There is some "reactive armor" which lessens the net-impact of an incoming projectile.
But, this is not simple, doesn't work the way you might think, is very expensive, and classified.

tiny-tim
Homework Helper
Hi Billy!
So based on the definition of an inelastic collision, in creating a small explosion that allows the two masses to meet and now travel at the same velocity as opposed to bouncing off each other would be an inelastic collision?

So the explosion pushes off mass A & B in opposite directions, slowing mass A to the same speed as mass B.

I believe the end result would be the same momentum of an inelastic collision plus the force of the explosion? (or is the force of the explosion simply nullified?)

I'm mystified as to why you're trying to do this.

An explosion would completely change the collision.

And mometum is conserved in every collision (and explosion) anyway, whether elastic, partially inelastic, or completely inelastic.

An explosion would completely change the collision.
QUOTE]

That's ONE of the key points. That's simple. Lateral ablation, however, is not.

If you would like we could think of the explosion as a loaded spring that pushes at the exact moment that A hits B. The power of the spring is set to exact amount of force required to push B forward and slow A to the speed of B.

I am just trying to understand in my mind how one could convert a collision from elastic to inelastic without the use of glue, magnets or any other medium that simply connects the two masses.

I do appreciate your help on this!

Last edited:
You said...

"And mometum is conserved in every collision (and explosion) anyway, whether elastic, partially inelastic, or completely inelastic."

This is correct but in an elastic collision mass A travels back at a velocity near equal to that it came at. I am trying to have it so it will not travel back.

Based on your statement being that momentum is conserverd in every collision/explosion that if the two mass are now traveling together than we should have achieved at the very least the momentum of a inelastic collision.

Is this right?

tiny-tim
Homework Helper
(just got up :zzz: …)
If you would like we could think of the explosion as a loaded spring that pushes at the exact moment that A hits B. The power of the spring is set to exact amount of force required to push B forward and slow A to the speed of B.

But then that's a collision between three bodies.
I am just trying to understand in my mind how one could convert a collision from elastic to inelastic without the use of glue, magnets or any other medium that simply connects the two masses.

Well, you can't … whether A and B end up with the same velocity depends on how their surfaces interact.
… in an elastic collision mass A travels back at a velocity near equal to that it came at.

No … if they have the same mass, then A stops dead, and B acquires the velocity that A had (like that desk toy with the five silver spheres ).
Based on your statement being that momentum is conserverd in every collision/explosion that if the two mass are now traveling together than we should have achieved at the very least the momentum of a inelastic collision.

Is this right?

Sorry, not following you.