Perfectly elastic/inelastic collision

ranon

Pendulum mass m released from stop h above bottom of arc.
within plane
m hits M at bottom of arc.--how high does m+M go?
Perfectly inelastic collision.
Kinetic energy is not conserved--work is done (deformation, noise, temp etc)
but momentum is conserved.
how is this possible?
If there is loss of energy, how can momentum be conserved?
What am I missing?

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cepheid

Staff Emeritus
Gold Member
Kinetic energy is not conserved--work is done (deformation, noise, temp etc)
but momentum is conserved.
how is this possible?
If there is loss of energy, how can momentum be conserved?
What am I missing?
Why not?

It's possible because energy and momentum are two totally different things. Take another look at the definitions for momentum and kinetic energy, and you'll see that the condition for conservation of momentum imposes a particular constraint on the system (represented by an equation relating the quantities before and after the collision) The condition for conservation of kinetic energy (if it were met) would impose a totally different constraint on the system (represented by another, different equation that relates the quantities before and after the collision).

ranon

In a perfectly inelastic collision, does
mv^2/2 + mgh = =(m+M)V + (m+M) gH with H the height after low point ?

If seems not if there is energy dissipated in inelastic colision.

If the post collision Velocity is effected by the loss, than why isn't momentum effected"

ranon

Hmmm, am I having difficulty as momentum is a vector quantity with energy a scaler?

cepheid

Staff Emeritus
Gold Member
In a perfectly inelastic collision, does
mv^2/2 + mgh = =(m+M)V + (m+M) gH with H the height after low point ?
I'm going to assume that you meant V2. No, this equation is not true. Mechanical energy is not conserved here.

If seems not if there is energy dissipated in inelastic colision.
Of course! That's the definition of an inelastic collision -- a collision in which kinetic energy is lost.

If the post collision Velocity is effected by the loss, than why isn't momentum effected"
The simple answer is, "because, in this situation, it is possible for the velocity to change while the momentum does not." If that still doesn't make sense, maybe this will help. Conservation of momentum before and after the collision says that:

$$mv = (m + M)V$$​

with the result that the velocity after is lower than the velocity before:

$$\left(\frac{m}{m + M}\right)v = V$$​

Now, the kinetic energy before the collision is just (1/2)mv2. The kinetic energy after the collision is:

$$\frac{1}{2}(m+M)V^2$$

If you substitute in the above expression for V, you end up with the result that the kinetic energy afterwards is:

$$\frac{1}{2} \left(\frac{m^2}{m+M}\right)v^2$$​

So, as you can see, as long as M is not zero, the final energy is always lower than the initial energy.

Summary: momentum is always conserved. But, as I have shown above, if momentum is conserved, and the two masses stick together, then kinetic energy cannot also be conserved -- it is impossible. The final kinetic energy will always be lower. Therefore, if the two masses stick together, you know that an inelastic collision has occured. In an elastic collision, the first mass would have bounced back.

Last edited:

cepheid

Staff Emeritus
Gold Member
Hmmm, am I having difficulty as momentum is a vector quantity with energy a scaler?
In general you have to be careful with that, although in this particular case, at the instant of collision, it's sort of a 1D problem, so it doesn't really matter too much.

ranon

cepheid--thank you for your reply--I have a better understanding.
It seems that one may have a whole range of inelastic collisions from near totally inelastic collisions with extensive energy loss and collisions with very little energy loss(hard billiard balls with a latch to connects them after collision)--assume no energy loss.
.5mv^2 + mgh =.5(m+M)V^2 + (m+M)gH
Would the above equation then be usable for calculating H(V=0)
without using momentum conservation?

ranon

From lack of response, I wonder if my model of elastic masses with frictionless silent rigid latches is acceptable in the context of this experiment. Any thoughts?

cepheid

Staff Emeritus
Gold Member
--assume no energy loss.
.5mv^2 + mgh =.5(m+M)V^2 + (m+M)gH
Would the above equation then be usable for calculating H(V=0)
without using momentum conservation?
I am not sure if it's really realistic to think that you can "latch together" the two masses without energy being lost. The math shows that if the final mass is m+M, AND momentum is conserved, then energy is lost. So, one of these two conditions (energy cons. or momentum cons.) must be relaxed if the situation is to be logically consistent - they cannot both be true. I'm not sure which one would be violated (maybe both). It is possible that momentum would not be conserved in that situation, since each ball would then have an outside force acting on it (from the latch). That's the best I can think of right now.

ranon

Thanx again.
It sounds like I may have been proposing something like Maxwell's demon.
I know it is not possible to latch the masses together without energy loss, but it may be possible to minimize this loss and therefore approach conservation of energy.
If I can design it, maybe I should try to do the experiment!!
In any event, be well.

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