Inelastic collisions -- how is momentum conserved but not energy?

In summary: Different standpoints give different answers to this question. My perspective is goal-oriented, and in order to maximize the loss of kinetic energy in an inelastic collision, the impact parameter and the coefficient of restitution must both be set to zero. That is what we call a head-on, perfectly inelastic collision.
  • #1
richengle
26
1
Homework Statement
in an INELASTIC collision, how is momentum conserved, but not energy?
Relevant Equations
Cons of Momentum
m1v1+m2v2=m1vs'+m2v2' , if car hits small fluffy object m2, initially v2=0, and v1'=v2' ... so
m1v1=[m1+m2](v2')
but why not energy? Why is there a KElost in Cons of Energy?
Cons of Energy
.5m1v1^2+.5m2v2^2=.5m1v1'^2+.5m2v'2^2 +KElost , and again v2=0, v1'=v2'
.5m1v1^2=.5[m1+m2]v2'^2+KElost
I know you could say deformation, but do you somehow have to take the derrivative of energy [wrt what i dont know] to find a maximum, which is dependent on the initial and final masses and velocities?
m1v1+m2v2=m1vs'+m2v2' , if car hits small fluffy object m2, initially v2=0, and v1'=v2' ... so
m1v1=[m1+m2](v2')

but why not energy? Why is there a KElost?
.5m1v1^2+.5m2v2^2=.5m1v1'^2+.5m2v'2^2 +KElost , and again v2=0, v1'=v2'
.5m1v1^2=.5[m1+m2]v2'^2+KElost
using consv of momentum...
KElost=.5[m1v1^2[1-m1/(m1+m2)] ?
.. and if m1>>m2, this is zero,,, how? Some energy should still be lost?
I know you could say deformation, but do you somehow have to take the derrivative of energy [wrt what i don't know] to find a maximum, which is dependent on the initial and final masses and velocities?
 
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  • #2
You don't have to take a derivative to find the maximum energy lost in an inelastic collision. Maximum kinetic energy is lost when the two masses are not moving relative to each other after the collision. That amount of energy is $$\Delta (KE)=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(v_2-v_1)^2$$where ##v_1## and ##v_2## are the velocities of the masses before the collision.
 
  • #3
kuruman said:
Maximum kinetic energy is lost when the two masses are not moving relative to each other after the collision.
That's certainly true for perfectly inelastic collisions in one dimension (head-on collisions), but unqualified "inelastic" is anything that is not perfectly elastic.

@richengle: That momentum is conserved follows from Newton's action/reaction law. If A exerts force F(t) on B then B exerts -F(t) on A. The change in momentum due to a force is its integral over time, so the changes in momentum from the collision are ∫F.dt and -∫F.dt, which clearly add to zero.
The work done on B is ∫F.dx, where the range is the displacement of B in the direction of F. So the work they do on each other also adds to zero, but this time there is no guarantee that the work will all reappear as KE.

For oblique collisions, I'm not sure perfectly inelastic implies sticking together. I would think it only means the coefficient of restitution is zero. So if the bodies are smooth, it is only the velocity components in the direction of the line of centres at collision that will become equal.
 
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  • #4
Just to help state the statement of the OP a bit more accurately, in inelastic collision momentum is conserved, and energy is conserved too, it is kinetic energy that is not necessarily conserved.
haruspex said:
@richengle: That momentum is conserved follows from Newton's action/reaction law. If A exerts force F(t) on B then B exerts -F(t) on A. The change in momentum due to a force is its integral over time, so the changes in momentum from the collision are ∫F.dt and -∫F.dt, which clearly add to zero.
The work done on B is ∫F.dx, where the range is the displacement of B in the direction of F. So the work they do on each other also adds to zero, but this time there is no guarantee that the work will all reappear as KE.
So the "moral" lesson of this is that impulse always transfers momentum between bodies, but work doesn't necessarily transfer kinetic energy, some work might go to heat or potential energy of deformation .e.t.c..
 
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  • #5
haruspex said:
That's certainly true for perfectly inelastic collisions in one dimension (head-on collisions), but unqualified "inelastic" is anything that is not perfectly elastic.
I agree, but I also think that we are looking at OP's question about a maximum
richengle said:
##\dots~## but do you somehow have to take the derrivative of energy [wrt what i don't know] to find a maximum, which is dependent on the initial and final masses and velocities?
from different standpoints. Mine is goal-oriented. Given that one is looking for maximum kinetic energy loss and since the kinetic energy of the CM is conserved, the inelastic collision must be head-on and perfectly inelastic. In order to achieve this maximum, the parameters to be varied are two: the impact parameter and the coefficient of restitution. When both are set equal to zero, one has a head-on, perfectly inelastic collision that maximizes the loss of kinetic energy.
 
  • #6
When I was at school I always thought the conservation of momentum was inferior, in some way, to the great conservation of energy law which applies to everything. It was only when I read Principia that I finally understood - that Aha! moment - what momentum is.

Newton defines it beautifully: Momentum is the quantity of motion a body possesses.

As momentum is mv, and kinetic energy is 1/2 mv^2 how can both be conserved when their equations are so different?

Because when two things collide we get sound , which takes energy away; the bodies deform and heat a little, which takes energy away. When you add up all those bits of energy loss it explains why momentum (the quantity of motion) is conserved and also why energy (kinetic energy of motion + sound loss + heating loss + whatever else ...) is also conserved, even though their equations are so different.

The conservation of momentum follows as Corollary III from Action and reaction are equal and opposite.

Newton.png
 
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Related to Inelastic collisions -- how is momentum conserved but not energy?

1. What is an inelastic collision?

An inelastic collision is a type of collision between two objects where the total kinetic energy of the system is not conserved. This means that some of the initial kinetic energy is lost and converted into other forms of energy, such as heat or sound.

2. How is momentum conserved in an inelastic collision?

Momentum is conserved in an inelastic collision because the total momentum of the system before and after the collision remains the same. This means that the sum of the momenta of the two objects involved in the collision is equal before and after the collision.

3. Why is energy not conserved in an inelastic collision?

Energy is not conserved in an inelastic collision because some of the initial kinetic energy is converted into other forms of energy, such as heat or sound. This is due to the fact that the objects involved in the collision deform or stick together, causing a loss of kinetic energy.

4. How is the coefficient of restitution related to inelastic collisions?

The coefficient of restitution is a measure of the elasticity of a collision. In an inelastic collision, the coefficient of restitution is less than 1, indicating that some of the initial kinetic energy is lost. The lower the coefficient of restitution, the more inelastic the collision is.

5. Can momentum and energy be conserved in an inelastic collision?

No, both momentum and energy cannot be conserved in an inelastic collision. While momentum is always conserved, energy is only conserved in elastic collisions where there is no loss of kinetic energy. In inelastic collisions, some of the initial kinetic energy is converted into other forms of energy, making it impossible for both momentum and energy to be conserved.

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