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## Conservation of momentum and energy contradiction?

Suppose we have two objects and we're only talking about rectilinear motion.
Initially, one object has mass m and is moving at velocity V. The other has mass M and is standing still.
Then they hit each other and suppose that all kinetic energy is conserved and they stick together and move at velocity v.

Then this follows from the conservation of energy:
1/2mV$^{2}$ = 1/2(m+M)v$^{2}$

And this follows from the conservation of momentum:
mV = (m+M)v

But when we divide the first equation by the second one we get:
1/2V = 1/2v
V = v

Which implies that M is zero but I haven't stated that anywhere, at least not directly. What did I do wrong?

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Mentor  If they stick together, then the collision must be inelastic ie kinetic energy cannot be conserved.   Remember that momentum is always conserved in a collision (provided there are no external forces). So, to see that what I'm saying is true, first apply conservation of momentum to figure out the final velocity in terms of the initial one. Then use these velocities to compare the final kinetic energy to the initial.

Quote by cepheid 					 				 				If they stick together, then the collision must be inelastic ie kinetic energy cannot be conserved.
Can you elaborate on this? This is kind of too abstract for me.

Does this mean that for these masses and velocity always the same amount of kinetic energy is lost? I'd imagine it would also depend on the materials or contact area etc.

Mentor
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## Conservation of momentum and energy contradiction?

Quote by Fyreth 					 				 				Can you elaborate on this? This is kind of too abstract for me.
Just do the calculation. Momentum is always conserved. And since they stick together, it's easy to calculate the final velocity. Then you can directly compare the final KE with the initial.

Does this mean that for these masses and velocity always the same amount of kinetic energy is lost?
Right. Since they stick together, the maximum amount of KE is lost.

I'd imagine it would also depend on the materials or contact area etc.
All you need to know is that they stick together.

Quote by Doc Al 					 				 				Just do the calculation.  Momentum is always conserved.  And since they stick together, it's easy to calculate the final velocity.  Then you can directly compare the final KE with the initial.  Right.  Since they stick together, the maximum amount of KE is lost.  All you need to know is that they stick together.
I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss?

Quote by Fyreth 					 				 				I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss?
Hey,

Try this. Do NOT assume that the balls stick together afterwards. Assume the initial conditions you gave. Conserve BOTH momentum and energy, and see what you expect the system to do.

Now, ask yourself: Why do the balls not naturally stick together. The answer is, you need extra energy to force the balls to stick together. This energy comes from the kinetic energy of the ball. Thus, some of the energy is lost there and we cannot account for it. Of course,

total energy = KE of both balls afterwards + Energy lost in forcing balls together

is still conserved. But since we cannot calculate the latter (atleast directly), we cannot use energy conservation.

You may ask of course, Why is MOMENTUM not lost in forcing the balls to stick together. The answer is that momentum is ALWAYS conserved if there are no net force. (Energy conservation is a more subtle issue). In this case, the balls experience forces that make it stick together, but all these forces are internal and effectively cancel out leaving zero net force.

Thus, you may use momentum conservation, but not energy conservation here.

Quote by Fyreth 					 				 				I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss?
When the balls stick together, some energy is converted to heat (assuming no other losses, like sound waves, etc.)

Suppose you had two balls one coming from the left with a speed of v, one coming from the right at the same speed. Suppose they have the same mass, and they stick together. Then when they stick together, they will stop dead. The total momentum is zero before and after, because their velocities were equal and opposite, no problem, but their kinetic energy has gone to zero. It all turned into heat, the balls are warmer after the collision.

This usually happens because the two balls deform when they collide, and the different parts of the ball moving past each other creates friction, which creates the heat. You could also have the collision create sound waves inside the balls, and the sound waves would bounce around inside the balls, eventually getting absorbed and turned into heat.

If there is a sound when they collide, and that sound gets radiated away, and carries some of the kinetic energy with it.

What I'm saying below is parasitic on Rap's post.  As Doc Al suggested, use momentum conservation to find v. As you wrote: mV = (m+M)v. So v = mV/(m+M).   Now imagine that you view the collision from a car which is moving with this just this velocity. In your new frame of reference, m and M collide and each becomes stationary. That's what makes sticking together so special: there is a frame of reference in which both bodies lose all their KE. This won't happen for all colliding bodies, but there are quite a few materials (the obvious one is putty) for which this does happen, owing to the nature of the forces between the particles making up the material.   But losing all KE in this special frame equates to sticking together in other frames of reference, and the loss of some of the original KE. The fact that only a certain fraction of the KE is lost in these frames, and that this fraction can be calculated with the aid of momentum conservation, makes it seem mysterious – as if the properties of the material aren't involved. But they are – as we saw by looking at the collision from the frame of reference (called the 'centre of mass frame') in which both bodies finish up stationary.

Mentor

Quote by Fyreth 					 				 				I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss?
Think of different kinds of collisions with the ground. If something collides with the ground and maintains its KE then it bounces back up. If something collides with the ground and loses its KE then it splats on the ground. So, just by experience you know that "bouncy" collisions maintain KE and "sticky" collisions do not.

Now, think about the result of a bouncy or a sticky collision on the bouncy or sticky object. A rubber ball is largely unaffected by a collision with the ground. It temporarily deforms, and then bounces right back to its original shape. A ball of mud or clay on the other hand deforms permanently. After the sticky collision it is no longer a ball. So, what happened to the KE of the clay?

"1/2mV2 = 1/2(m+M)v2"      This equation is wrong,because when they hit each other,some energy chenge into heat Q,which is about the movement of molecule in the body.So he have:      So, conservation of energy is also true,but conservation of kinetic energy is not satisfied.     With the equation which contain the heat ,you won't get the contradiction which make m=0.So conservation of momentum is also not broken.

Tags
conservation laws, contradiction

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