How is momentum conserved in an inelastic collision?

AI Thread Summary
Momentum is conserved in inelastic collisions because, despite the loss of kinetic energy, the total momentum before and after the collision remains constant. This is due to the absence of external forces acting on the colliding bodies, which means the rate of change of total momentum is zero. In a perfectly inelastic collision, two objects can come to rest, resulting in zero momentum after the collision, but the vector sum of their momenta before the collision is also zero. The conservation of momentum can be explained through Newton's laws, where the forces exerted during the collision are equal and opposite, leading to equal and opposite changes in momentum. Therefore, momentum conservation holds true even when kinetic energy is not conserved.
II Ziv II
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Homework Statement


This was a test question that I got wrong and it has been bothering me so much. How is momentum conserved in an inelastic collision?


Homework Equations


p = mv
KE = (.5)m(v^2)


The Attempt at a Solution


I know that kinetic energy is not conserved because some of the energy in the collision escape as heat, sound, etc but why is momentum conserved? Shouldn't the system lose momentum because it it losing speed?
 
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There are a number of ways of approaching this. It depends, among other things, on the fact that momentum is a vector while kinetic energy is a scalar.
Two identical objects with the same speed could collide head in and both come to rest and stick together.
In this case, all the kinetic energy has gone. It's also a perfectly inelastic collision.
The momentum situation is different.
The one object, moving from left to right, has momentum mv, but the other, moving right to left, has momentum -mv.
Momentum before is mv + (- mv) [the vector sum takes account of the direction]
the momentum after is zero. (both at rest)
Before = after (both zero) yet you have "lost speed".

You can also argue for conservation of momentum from Newtons 2nd and 3rd Laws.
 
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Hi II Ziv II! Welcome to PF! :smile:
II Ziv II said:
This was a test question that I got wrong and it has been bothering me so much. How is momentum conserved in an inelastic collision?

I think the answer they want in the exam is good ol' https://www.physicsforums.com/library.php?do=view_item&itemid=26" … total force = rate of change of total momentum.

Since there are no external forces on the colliding bodies, the rate of change of total momentum is zero, ie the total momentum is constant. :wink:
 
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II Ziv II said:
Shouldn't the system lose momentum because it it losing speed?

Speed? We do not work with speed in momentum, we work with velocity.
 
II Ziv II said:

Homework Statement


This was a test question that I got wrong and it has been bothering me so much. How is momentum conserved in an inelastic collision?

Homework Equations


p = mv
KE = (.5)m(v^2)

The Attempt at a Solution


I know that kinetic energy is not conserved because some of the energy in the collision escape as heat, sound, etc but why is momentum conserved? Shouldn't the system lose momentum because it it losing speed?
In my view, the best way to explain conservation of momentum is this:

1. F = dP/dT => dP = Fdt (definition)

2. Ball A and ball B collide. The change in momentum of A is the force of B on A multiplied by the time through which it is applied ie. time of contact (I'm over-simplifying - since the force is not constant, you have to take the integral of force x dt) .

3. The change in momentum of B is the force of A on B x time of contact.

4. BUT, the force of A on B is EXACTLY equal to the force of B on A but EXACTLY opposite in direction (Newton's 3rd law). AND, the time through which these forces last is EXACTLY the same for both. So the magnitude of the change in momentum of A is EXACTLY equal to the magnitude of the change of momentum of B and EXACTLY opposite in direction.

AM
 
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