Let’s say there are two identical blocks on a frictionless surface. One block has momentum P while the other is at rest. Block with…?

…momentum P collided with the second block. Now let’s say after the collision, the two blocks stick together with the half the speed of the first block before the collision. 

So we have

mv1i + 0 = 2mvf

2mvf equals mv1i divided by two because 2mvf speed is half and mv1i is also multiplied by two because the two blocks are identical. So we have the total momentum of the system initially equal to the total momentum of the system finally. How is this a perfectly inelastic collision if the total momentum of the system is conserved?

Update:

I feel stupid. I realized after a while that elastic means total kinetic energy is conserved. For some reason I confused inelastic collisions to mean total momentum isn’t conserved. Silly me lol total momentum is always conserved

1 Answer

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  • 1 month ago

    because momentum is always conserved, elastic or not.

    KE is not conserved, which means it is not an elastic collision.

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