# Collisions (elastic vs inelastic)?

If i take two objects...like two toy cars and roll one at the second one(at rest)

In an elastic collision, is the loss momentum in the first one equal to the gain of momentum in the second one (at rest)?

and imagining the same situation, say that the collision is inelastic..

Is the loss in momentum of car 1 equal to the gain in momentum in car 2?

I don't know if im over thinking this

Relevance

Conservation of Momentum:

total momentum in = total momentum out

In the elastic collison:

m1 = mass of toy car #1

v11 = initial velocity of toy car # 1 (prior to collision)

v12 = velocity of toy car # 2 (just after collision)

m2 = mass of toy car # 2

v22 = initial velocity of toy car # 2 (just after collision)

applying conservation of momentum:

m1v11 = m1v12 + m2v22 (vector equation)

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In the perfectly inelastic collision:

m1 = mass of toy car # 1

v11 = velocity of toy car # 1 (prior to collision)

vf = velocity after collision

m2 = mass of toy car # 2

applying conservation of momentum:

m1v11 = (m1 + m2)vf (vector equation)

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• the answer to the first question is yes, however much momentum is lost from car 1 is equal to the momentum gain in car 2 assuming it is a perfectly elastic collision.

assuming it is a perfectly inelastic collision for scenario two, then the momentum of car 1 before the impact is now equal to the momentum of car 1+ car 2 after the impact.

in any scenario where the collision is either not perfectly inelastic or elastic, then some much more complex math relying on many variables comes into play, just worry about perfectly elastic and inelastic.

Source(s): physics class.
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• Elastic means kinetic energy is conserved. Inelastic means it is not, some energy is lost in the noise, crunch, etc. Totally inelastic is a special case of inelastic, where you have the extra information that the two objects stick together after the collision. Totally inelastic is the easiest to solve, because there is only one v after the collision. Elastic is harder, because you have two equations for two velocities: conservation of momentum, and conservation of energy. In fact, a lot of people find the quadratic equation that results from substitution to be a pain, so there is actually a shortcut that works ONLY for elastic, head-on (i.e. one dimensional) collisions: v1 + v1' = v2 + v2' So if you combine that with m1v1 +m2v2 = m1v1' +m2v2' You get two easier equations in two unknowns. Good luck!

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