If momentum is conserved in a perfectly inelastic collision (no lost to heat or defor

DaleSpam

Quote by justquestions

So given
m1=10 kg v1i= 10 m/s
m2=10 kg v2i = 0

pi = 100 kg m/s = pf = (10+10)*vf -> vf = 5 m/s (perfectly inelastic)

Now from Energy's point of view.

KEi = (1/2)(10)(10)^2 = 500 J

KEf = (1/2)(20)(5)^2 = 250 J

so KEi > KEf

You are saying that the decrease in KE is totally due to heat?

Your calculations are all correct. The decrease in KE doesn't all have to go to heat, it can in principle go to any form of internal energy, but most of those other forms will themselves eventually go to heat.

Woopydalan

I think one of the issues is how we're allowed to assume there is no deformations with an elastic collision, yet with an inelastic we are not allowed to assume no deformation.

thebrainstorm

In an inelastic collision, KE is always lost as heat and the body gets deformed.

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Woopydalan	I think one of the issues is how we're allowed to assume there is no deformations with an elastic collision, yet with an inelastic we are not allowed to assume no deformation.

thebrainstorm	In an inelastic collision, KE is always lost as heat and the body gets deformed.