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# 2 particles are in a completely inelastic collision, calculate final Kinetic Energy? Watch

1. 2 particles collide completely inelastically
Particle A has mass m and momentum 2p.
Particle B has mass 2m and momentum p.
Initial kinetic energy is calculated as (3p^2)/(4m)
Momentum is conserved in competely inelastic collisions, whereas kinetic energy is not conserved here.
Thus, p(initial) = p(final)
Thus, (3p^2)/(4m) = p(final)
Total mass of system = m + 2m = 3m
Total momentum of system (using addition of vectors & Pythagors) = sqrt of (p)^2 + (2p)^2]
= sqrt of [p^2 + 4p^2]
= sqrt of 5p^2
= (sqrt of 5)*p
This momentum is the same before and after the inelastic collision.
to work out final kinetic energy, I use 0.5mv^2 = (0.5P^2)/M where the mass M and momentum P here are respectively the TOTAL mass and TOTAL momentum of the SYSTEM
M = 3m, so final KE = [0.5*(5p)^2]/(3m)
= (0.5*25p^2)/(3m)
= (12.5/3)*(mvmv/3m)
= (25/6)*(mvv/3)
= (25/18)*(mvv)
= (25mvmv)/18m
= (25p^2)/18
Where has this gone wrong?
2. What is the answer expected?

It's very difficult to read through the maths in the format you use.
It would help a lot if you used "Latex"

for example
3. 0.5mv^2 = (0.5P^2)/M ??????

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