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

    • Thread Starter
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    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?
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    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"

     \frac{1}{2}mv^2

    for example
    • Study Helper
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    0.5mv^2 = (0.5P^2)/M ??????
 
 
 
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