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Conservation of (Energy vs Momentum)

  1. May 21, 2014 #1
    1. A bullet (m1 = 0,01 kg) hits a ball hanging on a thread (m2 = 1kg) and stays in he ball (therefore new system = m1 + m2) and pushes the whole system into the height of 0,2m (max potential energy, Kinetic energy = 0) Calculate the speed initial speed of bullet (v) and the initial speed of whole system (v').

    2. Law of Conservation of Energy: Ek + Ep = const.
    Law of Conservation of momentum: p1 + p2 = const.

    3. So I've been doing the calculations with 2 different approaches and each gives a different result and I don't know why, so here they are:

    Approach #1: Let the whole system be an isolated system. Therefore the kinetic energy of bullet should equal the potential energy of the whole bullet+ball system at the max Ep state. Therefore:

    1/2 m1 v(bullet)^2 = (m1 + m2)gh

    (Note that for symplification we consider g = 10m/s^2). So I get

    v = sqrt{[2(m1+m2)gh]/m1}

    which gives the speed of bullet approximately 20,1 m/s. Now for calculation the speed of the system (at Ep of system = 0 and Ek is max) I used similiar approach (Ek1 = Ek2) and got v' = 2m/s . Energy is equal at all times.

    However here comes the second approach:

    Conservation of momentum says that: m1v1 = (m1+m2)v' so let's use the law of conservation of energy once more, however let's start with the initial kinetic energy of system (ball + bullet) should equal the potential energy at it's max state. Therefore:

    1/2(m1+m2)v'^2 = (m1+m2)gh

    which gives the same result of speed of the system as v' = 2m/s.
    However now I use this speed in the law of conservation of momentum and get:

    v(bullet) = ([m1+m2]v')/m1 which gives the speed of bullet as 202 m/s.

    So here's the conflict: Working with the conservation of energy purely the speed of bullet is aproximately 20,1 m/s. Working with the conservation of momentum the speed is 202 m/s + when I start calculating the energy values, it differs! So I don't know which solution is correct and why. Thanks for help in advance!
  2. jcsd
  3. May 21, 2014 #2
    The first method doesn't make sense because mechanical energy isn't conserved during the inelastic collision.
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