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Partialy elastic collisions

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A bullet with a mass of 4.0 g and a speed of 653 m/s is fired at a block of wood with a mass of 0.093 kg. The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 24 m/s.
    (a) What is the speed of the bullet when it exits the block?
    wrong check mark
    Your answer differs from the correct answer by orders of magnitude. m/s
    (b) Is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy?

    After collision speed is reduced for the sytem and the mass is the same therefore there is less KE than before collision.
    (c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.
    Ki =
    Kf = wrong check mark Check the syntax of your response.
    Your answer differs from the correct answer by 10% to 100%. J

    2. Relevant equations
    This is the problem I understand elastic and inelastic seperately. but this assumes it is a partially elastic problem and I can't seem to piece together the equation needed from those I have.

    Vf = (m1*v1,i + m2v2,i)/(m1 +m2) this assumes though (if im correct) that the bullet and block become fused.

    KE,i = .5*m1*v1^2 + .5*m2*v2^2

    KE,f = .5*m1*v1,f^2 + .5*m2*v2,f^2

    m1*v,i=m1*v1,f^2 + m2*v2,f

    v1,f = ((m1-m2)/(m1+m2))*v1,i

    v2,f = ((2*m1)/(m1+m2))v1,i

    P = m*v

    3. The attempt at a solution
    I know for a fact this problem will have some loss in K due to the inelastic portion of the problem. however, I cant find that until i am able to derive the final velocity of the bullet. To answer the homeworks problem in part 5 I need to use the values of velocity for both to find the KE,f.
    So the only thing I can't solve is how to find a velocity of an object which collided with and passed through an object.
    My attempt was to assume the bullet had "bounced" off in the positive direction. ((.004-.093)/(.004+.093))653 this gave me 599.144 m/s. However this was wrong due to the fact an elastic collision does not reduce KE of the bullet.

    I think what I need to do is figure the momentum given to the block by the impact and subtract that from the total initial momentum. This will give me the momentum of the bullet after impact from which I can use the mass to derive the velocity.
  2. jcsd
  3. Oct 19, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good! During the collision of bullet and block, momentum is conserved. Use that fact to find the speed of the bullet after it exits the block.
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