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Momentum conservation (ballistic pendulum)

  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data
    A .01kg bullet is fired into a 1.2kg block hanging from a 1m wire. The bullet exits the block with a speed of 200m/s and the block swings to a height of .2 meters. What is the original velocity of the bullet? What percentage of the original energy of the bullet is no longer in mechanical forms of energy?

    2. Relevant equations
    I know that initial momentum = final momentum but I can't seem to find the velocity of the block after the collision.

    3. The attempt at a solution
    I had this on a test and tried energy conservation. Obviously this isn't correct but I cant find the momentum of the block. I've been working this for about 5 hours now so i'm clearly not getting something so a step by step explanation would be great.

    GPE of block = (mgh) = (1.2 x 9.8 x 0.2m) = 2.353 Joules. This has resulted from KE of the bullet having swung the block.
    The bullet exited the block. Its KE after exit = 1/2 (m*v^2) = 1/2 (0.01*200^2) = 200 Joules.
    The total energy has come from the bullet, = (200 + 2.353) = 202.353 Joules.
    Original bullet V = sqrt.(2KE/m) = sqrt.(404.706/0.01) = 201.2 m/sec.
     
  2. jcsd
  3. Dec 10, 2016 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's the energy of the block after the bullet has passed through it. Use this to determine the speed of the block.
     
  4. Dec 11, 2016 #3
    Ok, so this is what I ended up with:

    mgh = 1/2mv2
    v=sqrt(2gh) = 1.98m/s

    Pi = Pf
    mpvpi = mbvb + mpvpf
    vpi = (mbvb + mpvpf) / mp
    vpi = 437.6 m/s

    Thanks for the help!
     
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