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Ballistic pendulum

  1. Mar 1, 2009 #1
    [solved]Ballistic pendulum

    1. The problem statement, all variables and given/known data
    A bullet of mass m = 0.024 kg is fired with a speed of vi = 97.00 m/s and hits a block of mass M = 2.61 kg supported by two massless strings. The bullet emerges from the right side of the block with a speed of vf = 48.50 m/s. Find the height to which the block rises.

    m1= 0.024 kg
    m2= 2.61 kg
    vi = 97 m/s
    vf = 48.5 m/s
    h = unknown

    prob06-1.gif

    2. Relevant equations
    Before+JA collision
    m1v1+ m2V2 = P
    Ki = 1/2m1v1^2+1/2m2vf^2
    1/2mv^2= mgh

    3. The attempt at a solution
    I know i need to use con of momentum before and just after bullet contact to find v. But thats where im having problems at, im not really sure how to set it up to find v

    I had been doing 1/2m1v1^2+1/2m2vf^2 = 3182.6 then trying to set that = mgh but it gives me 120.

    If i can find that v i know id need 1/2*m2*V = m2*g*h
    then solve from h there if im correct if someone could give me nudge into right direction would be great
     
    Last edited: Mar 1, 2009
  2. jcsd
  3. Mar 1, 2009 #2

    AEM

    User Avatar

    I don't notice where you take into account the energy lost during the collision.
     
  4. Mar 1, 2009 #3
    ok so would i do 1/2mvi^2 = 1/2 mVm2^2
    vm2= velocity of block just after bullet passes threw
    1/2*.024*97^2= 1/2*2.61*vm2^2

    vm2 = 9.3016

    1/2mvm2^2= mgh
    1/2*9.3016^2 = 9.8h (m's cancel out)

    i got h = 4.414 m but tells me thats wrong
     
  5. Mar 1, 2009 #4
    this is a complicated multi-step problem.

    first, use conservation of momentum to find the velocity of the block when the bullet is embedded (.884 m/s). Then, use conservation of momentum again using equation mass (bullet plus block)*velocity (bullet plus block)= Mbullet*Vfbullet+ Mblock*Vfblock

    subtract over the bullet final and divide by the mass of the block to obtain the velocity of the block after the bullet exits. Then, use conservation of kinetic energy to determine how high the block goes. 1/2mv^2=mgy Solve for y, and thats the answer. I got .1 m for the answer so if that's wrong I could be off.
     
  6. Mar 1, 2009 #5
    that didnt work but where are you getting the .1 when i do what you typed out i get
    (.024+2.61)*.884 = 2.61*48.5+2.61*Vb
    2.328456 = 126.585 + 2.61*VB
    -124.26 = 2.61*vb
    47.6 m/s = vb

    plugging that # in for v i get no where near .1, so you could be close but it wanted atleast 4 sig digits and adding 3 zeros wasnt it
     
  7. Mar 1, 2009 #6
    be careful, in your equation you entered 48.5 m/s as the block's velocity after impact. You should be solving for this variable, so insert 48.5m/s as the bullet's velocity. There was an error on my part; the height should be .0099 m
     
  8. Mar 1, 2009 #7
    thnx i put in wrong mass for bullet was the problem got it now was .01016
     
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