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Inelastic Collision, only masses known

  1. May 22, 2009 #1
    Inelastic Collision, only masses known need help

    1. The problem statement, all variables and given/known data
    A projectile (mass = 0.28 kg) is fired at and embeds itself in a target (mass = 2.35 kg). The target (with the projectile in it) flies off after being struck. What percentage of the projectile's incident kinetic energy does the target (with the projectile in it) carry off after being struck?


    2. Relevant equations
    m1V1i+m2V2i=m1V1f+m2V2f


    3. The attempt at a solution
    I just don't know how to solve without velocities. I know the KE+PE must equal the final KE+PE. No idea how to get percentages with this information.
     
    Last edited: May 22, 2009
  2. jcsd
  3. May 22, 2009 #2

    Cyosis

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    Kinetic energy isn't conserved in an inelastic collision and there isn't really any potential energy to speak of.

    What you do know is that the kinetic energy of the bullet is 1/2 mv^2 and the kinetic energy of the block+bullet 1/2 M u^2. You can solve u as a function of v through conservation of momentum. From that you can calculate the percentage.

    So the first step is to write down the equation for conservation of momentum and solve for u.
     
  4. May 22, 2009 #3
    Am I assuming the KE is 0 or that it is equal to the KE of just the bullet? If KE=0 then u=1.1467. If not,

    u= sqrt(KEb+B/(.5*(.28+2.35)))
     
  5. May 22, 2009 #4

    Cyosis

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    You can't calculate the velocities explicitly. Also Kinetic energy isn't conserved, the kinetic energy of the bullet will not be equal to the kinetic energy of the bullet+block after impact so you cannot use that equation. Momentum however is conserved so try to answer these two questions.

    What is the momentum before the impact in terms of the mass of the bullet and velocity of the bullet?

    What is the momentum of the bullet+block after the collision in terms of mass of the bullet, mass of the block and speed of the composite system?
     
  6. May 22, 2009 #5
    Ok, so conservation of momentum would mean that
    0.28V= before
    (2.35+0.28)Vf= after
    thus 0.28V should = (2.35+0.28)Vf
     
  7. May 22, 2009 #6

    Cyosis

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    Yep that is correct. Now solve that equation for Vf. You can then plug vf into the equation for kinetic energy. This way you will have both kinetic energies, for bullet and composite as a function of v. Calculating the percentage will then cause v to drop out.
     
  8. May 22, 2009 #7
    So, if I'm doing this right

    0.5*0.28V=0.28/2.63
    V=0.76

    plugged into the original KE, KE=0.081

    Do I then plug that number in for KEb+B then use that number to compute the percentage of the original KE that the larger mass has?
     
  9. May 22, 2009 #8

    Cyosis

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    You cannot calculate v nor is it necessary to do so. Two questions need to be answered.

    1) What is the kinetic energy of the bullet as a function of its mass and v
    2) What is the kinetic energy of the bullet+block composite as a function of the mass of the bullet the mass of the block and the speed vf.
     
    Last edited: May 22, 2009
  10. May 22, 2009 #9
    KEb=0.14V^2
    KEb+B= 1.315Vf^2

    the answer for the problem should be (KEb+B/KEb)*100 the V's cannot cancel because they are not equal. If they were, it'd be over 100% of the KEb. I took the liberty of multiplying out the 0.5
     
    Last edited: May 22, 2009
  11. May 22, 2009 #10

    Cyosis

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    The v's are not equal that is correct, but you have already calculated vf as a function of v. This will make the "v"s equal.

    [quote='BATBLady]Ok, so conservation of momentum would mean that
    0.28V= before
    (2.35+0.28)Vf= after
    thus 0.28V should = (2.35+0.28)Vf
    [/quote]
     
  12. May 22, 2009 #11
    plugging into equation:

    0.5(0.28)V^2=0.28V/2.36 ...V's cancel

    0.14^2 (does not)= 0.28/2.36

    0.0196 (does not)= 0.1186

    That's where I'm at. I think I'm putting the equations into the wrong things...but I'm not sure. Comparing these give over 100% of the KE from the original.
     
  13. May 22, 2009 #12
    I'm sorry this is taking so long. I missed two lectures because I was sick, which is probably why I'm having this issue.
     
  14. May 22, 2009 #13

    Cyosis

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    These v's don't cancel there is a square on one side and a first power on the other side. This means you will be left with one v.

    Use the momentum equation and the momentum equation alone to calculate vf as a function of v.

    Edit: There is no need to apologize we will get to the answer!
     
  15. May 22, 2009 #14
    Ok, that's

    0.28V/2.63=Vf
     
  16. May 22, 2009 #15

    Cyosis

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    That is correct now plug it into the kEb+B equation.
     
  17. May 22, 2009 #16
    =0.14V

    which is the 0.5(0.28)V, excluding a ^2
     
  18. May 22, 2009 #17

    Cyosis

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    I am not sure what this excluding a square means.

    We have KEb+B=1.1315 vf^2 and vf=0.28 v/2.63, combining these yields KEb+b=1.1315 (0.28v/2.63)^2. Now you have a v^2 in both kinetic energy equations. Can you see how to proceed from here?
     
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