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Bullet and target- conservation of momentum and energy

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data
    A man aims his gun vertically and shoots a 0.245 kg target, which is positioned on a telephone pole. The bullet has a mass of 0.055 kg and is traveling at a velocity of 850 m/s upwards just before it hits the target. The bullet passes through the target, emerging with a velocity of 395 m/s upwards. How high does the target rise after being hit by the bullet?


    3. The attempt at a solution

    m1v1+m2v2 = m1v1`+m2v2`
    (0.055kgx850m/s) = (0.055kgx395m/s) + (0.245kgxv2`)
    46.75kgm/s=21.725kgm/s + (0.245kg)(v2`)
    102.1428571m/s = v2`

    but now what should i do i just found the speed at which the target will be moving at.. so should i use the conservation of energy like this

    1/2mv2^2 = 1/2mv2`^2 + mgh
    1/2(0.245kg)(102.1428571m/s)^2 = 1/2(0.245kg)(0)^2 + (0.245kg)(9.8m/s^2)(h)
    1278.062499 J = (2.401 kgm/s^2)(h)
    532.3042478 m = h
     
  2. jcsd
  3. Nov 11, 2012 #2

    Nugatory

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    Staff: Mentor

    You can check your answer by solving the problem in a different way. Ask yourself:
    1) how much energy is transferred from the bullet to the target? Conservation of energy helps here.
    2) how high will that much energy carry the target? Again, conservation of energy will tell you, because the maximum height is reached when there's no kinetic energy left, just potential.
     
  4. Nov 11, 2012 #3
    How would u solve the first question??
     
  5. Nov 11, 2012 #4
    IS THIS RIGHT ANYONE?!?! :confused:
     
  6. Nov 11, 2012 #5

    TSny

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    Your work looks very good to me.
     
  7. Nov 11, 2012 #6
    Oh thankx :approve:
    But is there a way i can confirm the solution..?
     
  8. Nov 11, 2012 #7

    Nugatory

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    Staff: Mentor

    Well, you know the velocity of the bullet before it hit the target, and you can use that to calculate the kinetic energy of the bullet before it hit the target.

    You know the velocity of the bullet after it hit the target, so you also know the kinetic energy of the bullet after it hit the target.

    The difference between the two is the energy transferred to the target.
     
  9. Nov 11, 2012 #8
    so is this what you are trying to say..

    ET1- Wf = ET2
    19868.75 J - Wf = 4290.6875 J
    15578.06 = Wf
     
  10. Nov 11, 2012 #9

    TSny

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    Just be careful with how you interpret what type(s) of energy Wf represents.
     
  11. Nov 11, 2012 #10
    its the energy that is used to deform the target!
     
  12. Nov 11, 2012 #11

    TSny

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    Part of the energy goes into deforming it, some goes into tearing a hole in it, some goes into heat, some ends up as KE of the target just after the collision, etc.
     
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