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Conservation of energy and momentum

  1. Dec 5, 2014 #1
    xLCawVg.png

    this would be considered as an inelastic collision where (m1v1) + (m2v2) = (m1+m2)vf
    also the conservation of energy being (∆K +∆U)=0


    i tried using the conservation of energy to find v2 as the variable we are looking for is Vf, the speed the bullet leaves the block, and v2 (velocity of the block) is whats missing in the collision equation in order to solve for Vf, although, knowing that the correct answer is 100, these method doesn't the correct answer, it gives a very small value for Vf which doesnt't make sense.
     
  2. jcsd
  3. Dec 5, 2014 #2

    jbriggs444

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    Science Advisor

    The bullet penetrates the block and goes on through. The block and the bullet have different final velocities. So the equation you give for conservation of momentum cannot be correct.

    Try working backward from the 5 cm figure. How much energy must the block have had following the collision to have allowed it to compress the spring by 5 cm?
     
  4. Dec 5, 2014 #3
    the energy of the block giving the velocity of the block after the bullet hits it right? but then what collision formula could be used to find the velocity as the bullet leaves
     
  5. Dec 6, 2014 #4

    jbriggs444

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    Right, this gives you the velocity of the block after the bullet hits it. Now you just need conservation of momentum. Total momentum before = Total momentum after.

    You were almost there in your original post but just used the wrong formula for total momentum after the collision.
     
  6. Dec 6, 2014 #5
    But this equation would give (0.5)(m)(v)^2=-(0.5)(k)(x)^2
    because Ui=0 and Ki=0 and Uf=1/2kx^2 and Kf=1/2mv^2 , resulting in a negative velocity
     
  7. Dec 6, 2014 #6

    Doc Al

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    Staff: Mentor

    No, not right.

    After the bullet passes through the block, the block has just enough speed to compress the spring the given amount. Solve for that speed.
     
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