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Conservation of Momentum/Collision

  1. Dec 14, 2013 #1
    1. The problem statement, all variables and given/known data

    A 5-gram bullet is shot horizontally at a 2-kg wooden block resting against a relaxed 100 N/m spring (the other end of the spring is fixed against a wall, the spring and the block sitting on a horizontal, frictionless table). It is observed that the bullet bounces back at 50 m/s and that the spring shows a maximal compression of 5-cm.
    A) What was the initial speed of the bullet?
    B) What fraction of the initial kinetic energy of the bullet was lost during the collision?

    m of bullet = .005 kg
    M of wooden block = 2 kg
    Vfinal of bullet = 50 m/s
    Max compression (x) = 5-cm = .05 m

    2. Relevant equations
    Looks like elastic collision so .. total momentum is conserved as well as kinetic energy.
    KE = (1/2)m*v^2
    EPE = (1/2)k*x^2

    3. The attempt at a solution
    When spring is completely compressed:
    KEinitial(bullet) = PE(spring) + KEfinal(bullet)
    (1/2).005*v^2 = (1/2)*100*.05^2 + (1/2).005*50^2
    Would this be right? It seems too simple to me ..

    Any help is appreciated.
     
  2. jcsd
  3. Dec 14, 2013 #2

    Curious3141

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    Why would you assume that?

    This is not how you approach it. You shouldn't assume an elastic collision. The very fact that they ask what fraction of the bullet's KE is lost should tell you the collision is not perfectly elastic.

    Start with conservation of linear momentum to relate the initial velocity of the bullet to its final velocity and the velocity of the block at the moment it just starts moving. At this moment, the spring is uncompressed.

    When the spring is fully compressed, the block is at rest. At this point, all the block's KE has been converted to EPE.

    From this, you should be able to set up two equations to answer both parts in turn.
     
  4. Dec 14, 2013 #3
    Okay, so linear momentum would be:
    .005*Vi + 0 = 2.005*Vf + .005*50
    .005*Vi = 2.005*Vf + 0.25
    Vi = 401*Vf + 50

    KE:

    (1/2)(m+M)*Vf^2 = (1/2)100*.05^2

    then solve that for Vf^2 and plug that into Vi = 401*Vf + 50 to find the initial velocity?
     
  5. Dec 14, 2013 #4

    Curious3141

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    Wait, wait, wait. The bullet bounces back. There are errors in two of your terms here (one is a sign error, the other is numerical).

    Again, the bullet bounces back.
     
  6. Dec 14, 2013 #5
    .005*Vi = 2.005*Vf - .005*50

    not sure where second error is =(

    (1/2)(m+M)*Vf^2 - (1/2)m*50^2 = (1/2)100*.05^2

    ?
     
  7. Dec 14, 2013 #6

    Curious3141

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    If the bullet bounces back (off the block), why are you adding the masses of the block and bullet? Shouldn't that be just "2" instead of "2.005"?

    As for the energy equation, your first go in post #3 was correct, except that should be just M instead of (m+M). The bullet does not add to the block's mass.
     
  8. Dec 14, 2013 #7
    Ohhhhhh, for some reason I was thinking the bullet would travel with the block before bouncing off. Thank you so much!
     
  9. Dec 14, 2013 #8

    Curious3141

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    No worries, glad to help.
     
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