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Momentum and energy in collisions, bullet problem

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A bullet of mass 0.125 kg traveling horizontally at a speed of 250 m/s embeds itself in a block of mass 3.5 kg that is sitting at rest on a nearly frictionless surface.
    (a) What is the speed of the block after the bullet embeds itself in the block?

    v = ? m/s

    (b) Calculate the kinetic energy of the bullet plus the block before the collision:

    Ki = ? J

    (c) Calculate the kinetic energy of the bullet plus the block after the collision:

    Kf = ? J

    d) Was this collision elastic or inelastic?

    (e) Calculate the rise in thermal energy of the bullet plus block as a result of the collision:

    Ethermal,bullet + Ethermal,block = ? J

    (f) Which of the following statements are true?
    The bullet and block got hot because Q was large.
    Because the process was very fast, Q was negligible.
    Q was negative during the collision because the bullet and block were hotter than the surroundings.

    2. Relevant equations
    delta p1+delta p2=0
    delta K=0

    3. The attempt at a solution
    For a.) i tried 2(m/M)v but it was wrong.
  2. jcsd
  3. Mar 30, 2009 #2


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    What is the equation for an inelastic collision then?

    Does this help?
  4. Mar 31, 2009 #3
    yea thanks
  5. Mar 31, 2009 #4
    for c.) wouldn't the final kinetic energy be 1/2*m*v^2 of just the block since the final speed of the bullet is 0?
  6. Mar 31, 2009 #5


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    In an inelastic collision the bullet will have the same velocity as the block. It is embedded isn't it? So the final kinetic energy will be the sum of the masses times the final velocity.
  7. Mar 31, 2009 #6
    i tried (m1+m2)*vf...(.125+3.5)*8.62 and got 31.2475. It said it was wrong. I calculated the final velocity and got it right. What's wrong?
  8. Mar 31, 2009 #7
    Try 1/2 *(m1+m2)*vf^2.....this should work
  9. Mar 31, 2009 #8


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    Isn't c) asking for 1/2 m*v2?

  10. Mar 31, 2009 #9
    For part e) take the difference between the kinetic energy of the bullet plus the block before the collision and the kinetic energy of the bullet plus the block after the collision to get the answer.
    f) it happens so fast, that q is negligible.
  11. Mar 31, 2009 #10
    yea i got it now
  12. Mar 31, 2009 #11
    hey, are you in 172??
    is this on the webassign?
    what did you get for the first problem? the one with different masses? i have a posting on here, i tried many different things but cant get it.
    can you help me on it??
  13. Mar 31, 2009 #12
    Yea man, Im in 172 and on webassign, im stuck at the same spot you are.
  14. Mar 31, 2009 #13
    problem one answers are

    |1,final| < |1,initial|

    If m2 >> m1, then the final speed of object 1 is greater than the final speed of object 2.

    1,initial = 1,final + 2,final

    is that where you were stuck?
    i guess we had to assume it was elastic where M1 bounces off of M2.
    not sure how the expect us to know that with what is given but oh well.
  15. Mar 31, 2009 #14
    No, sorry im mean on question #3 where you got stuck...I got #1
  16. Mar 31, 2009 #15
    Question #3 i found the Vfinal by using
    P1(inital) + P2(inital) = P1(final) + P2(final)
    you know all but P2(final) just solve for it by doing m*v (find the P) for each
    remember your answer is in P so use P= mv to get v (just divide by mass)

    not sure how to find E internal. im still stuck on that.

    lastly the box at the bottom just click all the boxes with Q = 0 (should be 2 boxes)
  17. Mar 31, 2009 #16
    alrite man thanks!
  18. Mar 31, 2009 #17
    the only thing im stuck on is the e internal
  19. Mar 31, 2009 #18
    no problem.
    see you on here tomorrow or thursday for the hw due then lol
    its worth 67 points but not sure how hard it is.
    good luck on it lol
  20. Mar 31, 2009 #19
    yeah i did not get that. got everything else so 21/22 is fine with me
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