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Momentum problem : bullet and chair

  1. Oct 27, 2006 #1
    There is a gun in which the barrel length is 62cm and has a muzzle velocity of 450m/s. A wooden chair of 30 kg sits on a wooden floor. A bullet penetrates the chair and the bullet lodges itself into the chair 9 centimeters. How far back, in centimeters, did the chair fly back? The coefficient of kinetic friction between the chair and the floor is 0.2

    I'm not really sure how to approach this problem other than trying to use momentum conservation but do I need to find the force of the bullet on the chair to find the acceleration and then use momentum conservation to find the velocity, but i don't think that works. Any direction please?
     
  2. jcsd
  3. Oct 27, 2006 #2

    Doc Al

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    Staff: Mentor

    Seems like a lot of extraneous information was supplied, but not the mass of the bullet. Was that given?
     
  4. Oct 27, 2006 #3

    rsk

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    I thought about the KE of the bullet and the work done by the retarding force to stop it after 9cm in the chair. But we can only get ke in terms of m, and the retarding force is very big - so when considering the forces on the chair, friction is insignificant (although this force is still in terms of m, so maybe wouldn't look so big if we knew m).

    Hope someone else can help!
     
  5. Oct 27, 2006 #4
    im sorry, i left the mass of the bullet out, it is 10 grams
     
  6. Oct 27, 2006 #5

    Doc Al

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    That's key information! Attack it in two parts:
    (1) The collision (What's conserved?)
    (2) The sliding across the floor (You can find the acceleration, or use energy methods.)
     
  7. Oct 27, 2006 #6
    for m1=bullet and m2=chair:

    ok, so i do m1v1 + m2v2 (which is 0 since the chair is at rest)=(m1+m2)vf
    which gives me 0.15m/s

    i can then say that the kinetic friction force is equal to the mass of both the bullet and the chair.

    so if friction force is equal to -Fk=mu_k(m1 +m2)g
    then the acceleration is equal to -mu_k(m1+m2)g/(m1+m2)
    Actually,all of that it equal to -mu_k*g which is -0.2 *9.81 =-1.96m/s^2

    so then just use kinematics using vf^2 =v^2 +2ax and that will give me the distance it slid?

    (You were right, a lot of extraneous numbers given, i guess they love the trickery!)

    Much appreciated.
     
    Last edited: Oct 27, 2006
  8. Oct 27, 2006 #7
    This question sounds soo familiar. Is this from the RANDALL KNIGHT text book: physics for scientists and engineers ?

    If yes, I have the solution somewhere..


    P.S is the full question about you being a physist lawyer that's trying to prove your client innocent?
     
  9. Oct 27, 2006 #8
    yeah it is, i tried to simplify the question as much as possible. I think i got an answer of 0.57cm
     
  10. Oct 27, 2006 #9

    rsk

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    Yes, I got that same answer when I tried - and I recognise some of the other values you gave aswell. I dismissed it thinking it was a pathetically small distance for a chair to move when hit by a bullet. But then it is a 30kg chair and I don't know much about bullets.....

    And I was also worried about the barrel length.
     
  11. Oct 27, 2006 #10
    Yes, it was definately intimidating at first, but once u realize that they don't give u enough information along with the numbers that are already there, then u start to simplify the problem a little.
     
  12. Oct 27, 2006 #11
    The answer was 1.3 CM and it was corrected by our teacher.. do you have an e-mail address? I can e-mail you the response work detail if you want.
     
  13. Oct 27, 2006 #12

    rsk

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    Yikes - so where's that factor of 2 gone then?
     
  14. Oct 27, 2006 #13
    well i put in 0.57 and it was correct, so hopefully my teacher won't see this... what is this factor of 2 discretion?
     
  15. Oct 27, 2006 #14

    rsk

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    Just that the answer thinktank gave is twice what you (and I) came up with - I thought we must have lost a 2 somewhere.
     
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