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Bullet Impact

  1. Apr 19, 2005 #1
    Problem:

    A 20-g bullet is fired into a 4-kg block and becomes embeded in it. knowing that the bullet and block then move up the incline for 1.2 seconds before coming to a stop Determine; (a) the Magnitude Of the bullets initial velocity, (b) the Magnitude of the Impulsive force by the bullet on the block.

    Vb_1=? mk=4kg t=1.2s
    mb=.02kg

    I want to use the Princ. Impulse n' Momentum.
    I have tried:
    sum(mv_1)=sum(mv_2) but this does not help much

    I have also thought of using some sort of constant Accel. Eq. Again to no avail.

    Send me in the right direction.
     

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  2. jcsd
  3. Apr 19, 2005 #2
    use the conservation of momentum along the incline plane....

    here are some hints:
    what is the magnitude of v_0 along the incline plane?
    what is the initial velocity of the block? (conservation of momentum)
    after you have the initial velocity, the rest should be easy
     
  4. Apr 19, 2005 #3
    V_o along the incline is (V_o/2)
    Initial Vel. Of block is 0
    But how does this apply?

    Impulse & Momentum:
    sum(mv_1)+sum(Imp_1,2)=sum(mv_2)
    Right, So:
    sum(mv_1)=mass bullet *v_o
    sum(Imp_1,2)=.5F-sin(15)4kg*9.81m/s
    sum(mv_2)=(mass bullet+mass block)*v_2


    F being the force exerted by the bullet on the block
    v_o being velocity of the bullet
    v_2 being velocity of bullet and block

    F is what is asked for in (b) part
    but i need v_o first
     
  5. Apr 19, 2005 #4

    Doc Al

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    Staff: Mentor

    [itex]V_0[/itex] along the incline (if I read your diagram correctly) is [itex]V_0 \cos{30}[/itex].

    I would treat the problem in two steps:

    (1) The collision of bullet and block: The collision is governed by conservation of momentum parallel to the incline.
    (2) The slide up the incline: Assuming no friction, the "bullet + block" system starts with some intial speed and has an acceleration down the plane due to gravity.

    Do part 2 first.

    Since you are not given the time it takes for the bullet to imbed itself into the block, I don't see how you can calculate the impulsive force.
     
  6. Apr 19, 2005 #5
    Thats what I thought.

    So:
    V_0=?
    V=0
    a=9.81m/s verticly or 2.54m/s down incline
    m=4.02kg
    t=1.2s

    cos(30)V_o=velocity of the bullet slong incline
     
    Last edited: Apr 19, 2005
  7. Apr 20, 2005 #6

    Doc Al

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    Use the acceleration and time to solve for the initial speed of the system (post collision) using kinematics. (That's what I called part 2)
     
  8. Apr 20, 2005 #7
    As DocAl suggested:

    1. Determine the initial post-collision speed of the sytem. You know t, and acceleration ([tex]gsin\theta[/tex]), so find [tex]v_0[/tex]. Then with the law of conservation of momentum determine the magnitude of the bullet's initial velocity.

    2. Are you sure that the question doesn't ask for the Impulse J? Or is collision time given?
     
  9. Apr 20, 2005 #8
    Found (a):
    V=Vo+at
    0=Vo+2.54*1.2
    Vo=3.04m/s
    mv_1=mv_2
    .02[V_1/cos(30)]=(4+.02)3.04

    V_1=707m/s

    as for (b)
    i think they mean:
    Imp_1,2=F(delta)t
    so Im guessing they want to find F(delta)t as a whole using
    m_1*v_1+F(delta)t=m_2*v_2
     
    Last edited: Apr 20, 2005
  10. Apr 20, 2005 #9
    by the way would this problem be done in the same way?

    4oz ball @ 9ft/s strikes a 10oz plate on springs with no loss of energy. find Vel. of ball after hitting plate, force exerted by ball on plate.
     

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  11. Apr 20, 2005 #10

    OlderDan

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    Yes, it's the same idea except that in this case you have energy conservation (elastic collision). The ball bounces off the plate. You have the same issue with the force. All you can do is find the total impulse because you cannot know the time or F(t) involved.
     
  12. Apr 21, 2005 #11
    [tex]J = \Delta p_{block} = p_{final}[/tex]

    In the final momentum do not add the bullet's mass.
     
  13. Apr 21, 2005 #12
    Found (a):
    V=Vo+at
    0=Vo+2.54*1.2
    Vo=3.04m/s
    mv_1=mv_2
    .02[V_1/cos(30)]=(4+.02)3.04
    V_1=707m/s

    Sorry ramollari, but you do using this method 707m/s is correct.
     
  14. Apr 21, 2005 #13

    OlderDan

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    I think ramollari's comment is directed toward the new problem you asked about, not the original one.
     
  15. Apr 21, 2005 #14
    No OlderDan, I meant the original problem. Since VSCCEGR was looking for the impulse that the bullet applies on the block, then by definition it is the change of momentum of the block.

    [tex]J = \Delta p = p_{final} - 0 = p_{final}[/tex]
     
  16. Apr 21, 2005 #15

    OlderDan

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    Yes. of course you are right. Same thing applies to both the old problem and the new. Im just getting confused by the sequencing of comments with some of these these add on problems. Thanks for the clarification.
     
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