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Help Not sure what im doing wrong

  1. Oct 30, 2006 #1
    Help!!! Not sure what im doing wrong

    A 11.4 g bullet is fired horizontally into a 94 g wooden block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.650, what was the speed of the bullet immediately before impact? m/s

    first thing i did was find the block speed

    t = sqrt((7.5)(.65)/9.8)

    t = .7053

    then i did

    V = deltax/t = 7.5/.7053 = 10.6338

    mv_1 + Mv_2 = (m + M)V

    V = 98.31 which is incorrect but answer is within 10% of the correct value.

    I am not sure if im doing this correctly or im just completely off.
     
  2. jcsd
  3. Oct 30, 2006 #2

    OlderDan

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    Your V is on the order of 10% too high. It should be less than 10m/s

    t = sqrt(mu*s/g)????

    I assume that last V means v_1
     
    Last edited: Oct 30, 2006
  4. Oct 30, 2006 #3
    V is the velocity of the moving block.

    Also i thought to get "t" it is t = sqrt(deltaX/g) but i was not sure how to get it on here since friction play a part of this problem.

    t = sqrt(mu*s/g) what is "s"? i know that mu is the kinetic friction and g is gravity but what is s?
     
  5. Oct 30, 2006 #4

    OlderDan

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    That is not what it is here
    s is often used for distance. You called it deltaX here. So change my query to

    t = sqrt(mu*deltaX/g)??? There is no such equation.

    From V_f - V = a*t you get V = -a*t (zero final velocity)
    V is the initial velocity of the block after collision, as you defined it. The acceleration is negative (decleration). So

    t = -V/a

    apply Newton's second law to find a

    (M+m)a = F_f = force of friction = mu*N = mu*(M+m)g

    etc

    Alternatively, once you identify a from the above, instead of finding the time you can use the equation that relates the change of velocity squared to acceleration and distance moved.
     
  6. Oct 30, 2006 #5
    okay you just completely lost me there.

    but i seem to got the answer somehow by doing this way which is probably incorrect to do it and i just got lucky and got it correct (this is what i got from reading what you posted). Im trying to figure out what you are telling me but i cant seem to understand it. sorry if im giving you a much harder time on this, but my professor never went over this (she skip more than half of the chapter) and expect us to know it.

    m_bulletv_i = (m + M)g

    11.4v_i = (11.4 + 94)(9.8)

    11.4v_i = 1032.92
    v_i = 90.61 m/s
     
  7. Oct 30, 2006 #6

    PhanthomJay

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    Momentum is conserved at impact, but why do you say m_bulletv_i = (m +M)g? where does the g come from??
     
  8. Oct 30, 2006 #7

    OlderDan

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    Mass time velocity cannot possibly equal mass times acceleration.
    You had this correct earlier
    Just recognize that v_2= 0

    Solve the equation I started for you relating (M+m)a to mu*(M+m)*g to find a. Find the value of V from the known stopping distance and the acceleration. Then use V in the equation above to find v_1
     
  9. Oct 30, 2006 #8
    (94 + 11.4)a = .65(94 + 11.4) (9.8)
    105.4a = 671.398
    a = 6.37 is this correct?
     
  10. Oct 31, 2006 #9

    PhanthomJay

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    Yes, now proceed per Dan's earlier responses.
     
  11. Oct 31, 2006 #10
    that is where i am stuck how would i go from there if t = -V/a

    which is t = -V/6.37
     
  12. Oct 31, 2006 #11

    PhanthomJay

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    It's best now that you know what a is, to use v^2 = 2as, wher s = 7.5 meters. and a = 6.37m/s/s. This is the initial velocity of the block/bullet system after impact. Then use conservation of momentum to solve for v_bullet. Make sure to include units in your answers.
     
  13. Oct 31, 2006 #12
    Thanks a lot it was correct. Thanks for taking the time on explaining and giving a step by step on how to do this. I forgot about the kinetic equation. Both of you were great help on explaining and taking the time to help me solve this.
     
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