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Finding the energy lost in a collision as the bullet collides to the block

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data
    a 50 g bullet hits and becomes embedded in a wooden block of as 450 g which is at rest on a surface (μ = 0.41) and is attached to a spring bumper. The block and bullet compress the spring bumper a distance of 20 cm. Is the spring constant is 250 N/m
    find initial speed of bullet? how much energy has been lost in the collision?

    3. The attempt at a solution
    a)
    mbvb +mwvw=mtvc
    (0.05 kg)vb=(0.5kg)(4.47m/s)
    vb=44.7m/s

    b)
    ET2= 1/2(0.5 kg)(4.47 m/s)^2
    ET2= 4.995 J

    ET1= 1/2(0.05 kg)(44.7m/s)^2
    ET1= 49.95 J

    ET1-W=ET2
    W= 44.957 J

    How would i represent this in a percentage?
     
  2. jcsd
  3. Nov 10, 2012 #2
    Where did you get 4.47m/s?
     
  4. Nov 10, 2012 #3
    i calculated it...

    1/2mvc^2=1/2kx^2
    1/2(0.5kg)vc^2=1/2(250N/m)(0.2m)^2
    0.25vc^2=5J
    4.47m/s=vc^2
     
  5. Nov 11, 2012 #4
    You forgot about friction.
     
  6. Nov 11, 2012 #5
    so umm...

    1/2mvc^2 - Wf =1/2kx^2
    1/2(0.5kg)vc^2 - (0.5kgx9.8m/s^2x0.41)=1/2(250N/m)(0.2m)^2
    0.25vc^2 - (2.009J) =5J
    5.29m/s=vc

    IS THIS CORRECT??
    but the speed has increased????
     
  7. Nov 11, 2012 #6
    That's the force of friction, you need to multiply by the distance traveled.
    Yes, the speed has increased. Your original calculation only accounted for the energy in the spring, and not the energy lost to friction. More energy means a higher speed.
     
  8. Nov 11, 2012 #7
    1/2mvc^2 - Wf =1/2kx^2
    1/2(0.5kg)vc^2 - (0.5kgx9.8m/s^2x0.41x0.2m)=1/2(250N/m)(0.2m)^2
    0.25vc^2 - (0.4018J) =5J
    4.648m/s=vc

    This is the right answer right?
     
  9. Nov 11, 2012 #8
    Yes.
    Now, what is the momentum of the block and bullet together?
     
  10. Nov 11, 2012 #9
    i think you are trying to ask what is the final momentum well

    mtvc
    = (0.5kg)(4.648m/s)
    =2.324 kgm/s

    now when finding the speed of the bullet i am assuming you do this
    mbvb +mwvw=mtvc
    (0.05 kg)vb=(0.5kg)(4.648m/s)
    vb= 46.48 m/s
     
  11. Nov 11, 2012 #10
    Right, and now you can determine the energy of the bullet before it stuck the block.
     
  12. Nov 11, 2012 #11
    so would you just do this??

    ET2= 1/2(0.5 kg)(4.648 m/s)^2
    ET2= 5.400976 kgm/s

    ET1= 1/2(0.05 kg)(46.48m/s)^2
    ET1= 54.00976 kgm/s

    ET1-W=ET2
    W= 48.608784 kgm/s

    is this right??
    hwow ould i represent this in percentage?
     
  13. Nov 11, 2012 #12
    Yes, that's right, but why would you need to represent it as a percentage?
     
  14. Nov 11, 2012 #13
    Oh wait i don't need to represent it as a percentage..
    so this is right OMG!!! this is?? THANK YOU!!
     
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