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Inelastic Collision Calculate the Percent Change in KE

  1. Nov 25, 2013 #1
    1. The problem statement, all variables and given/known data

    An archer fires a 20g bolt with a speed of105 m/s through a 400 g apple initially at rest.
    The bolt exits the apple with a speed of 45 m/s. What is the percent change in kinetic energy of the apple/bullet system?

    m1= 20g
    Vi1= 105 m/s
    Vf2= 45 m/s
    m2= 400g
    Vi2= 0 m/s
    Vf2= ?


    2. Relevant equations

    (m1v2+m2v2)i=(m1v2+m2v2)f

    ΔKE= ((.5)(m1)(Vi1)2+(.5)(m2)(Vi2)2)i -((.5)(m1)(Vf1)2+(.5)(m2)(Vf2)2)f

    %ΔKE= 100(F-I/I)

    3. The attempt at a solution

    (20g)(105 m/s)+(400g)(0 m/s)=(20g)(45 m/s)+ (400g)(vf)
    2100(g)(m/s)=900(g)(m/s)+(400g)(vf)
    1200g(g)(m/s)=400g(vf)
    3 m/s = vf

    ΔKE= ((.5)(20g)(105m/s)2+(.5)(400g)(0)2)i -((.5)(20g)(45m/s)2+(.5)(400g)(3 m/s)2)f

    22050J-110250J=-88200J

    100x(-88200)/110250J= -80%

    Figured it out myself. :)

    :::::Lashawnda:::::
     
    Last edited: Nov 25, 2013
  2. jcsd
  3. Nov 25, 2013 #2
    You did 2100 = 900 + 400 vf and isolated it wrong at the next step.
     
  4. Nov 25, 2013 #3
    2100-900=1200
    1200/400= 3
     
  5. Nov 25, 2013 #4

    2100-900=1200
    1200/400= 3
     
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