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Elastic Collision: Calculate the Percent Change in KE

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

    A 0.26 kg cue ball with a velocity 1.2 m/s collides elastically with a 0.15 kg billiard ball at rest.
    What percentage of the initial kinetic energy is transferred to the billiard?

    m1= 0.26kg
    Vi1= 1.2 m/s
    Vf1= ?
    m2= 0.15kg
    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

    (0.26kg)(1.2m/s)+0=(0.26kg)(-vf1)+(0.15kg)(-vf2)
    (0.312 kg m/s)=(0.26kg)(-vf1)+(0.15kg)(-vf2)
    (0.312 kg m/s)-(0.26kg)(-vf1)= (0.15kg)(-vf2)
    (0.312 kg m/s)+(0.26kg)(vf1)= (-0.15kg)(vf2)
    ((0.312 kg m/s)+(0.26kg)(vf1))/(-0.15kg)=(vf2)

    (0.312 kg m/s)=(0.26kg)(-vf1)+(0.15kg)[[(0.312 kg m/s)+(0.26kg)(vf1)]/(-0.15kg)]
    (0.312 kg m/s)=(0.26kg)(-vf1)-[(0.312 kg m/s)]-[(0.26kg)(vf1)]
    (0.624 kg m/s)=(0.26kg)(-vf1)-[(0.26kg)(vf1)]
    (0.624 kg m/s)=(-0.52vf1)
    vf1= -1.2 m/s

    ((0.312 kg m/s)+(0.26kg)(-1.2 m/s))/(-0.15kg)=(vf2)
    ((0.312 kg m/s)+(-0.312 kg m/s))/(-0.15kg)=(vf2)
    0=vf2

    ΔKE= ((.5)(0.26kg)(1.2 m/s)2+(.5)(0.15kg)(0 m/s)2)i -((.5)(0.26kg)(Vf1)2+(.5)(0.15kg)(Vf2)2)f

    The answer is suppose to be 93%
     
    Last edited: Nov 25, 2013
  2. jcsd
  3. Nov 25, 2013 #2

    Doc Al

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    That's one of the equations you'll need. Since you have two unknowns, you'll need a second equation. Besides momentum, what else is conserved?

    You must have made a mistake somewhere. There's no way to solve for either final speed without using a second equation.
     
  4. Nov 25, 2013 #3
    Oh so I need to use ΔKE equation.

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

    and solve for either Vf1 or Vf2 then plug it back into:

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

    Find the other final velocity, and then:

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

    %ΔKE= 100(F-I/I)
     
  5. Nov 25, 2013 #4

    Doc Al

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    Good. That should do it.
     
  6. Nov 25, 2013 #5
    (0.26kg)(1.2m/s)+0=(0.26kg)(-vf1)+(0.15kg)(vf2)
    (0.312 kg m/s)=(0.26kg)(-vf1)+(0.15kg)(vf2)
    (0.312 kg m/s)-(0.26kg)(-vf1)= (0.15kg)(vf2)
    (0.312 kg m/s)+(0.26kg)(vf1)= (0.15kg)(vf2)
    ((0.312 kg m/s)+(0.26kg)(vf1))/(0.15kg)=(vf2)

    ((.5)(0.26kg)(1.2m/s)2+(.5)(0.15kg)(0 m/s)2)i = ((.5)(0.26kg)(-Vf1)2+(.5)(0.15)((0.312 kg m/s)+(0.26kg)(vf1))/(0.15kg))2)f
    (.1872 kg m/s)=((.5)(0.26kg)(-Vf1)2+(.5)(0.15)((0.312 kg m/s)+(0.26kg)(vf1))/(0.15kg))2)
    (.1872 kg m/s)=(-0.13kg)(vf1)2+(.13kg)[.0974kg m/s)+(0.0676)(vf1)2]/(.0225)

    Isn't there a simpler way? Lol
     
  7. Nov 25, 2013 #6
    ((.5)(0.26kg)(1.2 m/s)2+(.5)(0.15kg)(0)2)i = ((.5)(0.26kg)(vf1)2+(.5)(0.15kg)(Vf2)2)f
    (0.1872 kg m/s)+0=(0.13kg)(-vf1)2+(0.075kg)(vf2)
    (0.1872 kg m/s)+(0.13kg)(vf1)2=(0.075kg)(vf2)2
    SQR[(0.1872 kg m/s)+(0.13kg)(vf1)2]/(0.075kg)=vf
    (2.895 m/s)(vf1)=VF2

    (0.26kg)(1.2m/s)+0=(0.26kg)(-vf1)+(0.15kg)((2.895 m/s)(vf1))
    (0.312 KG M/S)+(0.26kg)(vf1)=(0.15kg)((2.895 m/s)(vf1))
    [(0.312 KG M/S)+(0.26kg)(vf1)]/(0.4343KG M/S)
    idk..
     
    Last edited: Nov 25, 2013
  8. Nov 25, 2013 #7

    vela

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    Yes, there is if the collision is one-dimensional, which it looks like you're assuming here. Check your textbook for a simpler equation you can use for one-dimensional elastic collisions, rather than the conservation-of-kinetic-energy equation. It's simpler because it doesn't have the squared terms, so the algebra is easier to handle.
     
  9. Nov 25, 2013 #8
    Providing the equation would have been less effort on both our parts. Thank you for the help. :)
     
  10. Nov 25, 2013 #9
    (0.26kg)(1.2m/s)+0=(0.26kg)(-vf1)+(0.15kg)(vf2)
    (0.312 kg m/s)=(0.26kg)(-vf1)+(0.15kg)(vf2)
    (0.312 kg m/s)-(0.26kg)(-vf1)= (0.15kg)(vf2)
    (0.312 kg m/s)+(0.26kg)(vf1)= (0.15kg)(vf2)
    ((0.312 kg m/s)+(0.26kg)(vf1))/(0.15kg)=(vf2)

    (2.08 m/s)+1.73(vf1)=(vf2)

    (1.2 m/s)+vf1=vf2
    (2.08 m/s)+1.73(vf1)=(1.2 m/s)(vf1)
    (2.08 m/s)=0.69(vf1)
    vf1= 3 m/s


    (2.08 m/s)+1.73(3m/s)=(vf2)
    vf2=7.27m/s
    ΔKE= (-(.5)(0.26 kg)(3 m/s)2+(.5)(0.15kg)(7.27 m/s)2)f-((.5)(0.26 kg)(1.2m/s)2+(.5)(0.15kg)(0)2)i
    =-1.17+3.96J-0.1872J
    =2.60J

    (100x2.60J)/-1.17J+3.96J= 93%

    This better be right. GRRRR.
     
    Last edited: Nov 25, 2013
  11. Nov 26, 2013 #10

    bobie

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    I might be mistaken, but I have studied that fv final velocity can never be more than 2iv
    if vi1 = 1.2 vf1 should not exceed 2.4, do you agree?
     
  12. Nov 26, 2013 #11
    Yeah I guess I need practice. I am having a hard time finding a good example. I haven't had math in a while so my algebra isn't very strong.
     
  13. Nov 26, 2013 #12

    vela

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    I figured you could use practice finding stuff in your book. :wink:

    I get ##v_{1f} = 0.322\text{ m/s}## and ##v_{2f} = 1.52\text{ m/s}##. Check your sign convention for ##v_{1f}##. You're not being consistent.
     
  14. Nov 26, 2013 #13
    How could both of my final velocities be positive if they are traveling in two different directions after collision?
     
  15. Nov 26, 2013 #14

    vela

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    How do you know they're traveling in different directions after the collision?
     
  16. Nov 26, 2013 #15
    (0.26kg)(1.2m/s)+0=(0.26kg)(-vf1)+(0.15kg)(vf2)
    (0.312 kg m/s)=(0.26kg)(-vf1)+(0.15kg)(vf2)
    (0.312 kg m/s)-(0.26kg)(-vf1)= (0.15kg)(vf2)
    (0.312 kg m/s)+(0.26kg)(vf1)= (0.15kg)(vf2)
    ((0.312 kg m/s)+(0.26kg)(vf1))/(0.15kg)=(vf2)

    (2.08 m/s)+1.73(vf1)=(vf2)

    (v1-v2)i=(v2-v1)f
    1.2-(vf1)=vf2
    (1.2 m/s)-vf1=vf2

    (2.08 m/s)+1.73(vf1)=(1.2 m/s)-(vf1)
    (2.08 m/s)+2.73(vf1)=(1.2 m/s)
    2.73(vf1)=-0.88 m/s
    2.73(vf1)/(2.73 m/s)=-0.88 m/s/(2.73m/s)
    vf1=-0.32 m/s)


    (2.08 m/s)+1.73(-0.32m/s)=(vf2)
    vf2=1.52 m/s

    ΔKE= (-(.5)(0.26 kg)(-0.32 m/s)2+(.5)(0.15kg)(1.52 m/s)2)f-((.5)(0.26 kg)(1.2m/s)2+(.5)(0.15kg)(0)2)i
    =-0.0133+.1732J-0.1872J
    = ???

    (100x2.60J)/-1.17J+3.96J= 93%
     
  17. Nov 26, 2013 #16
    How do I know that they are not? Lol

    I guess I just assumed that it's not a glancing collision. I thought the two balls bounce off each other in opposite directions, so the vf1 would be negative. No angles were given.
     
    Last edited: Nov 26, 2013
  18. Nov 26, 2013 #17
    AHA.:rofl:

    ΔKEcue= ((.5)(0.26 kg)(1.2m/s)2+(.5)(0.15kg)(0)2)i-((.5)(0.26 kg)(-0.32 m/s)2)f
    =0.1872J-0.0133J
    =.1739J

    100(.1739J)/.1872= 93%
     
  19. Nov 26, 2013 #18
    Finalllyyyyyyyy:!!)
     
  20. Nov 26, 2013 #19

    vela

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    Think about what happens when a bowling ball collides elastically with a ping pong ball at rest. The bowling ball is going to keep moving in the same direction as it was before because it's so much more massive than the ping pong ball. A collision doesn't have to cause the direction of its velocity to change.
     
  21. Nov 27, 2013 #20

    bobie

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    If the masses are equal the moving mass stops: fv1 = 0,
    if the moving mass is smaller it rebounds with max fv1 =≈ 2 iv1
    if it is greater it continues in the same direction with max fv1 = ≈ 1 iv1

    93.1% is not correct,
    you need not find fv1: Ef2/Ei1:
    the exact result is 92.8019036287 % (0.34745/0.3744) *100
     
    Last edited: Nov 27, 2013
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