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2 bumper cars collide (check my work please?)

  1. Apr 25, 2017 #1
    1. The problem statement, all variables and given/known data
    So my book doesn't have an answer for this problem at the back, was just wondering if someone could check my work..

    Two bumper cars at the county fair are sliding toward one another. Initially bumper car 1 is traveling to the east at 5.62 m/s and bumper car 2 is traveling 60 degrees south of west at 10 m/s. After they collide, bumper car 1 is observed to be traveling west with a speed of 3.14 m/s. Fiction is negligible between the cars and the ground. If the masses of bumper cars 1 and 2 are 596 kg and 625 kg respectively, what is the velocity of bumper car 2 immediately after the collision?

    2. Relevant equations
    Ki = Kf
    [itex] V_f = ((m_1 - m_2)/(m_1 + m_2)) (V_{im_1}) + ((2m_2)/(m_1 + m_2))(V_{im_2}) [/itex]

    3. The attempt at a solution
    So the x velocity of bumper car 2 according the the eq ^ up there.

    [itex] V_x = ((625 -596) / (625 + 596)) 10cos(240) + \frac {2(596)}{(625 + 596) }(5.62) [/itex]

    x velocity = 5.62 m/s

    [itex] V_y = ( (625-596)/(625+596) ) 10sin(240) + 0 [/itex]

    y velocity = -.2057 m/s

    I put 0 here because the y velocity of car 1 doesnt exist

    Now I find magnitude:

    [itex] ( (5.3682)^2 + (-.2057)^2 )^{1/2} [/itex]

    = 5.3721 m/s at 357 degrees (using arctan)

    Can anyone check my work and make sure its correct? Am I doing the right thing here?

    I also got another answer around 50 m/s using Kinetic energy initial = Ke final but that doesn't seem right so I will use this one since 5.37 m/s seems about right to me.
  2. jcsd
  3. Apr 25, 2017 #2
    I haven't gone through the math, but it does not seem reasonable to me that one of the bumper cars at the county fair would be traveling in excess of 100 miles per hour after the impact. Call me a stick in the mud, but that may be a bit too extreme for children on bumper cars.
  4. Apr 25, 2017 #3
    I know lol, thats what my answer was using the Kinetic energy initial = Ke final approach, but I dismissed it since it seemed too large to me.

    Instead I got v = 5.3721 m/s at 357 degrees which seems reasonable to me but I don't have the answer for this one in the back of my book unfortunately. :(
  5. Apr 25, 2017 #4
    I'll try to work through it, but some smarter person will probably beat me to it.
  6. Apr 25, 2017 #5
    I have faith in you and your abilities TomHart.
  7. Apr 25, 2017 #6
    I got a different answer than you did. Let me ask you a question. What is the y component of car 2's velocity before impact? And, since car 1 has no y component of velocity, either before or after the impact, what does that tell you about car 2's y component of velocity after impact?
  8. Apr 25, 2017 #7
    Car 2's y component = 10sin(240) degrees, right? That means car 2's y component of velocity after impact remains the same?
  9. Apr 25, 2017 #8
    Yes, that's right. So that is a magnitude of 8.66 m/s. And since your answer of 5.3721 m/s for the magnitude of the final velocity is less than the y component of 8.66, that means your answer has to be wrong.

    But since you now know the final y component of the velocity of car 2, it should now be a simple matter of equating before and after total x component momentum to find the x component of car 2's velocity.
  10. Apr 26, 2017 #9
    Sorry one question: Y initial was first 10sin(240) then it was the absolute value of 10sin(240), correct??
    Last edited: Apr 26, 2017
  11. Apr 26, 2017 #10
    Personally, I don't like the angle 240. It is kind of confusing to me. When the problem says, "60 degrees south of west", it means, from the west direction you rotate 60 degrees toward the south. It seems natural to me to define the north direction as positive y, and east direction as positive x. So starting from the +y direction, the initial angle of car 2 would actually be 210 degrees if you rotate clockwise from the starting point.
    So that means the initial y component of car 2's velocity would be (10)(cos30) = 8.66 m/s in the negative y (or south) direction. And because car 1 has no y component of velocity, either before or after impact, due to conservation of the y component of momentum, the final y component of car 2's velocity has to remain unchanged - specifically, 8.66 m/s in the negative y (or south) direction. In other words, the y component of car 2's velocity is unaffected by the collision.
  12. Apr 26, 2017 #11

    That is the picture that came along with the question. I thought y component was suppose to be using sin not cos??
  13. Apr 26, 2017 #12
    For the angle defined the way it is in that picture, yes, it should be sin instead of cos.
  14. Apr 26, 2017 #13
    Okay, so going by the picture I can assume the cars initial Y velocity to be 10sin(240), right?

    And after the collision its still 10sin(240) because car 1 doesn't have a y component, right?
  15. Apr 26, 2017 #14
    Exactly . . . But I guess to be more accurate we should really say, car 2's y component of velocity has to remain the same because we know that car 1's y component of velocity is unchanged.
  16. Apr 26, 2017 #15
    I see.

    Did you get [itex] v_{Car2} = 3.35 ihat m/s + 10sin(240) jhat m/s [/itex]?

    Sorry I forgot how to do the hat symbol again -_-
  17. Apr 26, 2017 #16
    Yes, that looks right. I got vx = +3.354 m/s and vy = -8.66 m/s
  18. Apr 26, 2017 #17
    Cool thanks TomHart
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