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Escort-Camaro Collision

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data
    An Escort and a Camaro traveling at right angles collide and stick together. The Escort has a mass of 1100 kg and a speed of 25 km/h in the positive x direction before the collision. The Camaro has a mass of 1500 kg and was traveling in the positive y direction. After the collision, the two move off at an angle of 53 degrees to the x axis. What was the speed of the Camaro?(in km/h)


    2. Relevant equations
    m1v1 + m2v2cos[tex]\theta[/tex] = (m1+m2)vfcos53 (??)


    3. The attempt at a solution
    I have tried using the unknown theta as 90 degrees because they are traveling at a right angle to eachother, but that does not seem to work. I know that in an inelastic collision, kinetic energy is not conserved, but momentum is conserved. I know that the total momentum before the collision is the individual momentum of the two cars added together. I feel like I know little bits and pieces of information about this problem, but am unsure of how it relates to the problem. I had a very similar problem to this one where the second car was at rest and we used an equation similar to the one above. I am not sure what to do when the second car is actually moving! Thanks for any help!
     
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  3. Mar 24, 2009 #2

    mgb_phys

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    1, Draw a diagram - so you know where everything is going.
    2, Consider the X and Y momentum separately.
     
  4. Mar 24, 2009 #3
    I have drawn a diagram of the two cars. I also used the following calculations to determine final velocity: (which I'm not completely sure is correctly)

    sin53 = (1100 x 25)/((1500 + 1100)Vf)
    2600Vf = (1100 x 25)/sin53
    Vf = (1100 x 25)/(2600 x sin53)
    Vf = 13.24 km/hr

    I'm unsure how to obtain the angles needed and find the initial speed of the camaro.

    Still stumped...
     
  5. Mar 24, 2009 #4
    Using the final velocity, I used the following calculations to try to calculate the initial velocity...

    Vi = (1500+1100)/1500 x (13.24)(cos53)
    Vi = 13.81 km/hr

    Unfortunately, this is not the answer, and I'm really not sure why! Someone please help!!!!
     
  6. Mar 24, 2009 #5

    LowlyPion

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    As suggested previously you need to consider the momentum in X and momentum in Y separately.

    Write your equation in X and then in Y. This will reveal the x,y components of the final velocity. Since you know 1 of the initial velocities you can determine through the angle, the other component of final velocity, and hence what the other initial velocity is.
     
  7. Mar 24, 2009 #6
    I guess my problem is trying to figure out the angles. When my professor did a fairly similar problem in class, he ended up with an angle that I'm unclear of how he got that angle. Can someone please tell me exactly how to figure out the angles. I can use my equations if I had the angles.
     
  8. Mar 24, 2009 #7

    LowlyPion

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    As stated you need to start with the momentum in X and momentum in Y expressions.

    If you do not separate the x,y momentum you cannot solve the problem. Don't worry about the angle yet. That's too many worries ahead when you don't have the equations that you need to use first.
     
  9. Mar 24, 2009 #8
    I guess I should have put all of the equations that I have in the first post, but they're a little difficult to type out...but I'll try! Hope this is more clear of the point that I'm at...

    P1xi = m1|V1i|cos[tex]\theta[/tex]
    P1yi = -m1|V1i|sin[tex]\theta[/tex]

    P2xi = m2|V2i|cos[tex]\theta[/tex]
    P2yi = m2|V2i|sin[tex]\theta[/tex]

    P1xf = m1|V1f|cos[tex]\theta[/tex]1
    P1yf = m1|V1f|sin[tex]\theta[/tex]1

    P2xf = m2|V2f|cos[tex]\theta[/tex]2
    P2yf = m2|V2f|sin[tex]\theta[/tex]2

    P1xi + P2xi = P1xf + P2xf
    m1|V1i|cos[tex]\theta[/tex] + m2|V2i|cos[tex]\theta[/tex] = m1|V1f|cos[tex]\theta[/tex]1 + m2|V2f|cos[tex]\theta[/tex]2

    P1yi + P2yi = Piyf + P2yf
    -m1|V1i|sin[tex]\theta[/tex] + m2|V2i|sin[tex]\theta[/tex] = m1|V1f|sin[tex]\theta[/tex]1 + m2|V2f|sin[tex]\theta[/tex]2

    This will give:

    tan[tex]\theta[/tex] = (m1V1fsin[tex]\theta[/tex]1 + m2|V2f|sin[tex]\theta[/tex]2)/(m1V1fcos[tex]\theta[/tex]1 + m2V2fcos[tex]\theta[/tex]2)

    which will give me the value of [tex]\theta[/tex] that I need to plug into:

    m1|V1icos[tex]\theta[/tex] + m2|V2i|cos[tex]\theta[/tex] = m1|Vif|cos[tex]\theta[/tex]1 + m2|V2f|cos[tex]\theta[/tex]2

    I am not sure how to figure out the various angles from my diagram...and I'm not sure if the value for Vf that I have is correct. I am also not completely sure if I have laid out my equations correctly. I hope that putting everything I have will help someone help me! :-)
     
  10. Mar 24, 2009 #9

    LowlyPion

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    Half are of no great interest are they? If the initial velocity in x is 0 as it is for the Camaro, then you don't need it.

    But in the case of the Camaro in Y, note that the final combined mass of both will move off together with the original momentum of just the Camaro alone right?

    So writing the Camaro equation that describes the Y momentum, you should have

    McVc j = (Mc + Me)*Vec_y j

    So ...

    Vc j = (1100 + 1500/1500)*Vec_y j

    Likewise then do the equation for X which initially involves the Ve of the Escort.

    Then you can relate the tanθ to the ratio of the final x,y velocities to determine the original Vc the velocity of the Camaro.
     
  11. Mar 26, 2009 #10
    So I've FINALLY got it!

    tan53 = (1500Vc)/(1100 x 25)
    Vc = 24.33 km/hr

    Thanks for the help!
     
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