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Conservation of energy help

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data

    A red car and a blue car collide at an in intersection. Prior to the collision the red car with mass 1000kg was heading North at 15m/s. The blue car with mass 1500 kg was heading East at 20m/s. The collision is completely inelastic with the cars sticking together and moving as one. At what angle measured North of East do the cars move off?

    2. Relevant equations
    I used Vf = √(Vfx2) + (Vfy2)

    I alsoUsed m1vf1x+ m2vf2x= m1vi1x + m2vix for x component and
    m1vf1y+ m2vf2y= m1vi1y + m2viy for y component

    and then θ = tan-1(vfy/vfx) to findthe angle but i keep geting it wrong
    3. The attempt at a solution
    I ended up with an angle of around 27 degrees
     
  2. jcsd
  3. Dec 15, 2008 #2

    PhanthomJay

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    This is a conservation of momentum problem, not conservation of energy as your title suggests. However, your method looks good and I agree with your answer. What's wrong with it?
     
  4. Dec 15, 2008 #3
    That is correct about 27 degrees north of east. at a velocity about 13.5 m/s (13 with 2 sig figs)
     
  5. Dec 16, 2008 #4
    Cars stick after collision. So they have common velocity. Let velocity = vf
    Take east as +ve x and north as +ve y

    For x component:-
    (m1+m2)vfx = m1vi1x + m2vix
    (1000+1500)vfx = 1000*0 + 1500*20
    2500 vfx = 30000
    vfx = 30000/2500 = 300/25
    vfx = 12 m/s

    For y component
    (m1+m2)vfy = m1vi1y + m2viy
    (1000+1500)vfy = 1000*15 + 1500*0
    2500 * vfy = 15000
    vfy = 15000/2500 = 150/25 = 6 m/s

    theta = tan-1(vfy/vfx) = tan-1(6/12) = tan-1(1/2) = 26.565 deg
    So, as approximation, this is same as your answer. Why do you think your answer is wrong?
     
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