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Momentum cars being stuck together

  1. Jun 8, 2010 #1
    1. The problem statement, all variables and given/known data
    A 1300kg car moving east collided with a 2600 kg SUV moving north at 28 m/s. The vehicles became stuck together . If the speed of the vehicles immediately after the collision was 30 m/s, what was their direction?

    A) 21 degrees E of N
    B) 52 degrees E of N
    C) 58 degrees E of N
    D) 69 degrees E of N

    Answer is suppose to be B

    2. Relevant equations

    P1 + P2 = P aka m1v1 + m2v2 = (m1 + m2) v

    3. The attempt at a solution

    I first drew vectors and then tried to add them together but i dont know how to draw them here so yea..

    I used the formula given ..... m1v1 + m2v2 = ( m1 + m2) v.

    Not given v1 so i attempted to solve that with algebra.... v1 = ((m1 +m2)v - (m2v2))/ m

    v1 = ((1300+2600) x 30) - (2600 x 28)) / 1300
    v1 = 34 m/s

    I drew a vector triangle eventually got tan ((2600 x 28)/(1300 x 24) = 58.74 degrees east of north

    Please help and Thank You.... Sorry if this was to messy to understand first time doing this :p
     
    Last edited: Jun 8, 2010
  2. jcsd
  3. Jun 8, 2010 #2

    collinsmark

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    The above equation is fine, as long as you treat the velocities as vectors.

    Sorry, but that's not going to work for you. You need to treat the velocities as vectors, and add them appropriately as vectors.

    In this problem it's probably easiest to break up this problem into separate directions. If you consider the East-West direction as being the 'x' direction, and the South-North direction as being the 'y' direction, then

    [tex] m_{1}v_{1x} + m_{2}v_{2x} = (m_1+m_2)v_x [/tex]

    [tex] m_{1}v_{1y} + m_{2}v_{2y} = (m_1+m_2)v_y [/tex]

    [tex] |v| = \sqrt{v_x^2 + v_y^2} [/tex]

    See if that helps. :wink:
     
  4. Jun 9, 2010 #3
    I think i got it..

    So for the [tex]m_{1}v_{1x} + M_{2}v_{2x} = (m_1+m_2) v_x[/tex] I wasnt able to solve it since i have two variables.

    But in the vertical direction i was able to solve it and i ended up getting a velocity of 18.66667 m/s

    Drew a vector additional triangle. 30 m/s was my hypotenuse and 18.66667 became my opposite side.

    Sin ( 18.6666666667/ 30 ) = 38.4786 degress.... I'm gonna assume that this value is North of East. which is also the same as 51.5212 degrees East of North.
     
  5. Jun 9, 2010 #4

    collinsmark

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    That's the way to do it. :approve:
     
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