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Car collision: decomposing momentum in x- and y-direction

  1. Oct 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Two cars collide at an intersection. Car A, with mass 2000 kg, is going from west to east, while car B, with mass 1500 kg, is going from north to south at 15 m/s. As a result of this collision, the two cars become enmeshed and move as one afterward. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65° south of east from the point of impact.

    How fast were the enmeshed cars moving just after the collision?

    2. Relevant equations
    I use the law of conservation of momentum (p1 = p2 in both the x- and y-direction.

    3. The attempt at a solution
    First I try to find the momentum in the x-direction with the equation p1x = p2x.
    I insert the values from the problem statement:

    x-dir:

    2000 kg * vA1x + 1500 kg * 0 m/s = (2000 kg + 1500 kg) * vABx * cos 65°

    y-dir:
    2000 kg * 0 m/s + 1500 kg * 15 m/s = 3500 kg * vABy * sin 65°
    vABy = (1500 kg * 15 m/s) / (3500 kg * sin 65°) = 7.1 m/s

    With the value for vABy, I try to find the find the value for the second side and the hypotenuse with trigonometry:
    vABx = vABy / sin ∅ = 7.1 m/s / sin 65° = 7.83 m/s.

    With the Pythagorean theorem, I can now find the hypotenuse, and the velocity that the two enmeshed cars have:
    vAB = √(7.12 + 7.832) = 10.6 m/s.

    Looking at the solution, this is clearly not the correct answer. Can anyone see where my mistake is?
     
  2. jcsd
  3. Oct 3, 2016 #2

    kuruman

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    What trig function relates the two right sides?
     
  4. Oct 3, 2016 #3
    The tangent of the angle? I still don't see how that will get me to the correct answer?
     
  5. Oct 3, 2016 #4

    kuruman

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    The same way you tried to get to the correct answer before except now you will be using the correct trig function. BTW, what is the correct answer?
     
  6. Oct 3, 2016 #5
    Okay, thanks for clarifying that :)

    I'm still not getting the correct answer, though. The correct answer, according to the solutions manual, is 7.1 m/s (which is the answer I get for vABy. When the solutions manual decomposes the x-direction, they use the values for car B, which I don't understand (since in my head, car B moves only in the y-direction). They use this equation:
    p1x = p2x
    (1500 kg)(15 m/s) = (3500 kg)vAB2 * sin 65° ⇒ vAB2 = 7.1 m/s

    Why do they use car B when considering the x-direction?
     
  7. Oct 3, 2016 #6

    kuruman

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    So what? Why can't 7.1 m/s be the correct answer? I got 7.09 m/s. The given quantities in web-delivered problems are randomly generated and it so happened that by dumb luck you got a close match here.
    Momentum conservation in the x-direction will give you the initial speed of car A. You need the mass of car B in the expression for pABx after the collision.
     
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