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Why is the momentum not conserved

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Two vehicles collide one has a velocity of .0246 m/s the other a velocity of 4.374 m/s at impact. After impact one has a velocity of .49 m/s and the other a velocity of .91 m/s.

    2. Relevant equations

    (p) = mv

    3. The attempt at a solution

    (p)1 = 13158 * -0.0246
    (p)1 = -323.68

    (p)2 = 1726 * 4.374
    (p)2 = 7549.52

    at impact momentum = 7225.84 = (p)1 + (p)2

    (p)3 = 13158 * -0.49
    (p)3 = -6447.42

    (p)4 = 1726 * 0.91
    (p)4 = 1570.66

    after impact momentum = -4876.76 = (p)3 + (p)4

    So Iam missing 12102.6 kg*m/s. Why is this like this...am I missing something?
     
  2. jcsd
  3. Apr 28, 2012 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    Your problem statement doesn't actually pose a question. It is literally just a statement.

    It also doesn't give the mass of either vehicle. So we have no way of knowing if the masses you assumed are what was intended.

    It also doesn't give any directions. Are the two vehicles initially travelling in the same direction or opposite directions?

    Basically, we don't have the info we need to evaluate if your work is sensible or not.
     
  4. Apr 28, 2012 #3
    Sorry about that

    Two vehicles collide one has a velocity of .0246 m/s the other a velocity of 4.374 m/s at impact. After impact one has a velocity of .49 m/s and the other a velocity of .91 m/s.



    veh1

    weight 13158kg
    heading 0 deg
    departure 179
    impact velocity 0.0246 ms
    departure velocity 0.49 ms

    veh2

    weight 1726 kg
    heading 190
    departure 245
    impact velocity 4.374 ms
    departure velocity 0.91 ms


    Veh 2 is heading toward veh1 slightly off of 180 deg.
     
    Last edited: Apr 28, 2012
  5. Apr 28, 2012 #4

    gneill

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    Staff: Mentor

    Momentum is a vector quantity. Compute the separate components of the momentum for each vehicle both before and after collision. What's the total momentum vector before? After?
     
  6. Apr 28, 2012 #5
    Okay,

    weight 13158kg
    heading 0 deg
    departure 179
    impact velocity 0.0246 ms
    departure velocity 0.49 ms

    Momentum-in Veh1

    p = mv
    p= 13158 * 0.0246
    p = 323.68

    Out

    p = mv
    p= 13158 * 0.49
    p = 6447.42


    veh2

    weight 1726 kg
    heading 190
    departure 245
    impact velocity 4.374 ms
    departure velocity 0.91 ms

    Momentum-in Veh2

    p = mv
    p= 1726 * 4.374
    p = 7549.52

    Out

    p = mv
    p= 1726 * 0.91
    p = 1570.66

    The total before is 7873.2 and after is 8018.08

    okay, but I still see a difference in momentum. Am I to include the approach and departure anges in this as well?
     
    Last edited: Apr 28, 2012
  7. Apr 28, 2012 #6

    gneill

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    Staff: Mentor

    Of course! Momentum is a vector: it has both magnitude and direction; it has components.

    Compute the momentum vector components for the vehicles before and after. Find the total momentum vector components for before and after. Then compare them.
     
  8. Apr 28, 2012 #7
    weight 13158kg
    heading 0 deg
    departure 179
    impact velocity 0.0246 ms
    departure velocity 0.49 ms

    Momentum-in Veh1

    p = mv
    p= 13158 * 0.0246 * (sin 0 = 0)
    p = 0

    Out

    p = mv
    p= 13158 * 0.49 * (sin 179 = .017)
    p = 109.60



    veh2

    weight 1726 kg
    heading 190
    departure 245
    impact velocity 4.374 ms
    departure velocity 0.91 ms

    Momentum-in Veh2

    p = mv
    p= 1726 * 4.374 * (sin 190 = -0.173)
    p = -1306.06

    Out

    p = mv
    p= 1726 * 0.91 * (sin 245 = -0.906)
    p = -1423.01

    Well... this is big time messed up.
     
    Last edited: Apr 28, 2012
  9. Apr 28, 2012 #8

    gneill

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    Staff: Mentor

    That's the y-component. What happened to the x-component? Your collision is happening the the x-y plane, so vectors will have x and y components.
    Again, there should be both x and y components. And where did this angle "49" come from?
    I thought the "departure" angle was 179°?

    So you still need to sort out the rest of the vector components...
     
  10. Apr 28, 2012 #9
    weight 13158kg
    heading 0 deg
    departure 179
    impact velocity 0.0246 ms
    departure velocity 0.49 ms

    Momentum-in Veh1

    p = mv
    p= 13158 * 0.0246 * (sin 0 = 0)
    p = 0

    Out

    p = mv
    p= 13158 * 0.49 * (cos 179 = -0.999)
    p = -6446.43



    veh2

    weight 1726 kg
    heading 190
    departure 245
    impact velocity 4.374 ms
    departure velocity 0.91 ms

    Momentum-in Veh2

    p = mv
    p= 1726 * 4.374 * (sin 190 = -0.173)
    p = -1306.06

    Out

    p = mv
    p= 1726 * 0.91 * (cos 245 = -0.422)
    p = -662.81

    The 49 was just a idiot mistake.
     
  11. Apr 28, 2012 #10

    gneill

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    Staff: Mentor

    Still waiting to see ALL the components for ALL the momenta. For example, I see a calculation for the y-component of the momentum of the first vehicle for before the collision, but I don't see the corresponding x-component... and I see an x-component calculation for the first vehicle's momentum after the collision, but I don't see its y-component....
     
  12. Apr 28, 2012 #11
    weight 13158kg
    heading 0 deg
    departure 179
    impact velocity 0.0246 ms
    departure velocity 0.49 ms

    Momentum-in Veh1

    p = mv
    p= 13158 * 0.0246 * (sin 0 = 0)
    p = 0

    Out

    p = mv
    p= 13158 * 0.49 * (sin 179 = 0.017)
    p = 112.52



    veh2

    weight 1726 kg
    heading 190
    departure 245
    impact velocity 4.374 ms
    departure velocity 0.91 ms

    Momentum-in Veh2

    p = mv
    p= 1726 * 4.374 * (cos 190 = -0.98)
    p = -7398.53

    Out

    p = mv
    p= 1726 * 0.91 * (cos 245 = -0.422)
    p = -662.81



    Sorry this is taking so long to grasp.
     
  13. Apr 28, 2012 #12

    gneill

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    Staff: Mentor

    Let's concentrate on one particular vector to begin with; How about the initial momentum of the first vehicle?

    You've got the:

    vehicle mass: m1 = 13158 kg
    Speed: v1 = 0.0246 m/s (meters per second --- you had written ms, which is meter seconds)
    direction: θ = 0°

    The speed is not a velocity; Speed has magnitude but no direction. Velocity has both magnitude AND direction. You're given the direction angle for the velocity, so you can determine the TWO components (x and y) for the velocity, and hence the TWO components for the momentum.

    What are the x and y components of the momentum for the vehicle?
     
  14. Apr 28, 2012 #13
    weight 13158kg
    heading 0 deg
    departure 179
    impact velocity 0.0246 m/s
    departure velocity 0.49 m/s


    P1x = m1v1 cos 0 = 13258kg * 0.0246 * 1 = 323.68 kg*m/s
    P1y = m1v1 sin 179 = 13158kg * 0.49 * 0.017 = 109.60 kg*m/s

    weight 1726 kg
    heading 190
    departure 245
    impact velocity 4.374 m/s
    departure velocity 0.91 m/s

    P2x = m2v2 cos 190 = 1726kg * .91 * -0.984 = -1545.52 kg*m/s
    P2y = m2v2 sin 245 = 1726kg * 4.374 * -.906 = 1731.28 kg*m/s

    I am confused about the x and y ...is the approach of each vehicle on the x and then the seperations on the y or is one veh totally y and the othe totally x ?
     
    Last edited: Apr 28, 2012
  15. Apr 28, 2012 #14

    gneill

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    Staff: Mentor

    Er, no. The initial velocity has only one angle, namely 0°. The angle 179° pertains to a different velocity, the departing velocity. We're only concerned with the initial velocity here.

    Use 0° for BOTH the x and y component calculations for the initial velocity.
    A vector in the plane has two components: an x-direction component and a y-direction component. Every velocity in the plane thus has two components. There is ONE angle associated with a given velocity; it's generally the angle between the velocity vector and the positive x-axis. Use that ONE angle to determine BOTH components for that velocity.

    If V is the speed and θ is the angle, then Vx = V cos(θ) and Vy = V sin(θ).
     
  16. Apr 28, 2012 #15
    weight 13158kg
    heading 0 deg
    departure 179
    impact velocity 0.0246 m/s
    departure velocity 0.49 m/s

    Vx = .0246 * cos 0 = .0246
    Vy = .49 * sin 179 = .0085
     
  17. Apr 28, 2012 #16

    gneill

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    Staff: Mentor

    No!!! The initial velocity HAS ONLY ONE ANGLE. Forget the 179° for now. That comes later, after the collision. Only the 0° angle exists at the moment. An object cannot be heading in two directions at the same instant in time!

    The initial velocity of the first vehicle has one angle: zero degrees. That's the ONLY ANGLE!

    The initial velocity vector is comprised of two components: One is its x-component, and one is its y-component. The ONE ANGLE along with the magnitude (speed) define the two components. There is the component that is parallel to the x-axis, and the component that is parallel to the y-axis. Use the trig functions cos and sin to extract those two components USING THE SAME ANGLE.
     
  18. Apr 28, 2012 #17
    weight 13158kg
    heading 0 deg
    departure 179
    impact velocity 0.0246 m/s
    departure velocity 0.49 m/s

    Vx = .0246 * cos 0 = .0246
    Vy = .49 * cos 0 = .49
     
  19. Apr 28, 2012 #18

    gneill

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    Staff: Mentor

    Why are you using the speed from AFTER the collision? Why are you using the cosine function for both components? You want to find BOTH components of the INITIAL velocity (before the collision) for the first vehicle. The initial velocity has ONLY speed 0.0246 m/s at angle 0°. That's all. No other angles or speeds should enter into consideration here.

    Perhaps you are unfamiliar with the use of vectors? In the following diagram a velocity vector V is shown on a set of Cartesian coordinate axes. The velocity vector has magnitude |V| which corresponds to the speed, angle θ with respect to the positive x-axis, and x and y components are depicted in red. Use the appropriate trig function of the angle to extract the individual x and y components of V.

    attachment.php?attachmentid=46765&stc=1&d=1335652087.gif
     

    Attached Files:

  20. Apr 28, 2012 #19
    weight 13158kg
    heading 0 deg
    departure 179
    impact velocity 0.0246 m/s
    departure velocity 0.49 m/s

    weight 1726 kg
    heading 190
    departure 245
    impact velocity 4.374 m/s
    departure velocity 0.91 m/s



    Vx = .0246 * cos 0 = .0246
    Vx = 4.374 * cos 190 = -4.30

    Vy = .49 * sin 179 =0.008
    Vy = .91 * sin 245 = -0.824
     
  21. Apr 28, 2012 #20

    gneill

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    Staff: Mentor

    Those would be the x-components of the initial velocities of vehicles 1 and 2. You should give them distinctive variable names in order to distinguish between them. Something like V1x and V2x.

    Now, where are the initial y-components of those same velocities?
    Those are the y-components of the final velocities of the of vehicles 1 and 2. You should give them distinctive variable names in order to distinguish between them. Something like U1y and U2y.

    Now, where are the final x-components of these same velocities?
     
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