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2 Collision questions from EC

Discussion in 'MCAT Study Question Q&A' started by iceman132, Jul 31, 2011.

  1. iceman132

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    For the first picture:

    Would the velocity of B equal mvsin60 deg (All the horizontal velocity) It is originally at rest.
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    92 (30 min exams)

    If the 2 objects shown below collide and remain together without spinning what will be their final velocity

    Answer: 5.8 m/s

    The horizontal momentum of the 2 rock system equals zero so there will be no horizontal velocity. The vertical momentum is 87 kg m/s We divide this by 15 kg to get 5.8 m/s
    ---------

    Photos of each are in an attachment.

    I really don't understand how horizontal (left to right) velocity is zero. Something is not clicking for me because I see that the ball is going in a diagonal direction.

    Do the 2 horizontal velocities cancel out? Or is there no horizontal velocity at all?
     

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    #1 iceman132, Jul 31, 2011
    Last edited: Jul 31, 2011
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  3. Majik

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    This is how I solved it:

    [​IMG]

    Keep in mind that momentum is a vector. One object (m=5kg) is traveling east. The other object (m=10kg) is traveling south west. The reason why there's no horizontal component after the two objects collide is because the eastward component and the westward component of momentum cancel each other out.

    In the picture, the eastward momentum is: 50kgm/s.
    You could of solved for westward momentum using trig:
    (-100kgm/s)(cos(60)) = 100x0.5 = -50kgm/s.

    By conservation of momentum (Horizontal Direction; X-axis):
    Momentum Inititally: 50kgm/s + (-50kgm/s) = 0
    Therefore the final momentum in the x-direction must also equal 0. This explains why the horizontal component of velocity is zero.

    Now, what about the vertical component. The 5kg object has no vertical component, so its momentum is zero in the y direction. The 10kg object does have momentum in the y-direction.

    Specifically, it has: (100kgm/s)(sin60) = 87.0kgm/s of momentum

    By conservation of momentum (Vertical Direction; Y-axis):
    Momentum Inititally: 0 kgm/s + (-87kgm/s) = -87kgm/s
    Therefore, the final momentum must equal -87kgm/s.

    Since the two masses stick together:

    -87kgm/s (final momentum) = (m1+m2)vf
    -87kgm/s / 15kg = -5.8m/s or 5.8m/s southward.

    There's two ways to solve this problem. The easy way (the picture above), or breaking them into components. Save yourself the misery and stick to the easier method. Hope this helps!
     
  4. iceman132

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    Thank you, it definitely helps.

    So my follow up question is this....

    In the picture explanation they DON'T even bother to show how you cancelled out the horizontal component. Do you NEED to figure that out first? (Or will the horizontal always be zero in these situations)

    If the horizontal component WAS NOT ZERO then would you just add it to the vertical velocity?
     
  5. Majik

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    No, the definetly aren't always equal. It just so happened in this example, the same magnitude of momentum traveling east equaled the same magnitude of momentum traveling west. They cancelled out. However, they could of just as easily made it where one object had slightly more momentum in the eastward direction for example. In the above example, if that was the case then when the objects stuck together, they would move south east. The way momentum vectors cancel out is similar to the way force vectors of equal magnitude cancel each other when they act in opposite directions on the same object. If one force is slightly greater than the other, then there's a net force in that direction. In fact, momentum itself is derived from one of newton's laws. F=ma=mv/t and Ft (Impulse) = mv (momentum). This is known as the impulse-momentum theorem, which you're probably familiar with.
     
  6. iceman132

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    Thank you, so if one directional force (horizontally) was slightly greater then the other.... To find the velocity would you add it to the vertical component?

    mvsin + mvcos = v(m+m)
     
  7. Majik

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    You'd add the net momentum in say the X-direction, and the net momentum in the Y-direction. However, you cannot add vectors together by simple addition unless they are pointing along the same axis. Otherwise, you'd have to use the tip to tail method to add vectors and use the Pythagorean Theorem. Once you found the net momentum, you can find the final velocity by dividing by the total mass of the two objects that stuck together.
     
  8. iceman132

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    In the picture, the eastward momentum is: 50kgm/s.
    You could of solved for westward momentum using trig:
    (-100kgm/s)(cos(60)) = 100x0.5 = -50kgm/s.

    By conservation of momentum (Horizontal Direction; X-axis):
    Momentum Inititally: 50kgm/s + (-50kgm/s) = 0
    Therefore the final momentum in the x-direction must also equal 0. This explains why the horizontal component of velocity is zero.


    ----
    Momentum Inititally: 50kgm/s + (-50kgm/s) = 0

    So let's say here that the momentum = 10 instead of zero. (horizontally)

    Vertically it is the same at 87.0kgm/s of momentum

    Would you do this?

    Vertical momentum + Horizontal momentum
    87+10
    _____ = final velocity
    15
    Total mass

    OR

    Would you do
    10 squared + 87 squared = x squared

    x
    __ = final velocity
    15
     
    #7 iceman132, Jul 31, 2011
    Last edited: Jul 31, 2011
  9. Majik

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    You can't add vectors by simple addition unless they are pointing in the same direction. Otherwise, you have to use the Pythag. Theorem, so you'd use the later (bold) to find the net momentum of both the horizontal and vertical component vectors combined. Then you could divide that by mass to find the final velocity.
     

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