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Momentum and degrees

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data

    On a very muddy football field, a 110 kg linebacker tackles an 85 kg halfback. Immediately before the collision, the linebacker is slipping with a velocity of 8.8 m/s north and the halfback is sliding with a velocity of 7.4 m/s east.

    A.) What is the magnitude of the velocity at which the two players move together immediately after the collision?

    Answer is 5.9 kg*m/s

    p1 = m1v1 = 110*8.8 = 968
    p2 = m2v2 = 85*7.4 = 629

    square root of (968)^2+(629)^2 = 5.9

    B.) What is the direction of this velocity?

    \theta = what degrees east of north(NOT north of east)?

    I think it is 33 degrees, but I'm not sure if it is actually 57 degrees. Wouldn't it be y/x*arctan and then 90 degrees minus that number?

  2. jcsd
  3. Nov 4, 2011 #2
    never mind. its 33 degrees I'm pretty sure.
  4. Nov 6, 2011 #3


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    Staff Emeritus
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    Yes. Or you could take arctan(x/y)

    To help clear up any doubts, here is another way to look at it:

    If the northward and eastward components of momentum were the same, the angle would be 45 degrees -- exactly northeastward.

    If the northward component is larger than the eastward component, then the total (and in this problem, final) momentum points closer to northward than eastward. In other words, the angle from the north is less than 45.

    Hope that helps.
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