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2 problems- center of mass & momentum

  1. Oct 29, 2005 #1
    1. The ratio of the mass of the earth to the mass of the moon is 77.6. Assume that the radius of the earth is about 6404.0 km and that the distance between the center of the earth and the moon is 380604.0 km. Determine the distance of the center of mass of the earth-moon system from the center of the earth.
    I treated the Earth as weighing 77.6 kg and the moon to be 1 kg. The diameter of the Earth= 40237 km
    I used the equation for center of mass
    [tex] x_c_m = m_1 x_1 + m_2 x_2 / m_1+ m_2 [/tex]
    so x = (77.6)(40237) + (1)(380604) / (77.6+ 1)
    This gave me an answer of 4.46 x 10^4 km, which wasn't right..
    2. In a pool game, the cue ball, which has an initial speed of 3.9 m/s, makes an elastic collision with the 8-ball, which is initially at rest. After the collision, the 8-ball moves at an angle q = 33.6° with respect to the original direction of the cue ball, as shown in the figure. At what angle f does the cue ball travel after the collision? Assume the pool table is frictionless and the masses of the cue ball and the 8-ball are equal.
    I tried using conservation of momentum, and since the masses are equal and don't really affect the answer, just making them 1 kg. The initial in momentum in the x-direction is equal to 3.9 kg*m/s, and in the y-direction it is 0.
    After the collison, The momentum in the x-direction is equal to mv cos theta + mv cos 33.6. Since both balls have momentum now.
    In the y-direction, the momentum = mv sin theta + mv sin 33.6.
    I tried using [tex] sqrt (mvcos theta + mv cos 33.6)^2 + (mvsin theta + mvsin 33.6)^2 [/tex] , but I'm not sure how to solve it without any information.
    Can anyone help me? Thanks.
  2. jcsd
  3. Oct 29, 2005 #2

    Chi Meson

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    One at a time. The radius of the earth is 6404 km, but so what?
    the "x's" used in the cm equation are not the radiuses or dimaters of the objects. (BTW, 40,000 km is the earth's circumference, but again, so what?)

    Make an axis that goes through the centers of the earth and the moon. Arbitrarily make the zero at one point (try the center of the earth). The "x" that you use is the point on this line where your masses are. IF earth is at the zero, then the "x" is zero. What's the moon's "x"?
  4. Oct 29, 2005 #3
    2. Find the initial net momentum vector before the collision (the momentum of the cue ball since the 8ball is not moving). Since the collision is elastic then there is no change in the total momentum, and the momentums of both the balls must both add up to equal the initial net momentum. You can figure out the momentum of the 8 ball, so subtract it from the net momentum to get the cue ball's.
  5. Oct 29, 2005 #4

    Chi Meson

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    and this equals zero, right?
  6. Oct 30, 2005 #5
    So the inital momentum is 3.9 kg*m/s
    The final momentum is still giving me some problems.
    in the x-direction, v1cos theta+ v2 = 4.68
    in the y-direction, v1sin theta = v sin 33.6

    I'm confused about the momentum of the 8-ball. I know the inital is 0, but I don't know how to figure out the final, without knowing the velocity.
  7. Oct 30, 2005 #6

    Chi Meson

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    You have three unknowns: v1' , v2', and the angle "f". This means you need three equations. You have two already, the third is found using conservation of kinetic energy (for this is an elastic collision).

    There is a triple-simultaneous equation coming up, and I think it will involve the quadratic solution.
  8. Oct 30, 2005 #7
    I don't think I'm doing this right, but here's what I did..
    (1/2)mv^2= (1/2)mv^2 + (1/2) mv^2
    (1/2)(3.9)^2 =(1/2)v1 ^2 + (1/2)(v2)^2
    7.61= (1/2) V1^2+ (1/2)(v2)^2
    15.22= (V1)^2+ (V2)^2
    3.9 = V1 + V2
    3.9 - V1= V2

    Then I plugged in what I got for V2 for momentum in the y-direction to, but I still get another equation that doesn't really get me anywhere.
  9. Oct 31, 2005 #8


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    First of all in this particular question, you could quickly solve it by useing the fact that when two balls with equal mass collide in an elastic collision, then there is always a 90 degree angle between there direction after the collision.

    After you got the third equation unsing the kinetic energy, you know how V2 is related to V1. using that information in the first two equations you should get two equations with two variables.
  10. Oct 31, 2005 #9
    I figured both of them out.. Thanks
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