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. http://psblnx03.bd.psu.edu/res/fsu/capalibrary/18Collisions/prob11.problem?symb=uploaded%2fpsuerie%2f6821119950c439fpsueriel2%2fdefault_1129569435%2esequence___15___fsu%2fcapalibrary%2f18Collisions%2fprob11%2eproblem 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.