1. The problem statement, all variables and given/known data Two identical billiard balls are at rest on a frictionless, level surface, touching each other at one common point. A third, identical ball, the cue ball, is approaching along the common tangent with a constant speed of 20m/s. Assuming a completely elastic collision, with no spin on any balls, and making(reasonable) assumptions about symetry, calculate the velocity of the cue ball after the collision. The graph is as shown below (the masses of the 3 balls are the same, but the value is unknown) o------->8 20m/s 2. Relevant equations I think two equations could be used in this question, one is Pt = (20m/s)M = Pt`= (v1 m/s)M + (v2 m/s)M + (v3 m/s)M the other one is The Conservation of Energy Ek= Ek` (M(20M/S)^2)/2 = (M(v1 m/s)^2)/2 + (M(v2 m/s)^2)/2 + (M(v3 m/s)^2)/2 3. The attempt at a solution The only thing i can do is assume that ball1 and ball2 have the same speed after the collision so that i got 2 equations no.1 20M= 2(V1)M + (Vc)M, where V1 is the velocity of ball1 and ball2,Vc is the value we r looking for no.2 (20^2)M/2 = 2M(V1^2)/2 + (Vc^2)M/2 and by substituting 20M-2V1=Vc in no.1 into no.2, I got V1 = 13.33, which will result in Vc= - 6.67m/s but the answer in the book says Vc=-4m/s ... Did i do something wrong with my equations? Helps/Suggestions are really appreciated, Thanks in advanced!!