1. The problem statement, all variables and given/known data a .060kg tennis ball, moving with a speed of 2.5 m/s, collides head-on with a .09kg ball initially moving away from it at a speed of 1.15m/s. Assuming a perfect collision, what are the speed and direction of each ball after the collision 2. Relevant equations MaVa + MbVb = MaV’a + MbV’b (conservation of momentum) ½MaV²a +½ MbV²b = ½MaV’²a + ½MbV’²b (?) 3. The attempt at a solution I plugged in the numbers into both equations and got.... 1. (.06kg)(2.5m/s) + (.09kg)(1.15m/s) = (.06kg)(V1f) + (.09kg)(V2f) and... 2. 1/2(.06kg)(2.5m/s)^2 + 1/2(.09kg)(1.15m/s)^2 = 1/2(.06kg)(V1f)^2 + 1/2(.09kg)(V2f)^2 i then started to solve the second equation... .25 = 1/2(.06kg)(V1f)^2 + 1/2(.09kg)(V2f)^2 this is where I became confused....I got rid of the first part (1/2(.06kg)(V1f)^2) and just solved the second part [ 1/2(.09kg)(V2f)^2] .25 = 1/2(.09kg)(V2f)^2 .25 = .05(V2f)^2 .25/.05 = (V2f)^2 5 = (V2f)^2 sqrt(5) = V 2.24 = V I then went back and plugged 2.24 into the first equation ......and then rearranged the equation: (.06kg)(2.5m/s) + (.09kg)(1.15m/s) = (.06kg)(V1f) + (.09kg)(V2f) = - (.06kg)(V1f) = (.09kg) (2.23) - (.2523) -.06kg (V1f) = -.0528 (-.0528) / -(.06) = V1f v= .88 I know the answers are supposed to be .88 and 2.23 (back of the book), but it seems like I just randomly solved this (I don't get exactly how i got this), is this right? I really need to pass physics (architecture major) but i don't understand it, any help would be greatly appreciated.