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
MaVa + MbVb = MaV’a + MbV’b (conservation of momentum)
½MaV²a +½ MbV²b = ½MaV’²a + ½MbV’²b (?)
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)
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
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.