need help ASAP

[viendong]

I've been trying for hours..... and couldn't get it right though

A 100 g steel ball and a 200 g steel ball each hang from 1.00 long strings. After rest, the balls hand side by side, barely touching. The 100 g ball is pulled to the left until its string is at a 45 degree angle. The 200 g ball is pulled to a 45 degree angle on the right. The balls are released so as to collide at the very bottom of their swings
Question: To what angle does each ball rebound.?
Answer: m =100 g >>> 79.3 degree
m = 200 g>>>>>14.7 degree
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1) find (y_0)A when the 100 ball is pulled to the left w/ 45 degree angle
(y_1)A = L(1-cos(theta))
=1(1-cos(45)
= 0.293
2) From that I find (V_1)A = sqr(2*g*(y_0)A)
(V_1)A = 2.40 m/s

3) Now I find the (V_2)A = (m1-m2)/(m1+m2)*(V_1)A
V_2)A = - 0.8 m/s

4) Now I find (Y_2)A = (V_2)A ^2/2*g

4) Now I need to find the angle that rebound for Ma=100 g

(theta) = arcosin(1-(Y_2)A /L) = 37 degree....

but the answer is 79.3 degree, please help me what i've done wrong?

[Andrew Mason]

A 100 g steel ball and a 200 g steel ball each hang from 1.00 long strings. After rest, the balls hand side by side, barely touching. The 100 g ball is pulled to the left until its string is at a 45 degree angle. The 200 g ball is pulled to a 45 degree angle on the right. The balls are released so as to collide at the very bottom of their swings
Question: To what angle does each ball rebound.?

Think of this in terms of an elastic collision between the two balls at the bottom. You can determine their speeds at the bottom using:
mgh = mv2/2

Since the collision speed is independent of m, v1 = -v2
Using conservation of momentum you can determine their respective velocities after collision. From that, determine their respective kinetic energies immediately after the collision. Using mgh = mv2/2 you can determine the maximum height the respective masses will reach.

AM