physics help?

Question

A 2.00-kg ball, A, is moving at a speed of 5.00 m/s. It collides with a stationary ball, B, of the same mass. After the collision, A moves off in a direction 30.0° to the left of its original direction. Ball B moves off in a direction 90.0° to the right of ball A's final direction. How fast is ball A moving after the collision?

Answer 1

let's say ball A is moving along x-axis before the collision.
after collision ball B is moving along y-axis.

conservation of linear momentum.
before collision: pxi = M*Vai; and pyi = 0
after collision: pxf = M*Vaf*cos(theta); pyf = M*Vbf - M*Vaf*sin(theta)

where:
- Vai: speed of ball A before collision (5 m/s)
- Vaf: speed of ball A after collision (?)
- Vbi: speed of ball B before collision (zero)
- Vbf: speed of ball B after collision (?)
- theta: direction of ball A after collision (30°)

pxi = pxf
M*Vai = M*Vaf*cos(theta)
Vaf = Vai/cos(theta) = (5m/s)/cos 30 = 5.77 m/s

Answer 2

The center of mass will continue to move to the right at the same momentum, which is m*v*=m*5

The vertical components of the momentums will be equal. Since the masses are equal

cos(30)v2f=cos(60)v1f

v2f=cos(60)*v1f/cos(30)

The horizontal momentum must equal the starting momentum

m*5=m*sin(30)*v2f+m*sin(60)*v1f
the m divides out
substitute v2f from above
5=sin(30)*cos(60)*v1f/cos(30)+sin(60)*v1f
v1f=5/(sin(30)*cos(60)/cos(30)+sin(60))
v1f=4.33 m/s

I checked this and found that v2f = 2.5m/s

this satisfies conservation of energy too.

j