Physics question: A fireworks rocket is moving at a speed of 49.4 m/s. The rocket suddenly breaks into two

A fireworks rocket is moving at a speed of 49.4 m/s. The rocket suddenly breaks into two pieces of equal mass, which fly off with velocities v1 and v2, as shown in the drawing. What is the magnitude of (a) v1 and (b) v2?

Image URL: http://edugen.wileyplus.com/edugen/courses/crs2216/art/qb/qu/c07/ch07p_19.gif

Can someone please help me with this problem? I have 2 more hours to do it and I can't seem to figure it out. Thanks

difficult to tell from the drawing, but I assume they are all moving in the same plane.

X and y components of momentum have to cancel to zero. Assume x axis is in the original direction of the rocket. Original momentum is 49.4m in the x direction, zero in the y direction.

piece 1 has x momentum of m1v1 cos30, and y momentum of m1v1 sin30
piece 1 has x momentum of m1v2 cos60, and y momentum of –m1v2 sin60
(negative direction because it is in the –y direction)

2m1 = m

m1v1 sin30 – m1v2 sin60 = 0
m1v1 cos30 + m1v2 cos60 = 49.4m = 49.4•2m1

v1 sin30 – v2 sin60 = 0
v1 cos30 + v2 cos60 = 49.4•2

v1 (0.5) – v2 (0.866) = 0
v1 (0.866) + v2 (0.5) = 98.8
multiply top by (0.5/0.866) and add
v1 (0.289) – v2 (0.5) = 0
v1 (0.866) + v2 (0.5) = 98.8
V1 (1.155) = 98.8
V1 = 85.5 m/s

(85.5) (0.5) – v2 (0.866) = 0
42.8 = v2 (0.866)
V2 = 49.4 m/s

but check the math, the last number is funny.
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