A billiard ball traveling at 3.00 m/s collides perfectly elastically with an identical billiard ball initially at rest on the level table. The initially moving billiard ball deflects 30.0° from its original direction. What is the speed of the initially stationary billiard ball after the collision?
I set up two relationships involving momentum as follows
Let $m$ be the mass of both identical billiard balls, $v_1$ be speed of the initially moving billiard ball after the collision, $v_2$ be speed of the initially stationary billiard ball after the collision, and $\theta$ be the angle between the velocity vector of the second billiard ball after the collision and the direction of the initial momentum vector
Based on conservation of momentum:
$$1) \quad(3m/s)m=mv_1\cos(30^\circ)+mv_2\cos(\theta)$$ $$(3m/s)=v_1\cos(30^\circ)+v_2\cos(\theta)$$ $$2) \quad v_1\sin(30^\circ)=v_2\sin(\theta)$$
However this is as far as I got and I don't know how to solve for $v_2$. Am I missing something?
Btw I know that the answer is 1.5m/s.