Answer
After the collision, puck B's speed is $1.73~m/s$ and it moves at an angle of $30.0^{\circ}$ below the +x-axis.
Work Step by Step
Let $M$ be the mass of each puck.
By conservation of momentum, the final momentum of the system is equal to the initial momentum.
We can find the horizontal component $v_x$ of the final velocity of puck B:
$M~v_x +M~(1.0~m/s)~cos~60.0^{\circ} = M~(2.0~m/s)$
$v_x = (2.0~m/s)-(1.0~m/s)~cos~60.0^{\circ}$
$v_x = 1.5~m/s$
We can find the vertical component $v_y$ of the final velocity of puck B:
$M~v_y + M~(1.0~m/s)~sin~60.0^{\circ} = M~(0)$
$v_y = -(1.0~m/s)~sin~60.0^{\circ}$
$v_y = -0.866~m/s$
We can find the magnitude of puck B's velocity after the collision:
$\sqrt{v_x^2+v_y^2} = \sqrt{(1.5~m/s)^2+(-0.866~m/s)^2} = 1.73~m/s$
We can find the angle $\theta$ below the +x-axis:
$tan~\theta = \frac{0.866}{1.5}$
$\theta = tan^{-1}(\frac{0.866}{1.5})$
$\theta = 30.0^{\circ}$
After the collision, puck B's speed is $1.73~m/s$ and it moves at an angle of $30.0^{\circ}$ below the +x-axis.
