Answer
The angle between the direction of motion of the second ball and the original direction of motion of the cue ball is $~~41.0^{\circ}$
Work Step by Step
Let $M$ be the mass of each ball.
Let the original direction of motion be along the positive x axis. Let's assume the cue ball moves at an angle of $22.0^{\circ}$ above the positive x axis after the collision.
We can consider the conservation of momentum in the y-direction to find the angle $\theta$ below the positive x axis of the second ball's motion after the collision:
$p_{yf} = p_{yi}$
$(M)(3.50~m/s)~sin~22.0^{\circ} - (M)(2.00~m/s)~sin~\theta = 0$
$(M)(3.50~m/s)~sin~22.0^{\circ} = (M)(2.00~m/s)~sin~\theta$
$sin~\theta = \frac{(3.50~m/s)~sin~22.0^{\circ}}{2.00~m/s}$
$sin~\theta = 0.65556$
$\theta = sin^{-1}~(0.65556)$
$\theta = 41.0^{\circ}$
The angle between the direction of motion of the second ball and the original direction of motion of the cue ball is $~~41.0^{\circ}$
