Answer
The original speed of the cue ball is $~~4.75~m/s$
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.
In part (a), we found that the angle between the direction of motion of the second ball and the original direction of motion of the cue ball is $~~41.0^{\circ}$
We can consider the conservation of momentum in the x-direction to find the original speed $v_i$ of the cue ball:
$p_{xi} = p_{xf}$
$M~v_i = (M)(3.50~m/s)~cos~22.0^{\circ} + (M)(2.00~m/s)~cos~41.0^{\circ}$
$v_i = (3.50~m/s)~cos~22.0^{\circ} + (2.00~m/s)~cos~41.0^{\circ}$
$v_i = 4.75~m/s$
The original speed of the cue ball is $~~4.75~m/s$
