Answer
$v_2=.693cos30\hat{i}+.693sin30\hat{j}$
$v_3=.693cos30\hat{i}-.693sin30\hat{j}$
$v_1==.386cos30\hat{i}$
Work Step by Step
We know that the angle that the balls will all go at will be 30 degrees. We also know that momentum is conserved. Thus, we know from the fact that the y momentum is conserved that the two balls that are struck will be moving with equal speeds. We also know:
$mv_0=mv_f + 2mv{23}$
$v_0=v_f + 2v_{23}cos30^{\circ}$
$v_f=v_0-1.73v_{23}$
We also know from conservation of kinetic energy that:
$\frac{1}{2}v_0^2 = \frac{1}{2}v_f^2+ v_{23}^2$
Thus, we find:
$\frac{1}{2}v_0^2 = \frac{1}{2}(v_0-1.73v_{23})^2+ v_{23}^2$
$v_{23}=.693v_0$
Thus, we can find the velocities:
$v_2=.693cos30\hat{i}+.693sin30\hat{j}$
$v_3=.693cos30\hat{i}-.693sin30\hat{j}$
$v_1=1.386cos(30)\hat{i}-1\hat{i}=.386cos30\hat{i}$
