Answer
After the collision, the direction of the velocity of ball 1 is at an angle of $30^{\circ}$ below the positive x axis.
Work Step by Step
Let $M$ be the mass of each ball.
We can use conservation of momentum to find the horizontal component of ball 1's velocity $v_x$ after the collision:
$p_{xf} = p_{xi}$
$(M)(1.1~m/s)~cos~60^{\circ}+M~v_{x} = (M)~(2.2~m/s)$
$(1.1~m/s)~cos~60^{\circ}+v_{x} = 2.2~m/s$
$v_x = (2.2~m/s)-(1.1~m/s)~cos~60^{\circ}$
$v_x = 1.65~m/s$
We can use conservation of momentum to find the vertical component of ball 1's velocity $v_y$ after the collision:
$p_{yf} = p_{yi}$
$(M)(1.1~m/s)~sin~60^{\circ}+M~v_y = 0$
$M~v_y = -(M)(1.1~m/s)~sin~60^{\circ}$
$v_y = -(1.1~m/s)~sin~60^{\circ}$
$v_y = -0.95~m/s$
We can find the direction of the velocity of ball 1 as an angle $\theta$ below the positive x axis:
$tan~\theta = \frac{0.95~m/s}{1.65~m/s}$
$\theta = tan^{-1}~(\frac{0.95~m/s}{1.65~m/s})$
$\theta = tan^{-1}~(0.576)$
$\theta = 30^{\circ}$
After the collision, the direction of the velocity of ball 1 is at an angle of $30^{\circ}$ below the positive x axis.
