Answer
Ball A moves at an angle of $26.6^{\circ}$ above the positive x-axis.
Work Step by Step
We can find an expression for the x-component of the velocity of ball A after the collision:
$m_A~v_x = m_B~v$
$v_x = \frac{m_B}{m_A}~v$
We can find an expression for the y-component of the velocity of ball A after the collision:
$m_A~v_y = m_B~\frac{v}{2}$
$v_y = \frac{m_B}{m_A}~\frac{v}{2}$
We can find the direction of ball A:
$tan~\theta = \frac{\frac{m_B}{m_A}~\frac{v}{2}}{\frac{m_B}{m_A}~v}$
$tan~\theta = \frac{1}{2}$
$\theta = tan^{-1}~\frac{1}{2}$
$\theta = 26.6^{\circ}$
Ball A moves at an angle of $26.6^{\circ}$ above the positive x-axis.
