Answer
After the collision, the velocity of the helium atom is $266~m/s$ to the right.
Work Step by Step
We can use conservation of momentum to find the helium atom's velocity after the collision:
$m_h~v_{h2}+m_o~v_{o2} = m_h~v_{h1}+m_o~v_{o1}$
$m_h~v_{h2} = m_h~v_{h1}+m_o~v_{o1}-m_o~v_{o2}$
$v_{h2} = \frac{m_h~v_{h1}+m_o~v_{o1}-m_o~v_{o2}}{m_h}$
$v_{h2} = \frac{(4.00~u)(618~m/s)+(32.0~u)(412~m/s)-(32.0~u)(456~m/s)}{4.00~u}$
$v_{h2} = 266~m/s$
After the collision, the velocity of the helium atom is $266~m/s$ to the right.
