Answer
The final speed of the alpha particle is $~~4.15\times 10^5~m/s$
Work Step by Step
Since the y-component of the initial momentum is zero, the y-component of the final momentum is zero after the collision.
After the collision, the y-component of the alpha particle's momentum is equal in magnitude to the y-component of the oxygen nucleus' momentum.
We can find the final speed of the alpha particle:
$m_1~v_{1f}~sin~64.0^{\circ} = m_2~v_{2f}~sin~51.0^{\circ}$
$v_{1f} = \frac{m_2~v_{2f}~sin~51.0^{\circ}}{m_1~sin~64.0^{\circ}}$
$v_{1f} = \frac{(16.0~u)~(1.20\times 10^5~m/s)~sin~51.0^{\circ}}{(4.00~u)~sin~64.0^{\circ}}$
$v_{1f} = 4.15\times 10^5~m/s$
The final speed of the alpha particle is $~~4.15\times 10^5~m/s$
