Answer
The initial speed of the alpha particle is $~~4.84\times 10^5~m/s$
Work Step by Step
In part (a), we found that the final speed of the alpha particle is $~~4.15\times 10^5~m/s$
The alpha particle's initial momentum is equal to the sum of the x-component of the alpha particle's momentum and the x-component of the oxygen nucleus' momentum after the collision.
We can find the initial speed of the alpha particle:
$m_1~v_{1,0} = m_1~v_{1f}~cos~64.0^{\circ} + m_2~v_{2f}~cos~51.0^{\circ}$
$v_{1,0} = \frac{m_1~v_{1f}~cos~64.0^{\circ} + m_2~v_{2f}~cos~51.0^{\circ}}{m_1}$
$v_{1,0} = \frac{(4.00u)~(4.15\times 10^5~m/s)~cos~64.0^{\circ} + (16.0u)~(1.20\times 10^5~m/s)~cos~51.0^{\circ}}{4.00u}$
$v_{1,0} = 4.84\times 10^5~m/s$
The initial speed of the alpha particle is $~~4.84\times 10^5~m/s$
