Answer
$7.65\space Mm/s$
$Direction =12.46^{\circ}$ to the alpha particle's direction
Work Step by Step
Please see the attached image first.
Here we use the principle of conservation of momentum.
$\vec P= Constant$
We can write,
$m_{Li}\vec V_{Li}= m_{P}\vec V_{P}+m_{\alpha}\vec V_{\alpha}$
Let's choose the x-axis along the direction of $\vec V_{Li}.$Then the two components of the momentum conservation equation become,
x component : $m_{Li}V_{Li}= m_{P}V_{Px}+m_{\alpha}V_{\alpha x}$
y component : $0= m_{P}V_{Py}+m_{\alpha}V_{\alpha y}$
$V_{px}= \frac{m_{Li}V_{Li}-m_{\alpha}V_{\alpha x}}{m_{p}}= \frac{m_{Li}V_{Li}-m_{\alpha}V_{\alpha}cos\theta}{m_{p}} $
Let's plug known values into this equation.
$V_{px}=\frac{5U\times2.25\space Mm/s-4U\times1.03\space Mm/s\times cos(23.6^{\circ})}{U}$
$V_{px}=7.47\space Mm/s$
$V_{py}= -\frac{m_{\alpha}V_{\alpha y}}{m_{p}}= -\frac{m_{\alpha}V_{\alpha }sin\theta}{m_{p}}$
$V_{py}=\frac{-4U\times1.03\space Mm/s\times sin(23.6^{\circ})}{U}=1.65\space Mm/s$
$V_{p}=\sqrt {V_{px}^{2}+V_{py}^{2}}= \sqrt {7.47^{2}+1.65^{2}}Mm/s$
$V_{p}=7.65\space Mm/s$
$tan\alpha=\frac{V_{py}}{V_{px}}=\frac{1.65\space Mm/s}{7.47\space Mm/s}$
$\alpha=12.46^{\circ}$
