Answer
(a) The nucleus recoils at an angle of $31.4^{\circ}$ above the -x-axis.
(b) The momentum of the nucleus is $9.60\times 10^{-19}~kg~m/s$
Work Step by Step
(a) We can assume that the electron moves away in the +x-direction and the neutrino moves away in the -y-direction.
We can assume that the initial momentum is zero. By conservation of momentum, the final momentum of the system is also zero.
We can find the horizontal component $p_x$ of the nucleus' final momentum:
$p_x + 8.20\times 10^{-19}~kg~m/s = 0$
$p_x = -8.20\times 10^{-19}~kg~m/s$
We can find the vertical component $p_y$ of the nucleus' final momentum:
$p_y - 5.00\times 10^{-19}~kg~m/s = 0$
$p_y = 5.00\times 10^{-19}~kg~m/s$
We can find the angle $\theta$ above the -x-axis that the nucleus recoils:
$tan~\theta = \frac{5.00\times 10^{-19}}{8.20\times 10^{-19}}$
$\theta = tan^{-1}(\frac{5.00\times 10^{-19}}{8.20\times 10^{-19}})$
$\theta = 31.4^{\circ}$
The nucleus recoils at an angle of $31.4^{\circ}$ above the -x-axis.
(b) We can find the magnitude of the momentum of the nucleus:
$p = \sqrt{p_x^2+p_y^2}$
$p = \sqrt{(-8.20\times 10^{-19}~kg~m/s)^2+(5.00\times 10^{-19}~kg~m/s)^2}$
$p = 9.60\times 10^{-19}~kg~m/s$
The momentum of the nucleus is $9.60\times 10^{-19}~kg~m/s$.
