College Physics (4th Edition)

(a) The nucleus recoils at an angle of $31.4^{\circ}$ above the -x-axis. (b) The momentum of the nucleus is $9.60\times 10^{-19}~kg~m/s$
(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 $9.60\times 10^{-19}~kg~m/s$.