# Chapter 9 - Impulse and Momentum - Exercises and Problems - Page 243: 65

The recoil speed of the nucleus is 1980 m/s

#### Work Step by Step

The initial momentum of the system is zero. By conservation of momentum, the final momentum of the system is also zero. Let's assume that the electron moves away along the positive x-axis. We can find $v_x$, the horizontal component of the recoil velocity of the nucleus. $(2.34\times 10^{-26}~kg)~v_x+(9.11\times 10^{-31}~kg)(5.00\times 10^7~m/s) = 0$ $(2.34\times 10^{-26}~kg)~v_x = -(9.11\times 10^{-31}~kg)(5.00\times 10^7~m/s)$ $v_x = \frac{-(9.11\times 10^{-31}~kg)(5.00\times 10^7~m/s)}{(2.34\times 10^{-26}~kg)}$ $v_x = -1947~m/s$ Let's assume that the neutrino moves away along the positive y-axis. We can find $v_y$, the vertical component of the recoil velocity of the nucleus. $(2.34\times 10^{-26}~kg)~v_y+(8.00\times 10^{-24}~kg~m/s) = 0$ $(2.34\times 10^{-26}~kg)~v_y = -(8.00\times 10^{-24}~kg~m/s)$ $v_y = \frac{-(8.00\times 10^{-24}~kg~m/s)}{(2.34\times 10^{-26}~kg)}$ $v_y = -341.9~m/s$ We can find the recoil speed of the nucleus. $v = \sqrt{(v_x)^2+(v_y)^2}$ $v = \sqrt{(-1947~m/s)^2+(-341.9~m/s)^2}$ $v = 1980~m/s$ The recoil speed of the nucleus is 1980 m/s.

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