Question

The nucleus of an atom has a mass of 3.8x10^-25 kg and is at rest. The nucleus is radioactive and suddenly ejects a particle of mass 6.6x10^-27 kg at a speed of 1.5x10^7 m/s. Find the recoil speed of the nucleus that is left behind.

Answer 1

The law of conservation of the momentum says

$$0=Mv-mu$$

Hence, the recoil speed of the nucleus that is left behind

$$v=\frac{mu}{M}=\frac{6.6\times 10^{-27}\times 1.5\times 10^7}{3.8\times 10^{-25}}=2.6\times 10^5\:\rm m/s$$