A nucleus of mass M+Δm is at rest and decays into two daughter nuclei of equal mass of M/2 each. The speed of light is c. Then speed of daughter nuclei is?

Conserving the momentum: 0 = (M/2)v1 - (M/2)v2
Since the two daughter nuclei have equal mass, their speeds must be equal and opposite to conserve momentum. Let v be the speed of each daughter nucleus. Then, 0 = (M/2)v - (M/2)v = 0. This equation doesn't help us find the speed. Let's conserve energy instead.

The initial energy is the rest energy of the parent nucleus: E_initial = (M + Δm)c²

The final energy is the sum of the kinetic and rest energies of the two daughter nuclei:
E_final = 2 * [(1/2)(M/2)v² + (M/2)c²]

By conservation of energy, E_initial = E_final:
(M + Δm)c² = 2 * [(1/2)(M/2)v² + (M/2)c²]
(M + Δm)c² = (M/2)v² + Mc²
(M + Δm)c² - Mc² = (M/2)v²
Δmc² = (M/2)v²
v² = (2Δmc²/M)
v = c√(2Δm/M)
Therefore, the speed of each daughter nucleus is c√(2Δm/M).