When a bomb explodes, the total momentum of the system before the explosion is equal to the total momentum of the system after the explosion. This is known as the conservation of momentum.

Total momentum before explosion = Total momentum after explosion

P(before) = P(after)



Before the explosion, the bomb was a single object with a mass of 12 kg. Therefore, the total momentum of the bomb before the explosion is given by:

P(before) = m(before) * v(before)

where,

m(before) = 12 kg (mass of the bomb before the explosion)

v(before) = 0 m/s (velocity of the bomb before the explosion)

P(before) = 12 kg * 0 m/s = 0 kg.m/s



After the explosion, the bomb breaks into two pieces with masses of 4 kg and 8 kg respectively. Let the velocity of the 8 kg mass be v(after).

The total momentum of the system after the explosion is given by:

P(after) = m(1) * v(1) + m(2) * v(2)

where,

m(1) = 4 kg (mass of the 4 kg mass after the explosion)

v(1) = 20 m/s (velocity of the 4 kg mass after the explosion)

m(2) = 8 kg (mass of the 8 kg mass after the explosion)

v(2) = v(after) (velocity of the 8 kg mass after the explosion)

P(after) = 4 kg * 20 m/s + 8 kg * v(after)

P(after) = 80 kg.m/s + 8 kg.v(after)



As per the conservation of momentum:

P(before) = P(after)

0 kg.m/s = 80 kg.m/s + 8 kg.v(after)

Solving for v(after):

-80 kg.m/s = 8 kg.v(after)

v(after) = -10 m/s

Therefore, the velocity of the 8 kg mass after the explosion is -10 m/s. The negative sign indicates that the mass is moving in the opposite direction to the 4 kg mass.