A bomb of mass 20 kg at rest explodes into two pieces of masses 12 kg and 8 kg . If the velocity of 8 kg mass is 6 ~ms ⁻¹ , then the kinetic energy of the other mass is:

Step-by-step solution

Let M = 20 kg be the mass of the bomb, m₁ = 12 kg, and m₂ = 8 kg be the masses of the two pieces. Let v₁ and v₂ be the velocities of the two pieces respectively. Initially, the bomb is at rest, so the initial momentum is zero. According to the law of conservation of linear momentum, the final momentum must also be zero: m₁ v₁ + m₂ v₂ = 0 Given m₂ = 8 kg and v₂ = 6 m/s, we have: 12 v₁ + 8 6 = 0 12 v₁ = -48 v₁ = -4 m/s. The kinetic energy of the 12 kg mass is given by: K₁ = 1 2 m₁ v₁^2 K₁ = 1 2 12 (-4)^2 K₁ = 6 16 =