A bomb initially at rest explodes into three pieces. Two pieces, each of mass m , fly off at right angles to each other with the same speed v . If the total kinetic energy generated in the explosion is 3 2 mv^2 , what was the total original mass of the bomb?

Step-by-step solution

Let the mass of the third fragment be m₃ and its velocity be v ₃ . The first two fragments have mass m and speed v , moving at right angles. Their momenta are m v i and m v j . By conservation of linear momentum: p _ initial = p _ final 0 = m v i + m v j + m₃ v ₃ m₃ v ₃ = -m v i - m v j The magnitude of the momentum of the third fragment is: p₃ = (mv)^2 + (mv)^2 = 2 mv The kinetic energy of the third fragment is: K₃ = p₃^2 2m₃ = ( 2 mv)^2 2m₃ = m^2v^2 m₃ The total kinetic energy released is the sum of the kinetic e