Question

A bomb of mass 12 kg at rest explodes into two pieces of masses in the ratio 1:3. If kinetic energy of smaller piece is 216 j, the momentum of larger piece will be

Answer 1

let the total mass of the bomb be "4m"

so 4 m = 12 kg

m = 12/4

m = 3 kg

since the ratio of pieces of masses is 1 :3

m₁ = mass of smaller piece = m = 3 kg

m₂ = mass of larger piece = 3 m = 3 x 3 = 9 kg

v₁ = velocity of smaller piece

v₂ = velocity of larger piece

initially the bomb was at rest , hence initial total momentum = 0

Using conservation of momentum

Final total momentum = initial total momentum

m₁ v₁ + m₂ v₂ = 0

m v₁ + (3 m) v₂ = 0

v₁ + 3 v₂ = 0

v₁ = - 3 v₂                                    eq-1

Given that : initial kinetic energy = 216

we know that : kinetic energy = (0.5) mv²  where m = mass , v = speed

so

(0.5) m₁ v₁² = 216

(3) v₁² = 432

v₁ = 12 m/s

using eq-1

v₁ = - 3 v₂

12 = - 3 v₂

v₂ = - 4 m/s

Momentum of the larger piece is given as

P₂ = m₂ v₂ = 9 (-4) = - 36 kgm/s