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Verified answer
Momentum of system is conserved.Momentum of bomb before explosion = Momentum after explosion
0 = P₁ + P₂ + P₃
P₃ = -(P₁ + P₂)
From above we can say that,
Vector addition of momentum of two perpendicular fragments is equal to momentum of third fragment
P₃ = -√[P²₁ + P²₂]
= -√[(2 × 12)² + (8 × 1)²]
= -25.3
m₃ = P₃ / v₃
= -(25.3 kgm/s) / (-20 m/s)
= 1.26 kg
Mass of third fragment is 1.26 kg