Question

please answer this question

Answer 1

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