Question

A 1.2-kg spring-activated toy bomb slides on a smooth surface along the x-axis with a speed of 0.50 m/s. At the origin O, the bomb explodes into two fragments. Fragment 1 has a mass of 0.40 kg and a speed of 0.90 m/s along the negative y-axis. у у V2 0.80 kg 0.50 m/s o 0 х 0 х 1.2 kg Before After 0.40 kg 0.90 m/s

a.) Find the angle θ, made by the velocity vector of fragment 2 and the x-axis.
b.) What is the energy released by the explosion?

Answer 1

This is the question of conservation of momentum.

Initial horizontal momentum oro mu=1.2x0.5 = 0.6 kg m/ Final horizontal momentum = M₂ Vq coso conserving momentum Mave cos o 0.6 m2 0-8 8 kg

Conserve vering vertical momentum m₂ V₂ sino - 0.4x0.9 M₂ V₂ sin o=0.36 0.6 .: tano = 0.36 0.6 o = tan-'(0.6)= 30.96 0.6 Jr 0.8 x cos (30.96) V2 = 0.875 m/s

Initial energy I mu² = 0.15 J 11 2 Final energy = 1 m2 v 2 + I mive 0.4x(0og)? Х 1) 0.8 x(0.875)2+ 2 0.46825 J released - 0.31825 J energy released