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A bomb of mass 9 kg explodes into 2 pieces of masses 3 kg and 6 kg .The velocity of mass 3 kg is 16 m/s.The kinetic energy of mass 6 kg in joule is what? A bomb of mass 9 kg explodes into 2 pieces of masses 3 kg and 6 kg .The velocity of mass 3 kg is 16 m/s.The kinetic energy of mass 6 kg in joule is what?
given mass of the bomb ,M=9kg initial value of the bomb=0m/s mass of smaller fragmet =3kg=m1 (let) velocity of the smaller part just after explosion=16m/s mass of the bigger fragmet =6kg=m2 (let) let, velocity of the bigger part just after explosion= v as there is no external force is acting on the bomb intially so the momentum is conserved in all direction i.e P(just after explosion)=P(just before explosion ) m1*16 + m2*v = M*0 => v = -8m/s so the kinetic energy of the fragment of mass 6kg = (1/2)*m2*v*v Joule =(1/2)*6*64 =192 Joule
Dear Student,Please find below the solution to your problem.Given, m1=3kg , m2 = 6kg , v1 = 16m/s , K.E 2 = ?According to law of conservation of momentum :(m1 × v1) + (m2 × v2) = (m1 + m2) ×v(3×16) + (6×(-v2)) = (3+6) × 0( -v2 because both are moving in opposite directions48 - (6 × v2) = 048 = 6 ×v2therefore, v2 = 8m/sK.E2 = 1/2 × m2 × v2 square= 1/2 ×6 x 8 square= 3 × 64= 192JThanks and Regards
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