A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms-1. The kinetic energy of the other mass is ?

According to me the answer is 486 J.By conservation of linear momentumM1v1=M2v"2v"=9m/sUsing this velocity we can calculate the kinetic energy of 12 kg mass which is =1/2*(12)(9)(9).

Hello everyone, From the given datam1v1 = m2v2KE1=1/2(m2 v22 )=1/2(4)(144)= 288JSo the kinetic energy of other mass is 288.This is the perfect answer.!

momentum before explosion ismv=0(since bomb is at rest)______1where m=16kgmomem. after explosion ism1v1+m2v2 _______2where m1=12kg,v1=4m/s m2=4kg,v2=?now by law of conservation of linear momentum the momentum before and after explosion will be same therefore from 1 and 2 we havem1v1+m2v2=mvm1v1+m2v2=0m1v1=-m2v212*4=4*v2(v2 is the velocity of mass fragment whose KE is to cal.)v2=12m/sthus kinetic energy of other part of mass isKE=1/2m2v2^2KE=1/2*4*144KE=288J

By conservation of linear momentum as in Hz dir the net ext force is zero 12×4=4×vV=12 so K.E=1/2×4×(12)^2=288

By conservation of momentum,We know,m1v1=m2v2Given.M1=12.V1=4M2=4V2=x(let)So.....m1v1=m2v2 or,12*4=4*x or,x=V2=12Now...KE=0.5*m2*(V2)² =0.5*4*12² =0.5*4*144 =288