Question

A bomb of mass 60 kg at rest explodes into two pieces of mass of 1 kg and 20 kg the velocity of the

Answer 1

Pi = Pf (conservation of linear momentum)

16×0 = 12×4 + 4×v

V = 12 m/s

K.E = 1/2 m v^2

= 1/2 × 4 × 12^2

= 288J

Since initially the bomb was at rest so its momentum(mv)= 0 and no external force is applied so we can use momentum conservationLet speed of 4kg particle is v m/s0= 12×4 + 4×vV=12 m/sSo its kinetic energy= 1÷2 ×4×12= 24 J