A bullet of mass 10 ~g pierces through a plate A of mass 500 ~g and then gets embedded into a second plate B of mass 1.49 ~kg as shown in the figure. Initially, the two plates A and B are at rest and move with same velocity after collision. The percentage loss in the initial kinetic energy of the bullet, when it is between the plates A and B is _______ (Neglect any loss of material of the plates during the collision)

Step-by-step solution

Let v₁ be the initial velocity of bullet and v₂ be the velocity with which each plate moves. So, applying law of conservation of momentum, aligned & m v₁=M₁ v₂+ (M₂+m ) v₂ & 0.01 v₁=0.5 v₂+(1.49+0.01) v₂ aligned Let v₃ be the velocity of bullet, when it comes out of plate M . So, momentum of bullet between plate M₁ and M₂ = Sum of momentum of plate M₂ and bullet. aligned m v₃ & = (M₂+m ) v₂ 0.01 v₃ & =(1.49+0.01) v₂=1.5 v₂ aligned aligned % loss in KE & = (1 / 2) m v₁^2-(1 / 2) m v₃^3 (1 / 2) m v₁^2 100 & = v₁^2-v₃