Two bodies A and B of mass m and 2 m respectively are placed on a smooth floor. They are connected by a spring of negligible mass. A third body C of mass m is placed on the floor. The body C moves with a velocity v₀ along the line joining A and B and collides elastically with A . At a certain time after the collision it is found that the instantaneous velocities of A and B are same and the compression of the spring i

Step-by-step solution

Initial momentum of the system block (C)=m v₀ . After striking with A , the block C comes to rest and now both block A and B moves with velocity v when compression in spring is x₀ . By the law of conservation of linear momentum m v₀=(m+2 m) v v= v₀ 3 By the law of conservation of energy K.E. of block C= K.E. of system + P.E. of system aligned & 1 2 m v₀^2= 1 2 (3 m) ( v₀ 3 )^2+ 1 2 k x₀^2 & 1 2 m v₀^2= 1 6 m v₀^2+ 1 2 k x₀^2 & 1 2 k x₀^2= 1 2 m v₀^2- 1 6 m v₀^2= m v₀^2 3 & aligned aligned & 1 2 k x₀^2= 1 2 m v₀^2-