A dumbbell consists of two balls `A` and `B` of mass `m=1` kg and connected by a spring. The whole system is placed on a smooth horizontal surface as shown in the figure. Initially the spring is at its natural length, the ball `B` is imparted a velocity `v_(0)=8/(sqrt(7))m//s` in the direction shown. The spring constant of the spring is adjusted so that the length of the spring at maximum elongation is twice that of the natural, length of the spring. Find the maximum potential energy stored (in Joule) in the spring during the motion.

Answer

2

Explanation

At the starting (with respect to COM frame)
At the instant of maximum elongation (with respect to the COM frame)

Conservation of angular momentum with respect to centre of mass (COM) frame
`2m (v_(0))/(2sqrt(2))=2mvl`
`:.v=(v_(0))/(4sqrt(2))`
From conservation of energy with respect to centre of mass frame
`2 1/2 ((v_(0)^(2))/8+(v_(0)^(2))/8)-2 12 m (v_(0)^(2))/32=U_("spring")`
`:. U_("spring")=mv_(0)^(2)(1/4-1/32)=(7mv_(0)^(2))/32=2` Joule