A dumbbell consists of two balls A and B each of mass m $$=$$ $$1$$ kg are 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 (assume$$ zero potential $$energy), the ball B is imparted a horizontal velocity $$v0=87$$ 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. The maximum potential energy stored in the spring during the motion is N joule then $$N=----------$$

Question

A  dumbbell  consists  of  two  balls  A  and  B  each of  mass  m  $$=$$  $$1$$  kg  are  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 (assume$$ zero potential $$energy), the ball B is imparted a horizontal velocity $$v0=87$$ 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.    The  maximum  potential  energy  stored  in  the  spring during the motion  is N  joule then $$N=----------$$

Infinity Learn · Accepted Answer

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 center of mass (COM) frame  $$2mv022$$ℓ$$2=2mv$$ℓ∴  $$v=v042$$ From conservation of energy with respect to center of mass frame $$212mv028+v028$$−$$212mv0232=Uspring$$​∴    $$Uspring=mv0214$$−$$132=7mv0232=2$$  Joule

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