A dumbbell consists of two balls A and B each 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 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 (= 1 meter) of the spring. Find the spring constant (in N/m) of spring.

Question

A dumbbell consists of two balls A and B each 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  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 (= 1 meter) of the spring. Find the spring constant (in N/m) of spring.

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 centre of mass (COM) frame2 mv022l2=2mvl v=v042 From conservation of energy with respect to centre of mass frame212 m(v028+v028)−212 mv0232=Uspring ∴Uspring =mv02(14−132)=7mv0232=2 Joule ⇒12K(l)2=2⇒K=4N/m

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