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Mechanical energy conservation

Two masses m and 2m are attached to two ends of an ideal spring and the spring is in the compressed state. The energy of spring is 60 joule. If the spring is released, then 


Fext =0 From momentum conservation  P1+P2=0 m1v1+(-m2v2)=0 m1m2=v2v1&v2v1=12 k1k2=m1v12m2v22=12212=2 Now k1+k2=60 k1+k12=60  =k1=40 J

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