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

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the energy of both bodies will be same

energy of smaller body will be 10J

energy of smaller body will be 20J

energy of smaller body will be 40 J

answer is D.

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Detailed Solution

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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