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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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Correct option is D
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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A simple pendulum is swinging in a vertical plane. The ratio of its potential energies when it is making angles and with the vertical is,
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