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Two masses m and 2m are attached to two ends of an ideal spring as' shown in figure. When the spring isin the compressed state, the energy of the spring is 60J, if the spring is released, then at it:- natural length, [BHU 2012]

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a

energy of smaller body will be 20 ]

b

energy of smaller body will be 40 ]

c

energy of smaller body will be 10 J

d

energy of both the bodies will be same

answer is B.

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Detailed Solution

According to law of conservation of momentum,mv1+2mv2=0 or v2=-v12 …(i)According to law of conservation of energy,12mv12+122mv22=60⇒ 12mv12+m-v122=60⇒ 12mv12+14mv12=60⇒ 3mv124=60⇒ 12mv12=40 J

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Two masses m and 2m are attached to two ends of an ideal spring as' shown in figure. When the spring isin the compressed state, the energy of the spring is 60J, if the spring is released, then at it:- natural length, [BHU 2012]