Two bodies of masses m and 2m are attached to | vedclass

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(D) Since the system is initially at rest,the momentum of the system is zero. By the law of conservation of linear momentum,the magnitudes of the mome

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The smaller body will have kinetic energy of $40 \ J$.

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(D) Since the system is initially at rest,the momentum of the system is zero. By the law of conservation of linear momentum,the magnitudes of the momenta of the two bodies must be equal at all times: $p_1 = p_2 = p$. Kinetic energy $K$ is related to momentum $p$ by $K = p^2 / (2m)$. Thus,$K_1 = p^2 / (2m)$ and $K_2 = p^2 / (2(2m)) = p^2 / (4m) = K_1 / 2$. Total energy $E = K_1 + K_2 = K_1 + K_1 / 2 = (3/2) K_1 = 60 \ J$. Solving for $K_1$: $K_1 = (2/3) 	imes 60 = 40 \ J$. Since $K_1$ corresponds to the mass $m$ (the smaller body),the smaller body has kinetic energy $40 \ J$.

Two bodies of masses $m$ and $2m$ are attached to the two ends of an ideal spring. The spring is compressed. The total energy stored in the spring is $60 \ J$. If the spring is released,then:

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