A body of mass m moving along a straight line | vedclass

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(B) Let the initial velocity of the body of mass $m$ be $v$. The initial kinetic energy is $K_i = \frac{1}{2}mv^2$.By the law of conservation of linea

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$2/3$

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(B) Let the initial velocity of the body of mass $m$ be $v$. The initial kinetic energy is $K_i = \frac{1}{2}mv^2$.By the law of conservation of linear momentum: $mv + (2m)(0) = (m + 2m)v'$,where $v'$ is the final common velocity.$mv = 3mv' \implies v' = v/3$.The final kinetic energy is $K_f = \frac{1}{2}(m + 2m)(v')^2 = \frac{1}{2}(3m)(v/3)^2 = \frac{1}{2}(3m)(v^2/9) = \frac{1}{6}mv^2$.The loss in kinetic energy is $\Delta K = K_i - K_f = \frac{1}{2}mv^2 - \frac{1}{6}mv^2 = \frac{3-1}{6}mv^2 = \frac{2}{6}mv^2 = \frac{1}{3}mv^2$.The fraction of kinetic energy lost is $\frac{\Delta K}{K_i} = \frac{\frac{1}{3}mv^2}{\frac{1}{2}mv^2} = \frac{1}{3} 	imes 2 = 2/3$.

$A$ body of mass $m$ moving along a straight line collides with a stationary body of mass $2m$. After collision,if the two bodies move together with the same velocity,then the fraction of kinetic energy lost in the process is

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