A body of mass m sliding along a frictionless surface collides with another body of mass m , which is stationary before impact. The two bodies stick together. If the kinetic energy of the two-body system is E , what is the initial velocity of the first mass before impact?

\sqrt{E / 2 m}

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\sqrt{2 E / 2 m}

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\sqrt{2 E / m}

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\sqrt{E / m}

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2 \sqrt{E / m}

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Solution

The correct option is C $2 \sqrt{E / m}$
$K i n e t i c e n e r g y, E = \frac{1}{2} (2 m) {v_{2}}^{2} T h u s, v_{2} = \sqrt{\frac{E}{m}} N o w, u s i n g l a w o f c o n s e r v a t i o n o f m o m e n t u m, I n i t i a l m o m e n t u m = F i n a l m o m e n t u m m v + m \times 0 = (m + m) v_{2} m v = 2 m v_{2} v = 2 v_{2} H e n c e, i n i t i a l v e l o c i t y, v = 2 \sqrt{\frac{E}{m}} m / s$

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