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A clay ball of mass $\mathrm{m}$ and speed $\mathrm{v}$ strikes another metal ball. of same mass $\mathrm{m}$, which is at rest. They stick together after collision. The kinetic energy of the system after collision is -
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The correct answer is:
$\mathrm{mv}^{2} / 4$
Applying the law of conservation of momentum, $\mathrm{mv}+0=(2 \mathrm{~m}) \mathrm{v}^{\prime}$
$\mathrm{v}^{\prime}=\mathrm{v} / 2$
$\mathrm{~K} . \mathrm{E}=\frac{1}{2}(2 \mathrm{~m}) \mathrm{v}^{\prime 2}$
$\mathrm{v}^{\prime}=\mathrm{v} / 2$
$\mathrm{~K} . \mathrm{E}=\frac{1}{2}(2 \mathrm{~m}) \mathrm{v}^{\prime 2}$
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