Question
Two identical particles are moving with same velocity $v$ as shown in figure. If the collision is completely inelastic then
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Answer
Velocity of approach
$=\operatorname{vsin}(\theta / 2)+\operatorname{vsin}(\theta / 2)=2 \operatorname{vsin}(\theta / 2)$
velocity of separation $=\mathrm{zero}$
$\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v}^{\prime}$
$m \operatorname{vcos}(\theta / 2)+m \operatorname{vcos}(\theta / 2)=2 m v^{\prime}$
$\mathrm{v}^{\prime}=\mathrm{v} \cos (\theta / 2)$
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