MCQ
Two identical particles are moving with same velocity $v$  as shown in figure. If the collision is completely inelastic then
  • A
    The velocity of separation is zero
  • B
    The velocity of approach is  $2\,v\,sin(\theta /2)$
  • C
    The common velocity after collision is $v\,cos(\theta /2)$
  • All of the above

Answer

Correct option: D.
All of the above
d
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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