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Question

Velocity of approach

$=\operatorname{vsin}(	heta / 2)+\operatorname{vsin}(	heta / 2)=2 \operatorname{vsin}(	heta / 2)$

velocity of separation

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

Velocity of approach

$=\operatorname{vsin}(	heta / 2)+\operatorname{vsin}(	heta / 2)=2 \operatorname{vsin}(	heta / 2)$

velocity of separation $=\mathrm{zero}$

$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}ight) v^{\prime}$

$m \operatorname{vcos}(	heta / 2)+m \operatorname{vcos}(	heta / 2)=2 m v^{\prime}$

$\mathrm{v}^{\prime}=\mathrm{v} \cos (	heta / 2)$

Two identical particles are moving with same velocity $v$ as shown in figure. If the collision is completely inelastic then

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