Two identical particles are moving with same velocity v as shown in figure. If the collis

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5.Work, Energy, Power and Collision

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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 $v \cos\,(\theta /2)$

D

All of the above

Solution

Velocity of approach

$=\operatorname{vsin}(\theta / 2)+\operatorname{vsin}(\theta / 2)=2 \operatorname{vsin}(\theta / 2)$

velocity of separation $=\mathrm{zero}$

$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v^{\prime}$

$m \operatorname{vcos}(\theta / 2)+m \operatorname{vcos}(\theta / 2)=2 m v^{\prime}$

$\mathrm{v}^{\prime}=\mathrm{v} \cos (\theta / 2)$

Standard 11

Physics

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