A particles of mass m with an initial velocity u\hat{i} collides perfectly elastically with a mass 3m at rest. It moves with a velocity v\hat{j} after collision, then, v is given by:
Option: 1 v=\sqrt{\frac{2}{3}}u
 
Option: 2 v=\frac{u}{\sqrt{3}}
Option: 3 v=\frac{u}{\sqrt{2}}  
Option: 4 v=\frac{1}{\sqrt{6}}u

Answers (1)

\begin{aligned} &\text { From momentum conservation }\\ &\overrightarrow{\mathrm{P}}_{\mathrm{i}}=\overrightarrow{\mathrm{P}}_{\mathrm{f}}\\ &\mathrm{m}(\mathrm{ui})+3 \mathrm{~m}(0)=\mathrm{mvj}+3 \mathrm{~m} \vec{\mathrm{v'}}\\ &\mathrm{mui}-\mathrm{mvj}=3 \mathrm{~m} \vec{\mathrm{v'}} \\ & \vec{\mathrm{v'}}=\frac{u i-v j}{3}\\ &\text { or }\left|v'\right|=\frac{\sqrt{u^{2}+v^{2}}}{3}\\ &\text { or } v'^{2}=\frac{u^{2}+v^{2}}{9} \ldots . .(1) \end{aligned}

As collision is perfectly elastic hence

k_{i}=k_{j} \\ \frac{1}{2} m u^{2}+\frac{1}{2} 3 m* 0^{2}=\frac{1}{2} m v^{2}+\frac{1}{2}* 3 m* v'^{2} \\ \Rightarrow \mathrm{u}^{2}=\mathrm{v}^{2}+3 v'^{2}

Using equation (1)

\begin{array}{l} u^{2}=v^{2}+3 \frac{\left(u^{2}+v^{2}\right)}{9} \\ \\ \Rightarrow 3 u^{2}=3 v^{2}+u^{2}+v^{2} \\ \\ \Rightarrow 2 u^{2}=4 v^{2} \\ \\ v=\frac{u}{\sqrt{2}} \end{array}


 

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