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Question

By momentum conservation along $y$

$m _{1} u _{1} \sin 	heta_{1}= m _{2} u _{2} \sin 	heta_{2}$

i.e. $mu _{1} \sin 	heta_{1}=10 mu _{2} \sin

Vedclass · Accepted Answer

$10$

Vedclass · Answer

By momentum conservation along $y$

$m _{1} u _{1} \sin 	heta_{1}= m _{2} u _{2} \sin 	heta_{2}$

i.e. $mu _{1} \sin 	heta_{1}=10 mu _{2} \sin 	heta_{2}$

$\Rightarrow u _{1} \sin 	heta_{1}=10 u _{2} \sin 	heta_{2}$ $...(i)$

$kf _{ m _{ 1 }}=\frac{1}{2} ki _{ m _{ 1 }}$ i.e. $\frac{1}{2} mu _{1}^{2}=\frac{1}{2} 	imes \frac{1}{2} mu ^{2}$

i.e. $\sqrt{ u _{1}=\frac{ u }{\sqrt{2}}}$ $...(ii)$

Also collision is elastic : $k _{ i }= k _{ f }$

$\frac{1}{2} m u^{2}=\frac{1}{2} m u_{1}^{2}+\frac{1}{2} \cdot 10 m \cdot u_{2}^{2}$

$\frac{1}{2} mu ^{2}=\frac{1}{2} 	imes \frac{1}{2} mu ^{2}+\frac{1}{2} 	imes 10 m \cdot u _{2}^{2}$

$\frac{1}{4} mu ^{2}=\frac{1}{2} 	imes 10 	imes mu _{2}^{2}$

$u _{2}=\frac{ u }{\sqrt{20}}$$...(iii)$

Putting $(ii)$ $\&$ $(iii)$ in $(i)$

$\frac{ u }{\sqrt{2}} \sin 	heta_{1}=10 \cdot \frac{ u }{\sqrt{20}} \sin 	heta_{2}$

$\sin 	heta_{1}=\sqrt{10} \sin 	heta_{2} \quad ightarrow$ Hence $n =10$

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