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A mass ' $m$ ' moves with a velocity ' $v$ ' and collides inelastically with another identical mass. After collision the Ist mass moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to the initial direction of motion. Find the speed of the $2^{\text {nd }}$ mass after collision
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The correct answer is:
$\frac{2}{\sqrt{3}} v$
Let mass $A$ moves with velocity $v$ and collides inelasticalfy with mass B, which is at rest.
According to problem mass $A$ moves in a perpendicular direction and let the mass $B$ moves at angle $\theta$ with the horizontal with velocity $v$.
Initial horizontal momentum of system (before collision) $=\mathrm{mv}$
Final horizontal momentum of system (after collision) $\quad=m V \cos \theta$
...(ii)
From the conservation of horizontal linear momentum $\quad m v=m V \cos \theta \Rightarrow V=V \cos \theta$ ....(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system $\frac{m v}{\sqrt{3}}-m V \sin \theta$
From the conservation of vertical linear momentum $\frac{m v}{\sqrt{3}}-m V \sin \theta=0 \Rightarrow \frac{v}{\sqrt{3}}=V \sin \theta$ ....(iv)
By solving (iii) and (iv)
$\begin{aligned}
& v^2+\frac{v^2}{3}=V^2\left(\sin ^2 \theta+\cos ^2 \theta\right) \\
& \Rightarrow \frac{4 v^2}{3}=V^2 \Rightarrow V=\frac{2}{\sqrt{3}} v .
\end{aligned}$
According to problem mass $A$ moves in a perpendicular direction and let the mass $B$ moves at angle $\theta$ with the horizontal with velocity $v$.
Initial horizontal momentum of system (before collision) $=\mathrm{mv}$
Final horizontal momentum of system (after collision) $\quad=m V \cos \theta$
...(ii)
From the conservation of horizontal linear momentum $\quad m v=m V \cos \theta \Rightarrow V=V \cos \theta$ ....(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system $\frac{m v}{\sqrt{3}}-m V \sin \theta$
From the conservation of vertical linear momentum $\frac{m v}{\sqrt{3}}-m V \sin \theta=0 \Rightarrow \frac{v}{\sqrt{3}}=V \sin \theta$ ....(iv)
By solving (iii) and (iv)
$\begin{aligned}
& v^2+\frac{v^2}{3}=V^2\left(\sin ^2 \theta+\cos ^2 \theta\right) \\
& \Rightarrow \frac{4 v^2}{3}=V^2 \Rightarrow V=\frac{2}{\sqrt{3}} v .
\end{aligned}$
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