A mass $'m'$ moves with a velocity $'v'$ and collides inelastically with another identical mass. After collision the $1^{st}$ mass moves with velocity $\frac{v}{{\sqrt 3 }}$ in a direction perpendicular to the initial direction of motion. Find the speed of the $ 2^{nd}$ mass after collision
A mass $'m'$ moves with a velocity $'v'$ and collides inelastically with another identical mass. After collision the $1^{st}$ mass moves with velocity $\frac{v}{{\sqrt 3 }}$ in a direction perpendicular to the initial direction of motion. Find the speed of the $ 2^{nd}$ mass after collision
$\frac{2}{{\sqrt 3 }}v$
$\frac{v}{{\sqrt 3 }}$
$v$
$\sqrt 3 \,v$
A mass $'m'$ moves with a velocity $'v'$ and collides inelastically with another identical mass. After collision the $1^{st}$ mass moves with velocity $\frac{v}{{\sqrt 3 }}$ in a direction perpendicular to the initial direction of motion. Find the speed of the $ 2^{nd}$ mass after collision
Let mass $A$ moves with velocity $v$ and collides inelastically with mass $B$, which is
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) $= mv$ $....(i)$
Final horizontal momentum of system
(after collision) $= mV\, cos$ $\theta$$ ....(ii)$
From the conservation of horizontal linear momentum $mv = mV $ $cos$$\theta$$⇒$ $ v = V cos$$\theta$ $...(iii)$
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system $\frac{{mv}}{{\sqrt 3 }} - mV\sin \theta $
From the conservation of vertical linear momentum $\frac{{mv}}{{\sqrt 3 }} - mV\sin \theta = 0$==>$\frac{v}{{\sqrt 3 }} = V\sin \theta $$...(iv)$
By solving $(iii)$ and $(iv)$
${v^2} + \frac{{{v^2}}}{3} = {V^2}({\sin ^2}\theta + {\cos ^2}\theta )$
$⇒$ $\frac{{4{v^2}}}{3} = {V^2}$ $⇒$$V = \frac{2}{{\sqrt 3 }}v$.
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