First slide

Collisions

Question

A mass 'm' moves with a velocity 'v' and collides inelastically with another identical mass. After collision the I^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

Moderate

A

$\frac{2}{\sqrt{3}} v$
B

$\frac{v}{\sqrt{3}}$
C

v
D

$\sqrt{3} v$

Solution

Let mass A moves with velocity v and collides inelastically 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 $θ$ with the horizontal with velocity v.
Initial horizontal momentum of system
(before collision) = mv ....(i)
Final horizontal momentum of system
(after collision) = mV cos $θ$ ....(ii)
From the conservation of horizontal linear momentum mv = mV cos $θ$ $\Rightarrow$ v = V cos $θ$ ...(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system $\frac{mv}{\sqrt{3}} - mVsinθ$
From the conservation of vertical linear momentum $\frac{mv}{\sqrt{3}} - mVsinθ = 0 \Rightarrow \frac{v}{\sqrt{3}} = Vsinθ$ ...(iv)
By solving (iii) and (iv)
$v^{2} + \frac{v^{2}}{3} = V^{2} (\sin^{2} θ + \cos^{2} θ)$
$\Rightarrow \frac{4 v^{2}}{3} = V^{2} \Rightarrow V = \frac{2}{\sqrt{3}} v$

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