A mass .m. moves with a velocity .v. and collides inelastically with another identical mass. After collision the 1st mass moves with velocity `(v)/(sqrt(3))` in a direction perpendicular to the initial direction of motion. Find the speed of the `2^(nd)` mass after collision

Options

A. `(2)/(sqrt(3))`
B. `(v)/(sqrt(3))`
C. v
D. `sqrt(3)v`

Answer

A

Explanation

Let mass A moves with velocity v and collides inelastically with mass B, which is at rest.

According to problem mass A moves in 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 rArr v= V cos theta` ...(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system `(mv)/(sqrt3)- mV sin theta`
From the conservation of vertical linear momentum `(mv)/(sqrt3)-mV sin theta=0 rArr (v)/(sqrt3)= V sin theta` ...(iv)
By solving (iii) and (iv)
`v^(2) + (v^(2))/(3)= V^(2) (sin^(2) theta + cos^(2) theta)`
`rArr (4v^(2))/(3)= V^(2) rArr V= (2)/(sqrt3)v`