A mass 'm' moves with a velocity 'v' and collides inelastieally with another identical mass . After collision the `1^(st)` mass moves with velocity `v/sqrt3` in a direction perpendicular to the initial direction of motion. Find the speed of the `2^(nd)` mass after collision.
`undersetunderset ((before),(collision))(m)(*rarr) underset(m)* uparrow v//sqrt3 underset((after),(collision):)`

Options

A. `(2)/(sqrt3) v`
B. `(v)/(sqrt3)`
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 a perpendicular direction and let the mass `B` moves at angle `theta` with the horizontal with velocity `v`.
Initical 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 implies V cos theta` (iii)
Initial vertical momentum of system (bsfore 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 implies (v)/(sqrt3) = V sin theta` (iv)
By solving (iii) and (iv)
`v^(2) + (v^(2))/(3) = V^(2) (sin^(2) theta + cos^(2) theta)`
`implies (4v^(2))/(3) = V^(2) implies V = (2)/(sqrt3) v`.