A particle `(A)` of mass `m_(1)` elastically collides with another stationary particle `(B)` of mass `m_(2)`. Then :

Options

A. `(m_(1))/(m_(2)) = (1)/(2)` and the particles fly a part in the opposite direction with equal velocities.
B. `(m_(1))/(m_(2)) = (1)/(3)` and the particles fly apart in the opposite direction with equal velocities.
C. `(m_(1))/(m_(2)) = (2)/(1)` and the collision angle between the particles is `60^(@)` symmetrically.
D. `(m_(1))/(m_(2)) = (2)/(1)` and the particles fly apart symmetrically at an angle `90^(@)`

Answer

B, C

Explanation

`m_(1)u + 0 = -m_(1)v+m_(2)v` ……….(1)
From kinetic energy conservation
`(1)/(2)m_(1)u^(2) = (1)/(2)m_(1)v^(2) + (1)/(2)m_(2)v^(2)` ………..(2)

From eq. (1) and and eq. (2) we get , `(m_(1))/(m_(2)) = (1)/(3)`
So, choice (b) is correct.
`P^(2) = P_(1)^(2) + P_(2)^(2) +2P_(1)P_(2) cos theta` .........(1)
Also from conservation of kinetic energy
`(P^(2))/(2m_(1)) = (P_(1)^(2))/(2m_(1)) + (P_(2)^(2))/(2m_(2))` ............(2)
Since particles fly symmetrically
`P_(1) sin "(theta)/(2) = P_(2) sin "(theta)/(2)` ..........(3)
Solving eq. (1), (2) and (3) we get
`(m_(1))/(m_(2)) = (2)/(1)`