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

A particle (A) of mass m_(1) elastically collides with another station

1.	A particle (A) of mass m_(1) elastically collides with another stationary particle (B) of mass m_(2). Then :
Answer» `(m_(1))/(m_(2)) = (1)/(2)` and the particles fly a part in the opposite direction with equal velocities. `(m_(1))/(m_(2)) = (1)/(3)` and the particles fly apart in the opposite direction with equal velocities. `(m_(1))/(m_(2)) = (2)/(1)` and the collision angle between the particles is `60^(@)` symmetrically. `(m_(1))/(m_(2)) = (2)/(1)` and the particles fly apart symmetrically at an angle `90^(@)` SOLUTION :`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)`