For example 32 consider target body B initially at rest, get the expressin for velocities of bodies after the collision `(a)m_(1)=m_(2)(b)m_(2)gt gtm_(1)(c)m_(2)ltltm_(1)`

Explanation

Put `m_(1)=m_(2)and u_(2)=0` in equations (x) & (xi)
(b) Put `m_(1)=0` in equations (x) &n (xi)
(c ) Put `m_(2)=0` in equations (x) & (xi)
`u_(2)=0` (given)
From equation (10)
`v_(1)=((m_(1)-m_(2))u_(1))/(m_(1)+m_(2))" "...(12)`
From equation (11)
`v_(2)=(2m_(1)u_(1))/(m_(1)+m_(2))" "...(13)`
(a) Put `m_(1)=m_(2)=m` (say) in equations (12) & (13)
`v_(1)=((m-m)u_(1))/(m+m)=0`
`v_(2)=(2m u_(1))/(m+m)=u_(1)`
i.e., body A comes to rest and body B starts mobing with the initial velocity of A `100%` KE of A is transferred to the body B.
(b) `m_(2)gt gtm_(1)`
`m_(1)` can be ignored
Put `m=0` in equations (12) & (13)
`v_(1)=(-m_(2))/(0+m_(2))u_(1)=-u_(1)`
`v_(2)=(2xx0xxu_(1))/(0+m_(1))=0`
When a light body a collides against a heavy body B at rest. A rebounds with its own velocity and B continous to be at rest e.g., a ball rabounds with same speed (direction of velocity is opposite) on striking a floor.
(c ) (When target body B at rest has negligible mass) `m_(2) lt lt m_(1),m_(2)` can be ignored.
Put `m_(2)=0` in equations (12) and (13)
`v_(1)=-((m_(1)-0))/(m_(1)+0)u_(1)=u_(1)`
`v_(2)=(2m_(1)u_(1))/(m_(1)+0)=2u_(1)`
When a heavy body AS undergoes an elasticv collision witha light body at rest, A keeps on moving with the same velocity of its own and B starts moving with double the initial velosity of A.