A ball of mass m`m`, moving with a speed `2upsilon_(0)`, collides inelasticaly `(egt0)` with an identical ball at rest. Show that `(a)` For head - on collision, both the balls move forward.
(b) For a genergcollision, the angle between the two velocities of scattered balls is less that `90^(@)`.

Explanation

For head on collision . Let `v_(1)` and `v_(2)` be velocities of the two balls after collision
From law of conservation of linear momentum
` m xx 2v_(0) = mv_(1) +mv_(2)`
or `2v_(0) = v_(1) +v_(2) " " ….(i)`
Now
` e = ("relative velocity of separation after collision ")/(" relative velocity of approach before collision ") `
or ` e = (v_(2) - v_(1))/(2v_(0))`
`v_(2) - v_(1) = 2v_(0)^(e) " " ...(ii)`
[ From equation (i) `v_(2) = 2v_(0) - v_(1) ] `
`2v_(0) - 2v_(1) = 2v_(0)e`
or `2v_(1) = 2v_(0) (1-e)`
or `v_(1) = v_(0) (1-e)`
As ` e lt 1 ` , so `v_(1)` has the same sign as `v_(0)` , so the ball moves forward after collision .
(b) For general collision
` vec(p) = vec(p)_(1) +vec(p)_(2)`
Since in inealstic collision , some KE is lost , so
`(p^(2))/(2m) gt (p_(1)^(2))/(2m) +(p_(2)^(2))/(2m)`
` [ :. E = (mv^(2))/2 = (m^(2)v^(2))/(2m) = (p^(2))/(2m) ] `
or `p^(2) gt p_(1)^(2) +p_(2)^(2),` it will be possible when `theta ` is acute or less then `90^(@)` .

Thus `vec(p), vec(p)_(1) and vec(p)_(1)` are related on shown in figure .Here `theta ` is acute `( theta lt 90^(@)) ` ( if `p^(2) = p_(1)^(2)+p_(2)^(2)` then it would five `theta = 90^(@)`).