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

(a) Let `v_(1) and v_(2)` are velocities of the two balls after collision. Now by the principle of conservation of linear momentum,
`2mv_(0)=mv_(1)+mv_(2)`
`or 2v_(0)=v_(1)+v_(2)`
`and e=(v_(2)-v_(1))/(2v_(0))`
`Rightarrow v_(2)=v_(1)+2v_(0)e`
`therefore `2v_(1)=2v_(0)-2ev_(0)`
`therefore v_(1)=v_(0)(1-e)`
Since, `e lt 1 Rightarrow v_(1)` has the same sign as `v_(0)` therefore, the ball moves on after collision
(b) Consider the diagram below for a general collision

By principal of conservation of linear momentum, `P=P_(1)+P_(2)`
For inelastic collision some KE is lost, hence `(p^(2))/(2m) lt (p_(1)^(2))/(2m)+(p_(2)^(2))/(2m)`
Thus p, `p_(1) and p_(2)` are releated as shown in the figure
`theta` is acute (less than `90^(@))(p^(2)=p_(1)^(2)+p_(2)^(2)) "would given" `theta=90^(@)`