$A$ ball of mass $m$,moving with a speed $2v_0$,collides inelastically $(e > 0)$ with an identical ball at rest. Show that
$(a)$ For head-on collision,both the balls move forward.
$(b)$ For a general collision,the angle between the two velocities of scattered balls is less than $90^o$.

(A-D) Let $v_{1}$ and $v_{2}$ be the velocities of the two balls after the collision.
From the law of conservation of linear momentum:
$2mv_0 = mv_1 + mv_2$
$\therefore 2v_0 = v_1 + v_2$
From the definition of the coefficient of restitution $e$:
$e = \frac{v_2 - v_1}{2v_0 - 0} \Rightarrow v_2 - v_1 = 2v_0e$
Adding the two equations:
$(v_1 + v_2) + (v_2 - v_1) = 2v_0 + 2v_0e$
$2v_2 = 2v_0(1 + e) \Rightarrow v_2 = v_0(1 + e)$
Subtracting the equations:
$(v_1 + v_2) - (v_2 - v_1) = 2v_0 - 2v_0e$
$2v_1 = 2v_0(1 - e) \Rightarrow v_1 = v_0(1 - e)$
Since $0 < e < 1$,both $v_1$ and $v_2$ are positive,meaning both balls move forward.
$(b)$ For a general collision,by the law of conservation of linear momentum:
$\vec{p} = \vec{p_1} + \vec{p_2}$
For an inelastic collision,kinetic energy is lost,so:
$\frac{p^2}{2m} > \frac{p_1^2}{2m} + \frac{p_2^2}{2m} \Rightarrow p^2 > p_1^2 + p_2^2$
Using the law of cosines for the vector triangle formed by $\vec{p}, \vec{p_1},$ and $\vec{p_2}$:
$p^2 = p_1^2 + p_2^2 - 2p_1p_2 \cos(180^o - \theta) = p_1^2 + p_2^2 + 2p_1p_2 \cos \theta$
Since $p^2 > p_1^2 + p_2^2$,it follows that $2p_1p_2 \cos \theta > 0$,which implies $\cos \theta > 0$.
Therefore,$\theta < 90^o$.