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 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 $90^o$.
$(a)$ Let $v_{1}$ and $v_{2}$ are velocities of the two balls after collision.
From law of conservation of linear momentum,
$ 2 m v_{0}=m v_{1}+m v_{2}$
$\therefore 2 v_{0}=v_{1}+v_{2}$
and $e=\frac{v_{2}-v_{1}}{2 v_{0}} \Rightarrow v_{2}=v_{1}+2 v_{0} e$
$\therefore 2 v_{1}=2 v_{0}-2 e v_{0}$
$\therefore v_{1}=v_{0}(1-e)$
Since, $e<1 \Rightarrow v_{1}$ has the same sign as $v_{0}$, therefore, the ball moves on after collision.
$(b)$ Consider the diagram below for a general collision.
From the law of conservation of linear momentum,
$P=P_{1}+P_{2}$
For inelastic collision some $\mathrm{KE}$ is lost, hence $\frac{p^{2}}{2 m}>\frac{p_{1}^{2}}{2 m}+\frac{p_{2}^{2}}{2 m}$ $p^{2}>p_{1}^{2}+p_{2}^{2}$
Thus, $p, p_{1}$ and $p_{2}$ are related as shown in the figure. $\theta$ is acute (less than $90^{\circ}$ )
$\left(p^{2}=p_{1}^{2}+p_{2}^{2}\right.$ would be given if $\left.\theta=90^{\circ}\right)$
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