Question
A ball of mass m, moving with a speed $2v_0$, collides inelastically $(e > 0)$ with an identical ball at rest. Show that: For head-on collision, both the balls move forward.
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Answer
Let the $v_1, v_2$_ are the velocities of the two balls after the collision. Now by the principle of law of conservation of momentum. $\text{mv}_0=\text{mv}_1+\text{mv}_2$
$2\text{v}_0=\text{v}_1+\text{v}_2\ ...(\text{i})$
$\text{e}=\frac{\text{v}_2-\text{v}_1}{\text{v}_0+\text{v}_0}$
$\text{v}_2-\text{v}_1=2\text{ev}_0$ From (i) $\text{v}_1=-\text{v}_2+2\text{v}_0$
$\text{v}_1=-\text{v}_1-2\text{ev}_0+2\text{v}_0$
$2\text{v}_1=2\text{v}_0-2\text{ev}_0$
$\text{v}_1=\text{v}_0(1-\text{e})$
$\because\ \text{e}<1 v_0$_ is positive so the direction of $v_1$ is same as $v_0$ or $v_1$ is in forward direction. Hence proved.
$2\text{v}_0=\text{v}_1+\text{v}_2\ ...(\text{i})$
$\text{e}=\frac{\text{v}_2-\text{v}_1}{\text{v}_0+\text{v}_0}$
$\text{v}_2-\text{v}_1=2\text{ev}_0$ From (i) $\text{v}_1=-\text{v}_2+2\text{v}_0$
$\text{v}_1=-\text{v}_1-2\text{ev}_0+2\text{v}_0$
$2\text{v}_1=2\text{v}_0-2\text{ev}_0$
$\text{v}_1=\text{v}_0(1-\text{e})$
$\because\ \text{e}<1 v_0$_ is positive so the direction of $v_1$ is same as $v_0$ or $v_1$ is in forward direction. Hence proved.
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