In two separate collisions, the coefficient of restitutions $e_1$ and $e_2$ are in the ratio $3: 1$. In the…
In two separate collisions, the coefficient of restitutions $e_1$ and $e_2$ are in the ratio $3: 1$. In the first collision the relative velocity of approach is twice the relative velocity of separation, then the ratio between relative velocity of approach and the relative velocity of separation in the second collision is
$1: 6$
$2: 3$
$3: 2$
$6: 1$
Solution
Coefficient of restitution is given by
$e=\frac{\text { relative velocity of separation }}{\text { relative velocity of approach }}$
we have
$e_1=\frac{1}{2}$
and $e_2=\left(\frac{\text { relative velocity of separation }}{\text { relative velocity of approach }}\right)_2$
Given, $\quad \frac{e_1}{e_2}=\frac{3}{1} \Rightarrow \frac{e_2}{e_1}=\frac{1}{3}$
$\Rightarrow \quad\left(\frac{\text { relative velocity of separation }}{\text { relative velocity of approach }}\right)_2$
$=\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}$
$\Rightarrow \quad\left(\frac{\text { relative velocity of approach }}{\text { relative velocity of separation }}\right)_2=\frac{6}{1}$