Question

In two separate collisions, the coefficients of restitution are $e_1$ 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 :

• A. $1:6$
• B. $2:3$
• C. $3:2$
• D. $6:1$

Answer 1

The correct option is D $6:1$

We know that: $e=\frac{relativevelocityaftercollision}{relativevelocitybeforecollision}$
Given $:\frac{e_1}{e_2}=\frac{3}{1},e_1=\frac{1}{2}$
$\Rightarrow \frac{\left(\frac{1}{2}\right)}{e_2}=\frac{3}{1}$
$\Rightarrow e_2=\frac{1}{6}$

So, the ration between relative velocity of approach and the relative velocity of separation in the second collision is $6:1$.