Q 11-05-012NEETJEE MainEAMCET 2007 (Engineering)Medium
In two separate collisions, the coefficients of restitution $e_1$ 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 the relative velocity of approach and the relative velocity of separation in the second collision is
Answer: (D) $6 : 1$
Concept. $e = \dfrac{\text{separation speed}}{\text{approach speed}}$.
Step 1. First. $e_1 = \dfrac12$, so $e_2 = \dfrac16$.
Step 2. Second. Approach : separation $= 6 : 1$.
Answer. $6 : 1$ (Option D).
Solution by Sreeraj P, M.Sc Physics