Question

In two separate collisions, the coefficients 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 the relative velocity of approach and relative velocity of separation in the second collision is:

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

Answer 1

Answer: (D)

The coefficient of restitution $ e=\left( \frac{Velocity\,of\,separation}{Velocity\,of\,approach} \right) $ $ \therefore $ $ {{e}_{1}}=\frac{{{({{v}_{separation}})}_{1}}}{{{({{v}_{apporach}})}_{1}}}=\frac{1}{2} $ and $ {{e}_{2}}=\frac{{{({{v}_{separation}})}_{2}}}{{{({{v}_{apporach}})}_{2}}} $ but $ \frac{{{e}_{1}}}{{{e}_{2}}}=\frac{3}{1} $ $ \Rightarrow $ $ \frac{{{e}_{2}}}{{{e}_{1}}}=\frac{1}{3} $ $ \therefore $ $ \frac{{{({{v}_{separation}})}_{2}}}{{{({{v}_{apporach}})}_{2}}}=\frac{1}{2}\times \frac{1}{3} $ $ \Rightarrow $ $ \frac{{{({{v}_{separation}})}_{2}}}{{{({{v}_{apporach}})}_{2}}}=\frac{6}{1} $