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Question

Physics Question on work, energy and power

In two separate collisions, the coefficients of restitutions e1{{e}_{1}} and e2{{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

1:06

B

2:03

C

3:02

D

6:01

Answer

6:01

Explanation

Solution

The coefficient of restitution e=(VelocityofseparationVelocityofapproach)e=\left( \frac{Velocity\,of\,separation}{Velocity\,of\,approach} \right) \therefore e1=(vseparation)1(vapporach)1=12{{e}_{1}}=\frac{{{({{v}_{separation}})}_{1}}}{{{({{v}_{apporach}})}_{1}}}=\frac{1}{2} and e2=(vseparation)2(vapporach)2{{e}_{2}}=\frac{{{({{v}_{separation}})}_{2}}}{{{({{v}_{apporach}})}_{2}}} but e1e2=31\frac{{{e}_{1}}}{{{e}_{2}}}=\frac{3}{1} \Rightarrow e2e1=13\frac{{{e}_{2}}}{{{e}_{1}}}=\frac{1}{3} \therefore (vseparation)2(vapporach)2=12×13\frac{{{({{v}_{separation}})}_{2}}}{{{({{v}_{apporach}})}_{2}}}=\frac{1}{2}\times \frac{1}{3} \Rightarrow (vseparation)2(vapporach)2=61\frac{{{({{v}_{separation}})}_{2}}}{{{({{v}_{apporach}})}_{2}}}=\frac{6}{1}