A) (5/6) times
B) Equal
C) Not related
D) Double for the first ball and half for the second ball
Correct Answer: A
Solution :
[a] Let masses of the two ball are 2 m and m and their speeds are u and 2u, respectively. |
By conservation of momentum, |
\[{{m}_{1}}{{\vec{u}}_{1}}+{{m}_{2}}{{\vec{u}}_{2}}={{m}_{1}}{{\vec{v}}_{1}}+{{m}_{2}}{{\vec{v}}_{2}}\] |
\[\Rightarrow \,\,\,2mu-2mu=m{{v}_{2}}-2m{{v}_{1}}\Rightarrow {{v}_{2}}=2{{v}_{1}}\] |
Coefficient of restitution |
\[=-\frac{({{v}_{2}}-{{v}_{1}})}{({{u}_{2}}-{{u}_{1}})}=-\frac{(2{{v}_{1}}+{{v}_{1}})}{(-2u-u)}=\frac{-3{{v}_{1}}}{-3u}=\frac{-3{{v}_{1}}}{-3u}=\frac{{{v}_{1}}}{u}=\frac{5}{6}\] [as given] |
\[\Rightarrow \frac{{{v}_{1}}}{{{u}_{1}}}=\frac{5}{6}=\]ratio of the speed of first ball before and after collision. |
Similarly, we can calculate the ratio of second ball before and after collision, |
\[\frac{{{v}_{2}}}{{{u}_{2}}}=\frac{2{{v}_{1}}}{2u}=\frac{{{v}_{1}}}{u}=\frac{5}{6}\] |
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