A) 0.1
B) 0.2
C) 0.5
D) 0.64
Correct Answer: A
Solution :
The loss in kinetic energy which is transformed into heat \[=\frac{1}{2}\left( \frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{({{u}_{1}}+{{u}_{2}})}^{2}}\] Here, \[{{m}_{1}}=m,\,{{m}_{2}}=\frac{m}{9},{{u}_{1}}=u,{{u}_{2}}=0\] Loss \[\Delta \Epsilon =\frac{1}{2}\left( \frac{m\times \frac{m}{9}}{m+\frac{m}{9}} \right)\times {{(u+0)}^{2}}=\frac{1}{2}\frac{m}{10}{{u}^{2}}\] Now, initial kinetic energy \[=\frac{1}{2}m{{u}^{2}}\] \[\text{Required}\,\text{fraction =}\frac{\text{loss}\,\text{in kinetic}\,\text{energy}}{\text{initial}\,\text{kinetic}\,\text{energy}}\] \[=\frac{\frac{1}{2}\frac{m}{10}{{u}^{2}}}{\frac{1}{2}m{{u}^{2}}}\] \[=\frac{1}{10}=0.1\]
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