A) \[\frac{1}{2}\]of its initial kinetic energy
B) \[\frac{1}{9}\]of its initial kinetic energy
C) \[\frac{8}{9}\]of its initial kinetic energy
D) \[\frac{1}{4}\]of its initial kinetic energy
Correct Answer: C
Solution :
Final velocity of first body in collision \[{{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{1}}+\left( \frac{2{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{2}}\] \[{{v}_{1}}=\left( \frac{m-2m}{m+2m} \right){{u}_{1}}+\left( \frac{2\times 2m}{m+2m} \right)\times {{u}_{2}}\] \[{{v}_{1}}=-\frac{v}{3}\] \[(\because \,\,{{u}_{2}}=0)\] Initial kinetic energy \[=\frac{1}{2}m{{v}^{2}}\] Kinetic energy after collision is given by \[=\frac{1}{2}m{{\left( \frac{v}{3} \right)}^{2}}\] \[=\frac{1}{2}m\frac{{{v}^{2}}}{9}\] Change in kinetic energy \[=\frac{1}{2}m\left( {{v}^{2}}-\frac{{{v}^{2}}}{9} \right)\] \[=\frac{8}{9}\left( \frac{1}{2}m{{v}^{2}} \right)\] \[=\frac{8}{9}\] (initial kinetic energy)
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