A) 1 : 1
B) 2 : 1
C) 4 : 1
D) 9 : 1
Correct Answer: D
Solution :
K.E. of colliding body before collision \[=\frac{1}{2}m{{v}^{2}}\] After collision its velocity becomes \[{v}'=\frac{({{m}_{1}}-{{m}_{2}})}{({{m}_{1}}+{{m}_{2}})}v=\frac{m}{3m}v=\frac{v}{3}\] \ K.E. after collision\[\frac{1}{2}mv{{'}^{2}}\]\[=\frac{1}{2}\frac{m{{v}^{2}}}{9}\] Ratio of kinetic energy =\[\frac{\text{K}\text{.E}{{\text{.}}_{\text{before}}}}{\text{K}\text{.E}{{\text{.}}_{\text{after}}}}=\frac{\frac{1}{2}m{{v}^{2}}}{\frac{1}{2}\frac{m{{v}^{2}}}{9}}=9:1\]
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