A) \[4:1\]
B) \[1:4\]
C) \[1:2\]
D) \[2:1\]
Correct Answer: C
Solution :
\[\text{E=}{{\text{W}}_{\text{0}}}\text{+}\frac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}\] \[\text{h(2}{{\text{v}}_{\text{0}}}\text{)=h}{{\text{v}}_{\text{0}}}\text{+}\frac{\text{1}}{\text{2}}\text{mv}_{\text{1}}^{\text{2}}\] \[\text{h}{{\text{v}}_{\text{0}}}\text{=}\frac{\text{1}}{\text{2}}\text{mv}_{\text{1}}^{\text{2}}\] ?(i) \[\text{h(5}{{\text{v}}_{0}})=h{{v}_{0}}+\frac{1}{2}mv_{2}^{2}\] \[4h{{v}_{0}}=\frac{1}{2}mv_{2}^{2}\] ?(ii) Divide (i) by (ii) \[\frac{1}{4}=\frac{v_{1}^{2}}{v_{2}^{2}}\] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{1}{2}\]
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