A) 2 : 1
B) 4 : 1
C) 1 : 8
D) 8 : 1
Correct Answer: D
Solution :
\[K=\frac{1}{2}m{{v}^{2}}=\frac{{{p}^{2}}}{2m}\] \[\therefore \] \[\frac{{{K}_{1}}}{{{K}_{2}}}={{\left( \frac{{{p}_{1}}}{{{p}_{2}}} \right)}^{2}}.\left( \frac{{{m}_{2}}}{{{m}_{1}}} \right)\] \[\frac{{{K}_{1}}}{{{K}_{2}}}={{\left( \frac{2p}{p} \right)}^{2}}.\left( \frac{2m}{m} \right)=\frac{8}{1}\]
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