A) 1 : 2
B) 2 : 1
C) 1 : 4
D) 4 : 1
Correct Answer: B
Solution :
[b] \[{{E}_{2}}={{E}_{1}}\] or \[\frac{1}{2}{{m}_{2}}\,{{v}_{2}}^{2}=\frac{1}{2}\,{{m}_{1}}{{v}_{1}}^{2}\] or \[\frac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\frac{{{m}_{2}}}{{{m}_{1}}}}\] So, \[\frac{{{p}_{1}}}{{{p}_{2}}}=\frac{{{m}_{1}}{{v}_{1}}}{{{m}_{2}}{{v}_{2}}}\] \[=\frac{4}{1}\times \sqrt{\frac{1}{4}}=\frac{2}{1}\]
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