A) \[1:3\]
B) \[3:1\]
C) \[1:\sqrt{3}\]
D) \[\sqrt{3}:1\]
Correct Answer: C
Solution :
Kinetic energy of body is given by \[{{E}_{K}}=\frac{1}{2}\,m{{v}^{2}}\] ... (i) and linear momentum of particle is given by \[p=mv\] ... (ii) From Eqs. (i) and (ii), we get \[{{E}_{K}}=\frac{{{m}^{2}}{{v}^{2}}}{2m}=\frac{{{p}^{2}}}{2m}\] \[\therefore \] \[p=\sqrt{2\,m{{E}_{K}}}\] When \[{{E}_{{{K}_{1}}}}={{E}_{{{K}_{2}}}}\] \[\frac{p_{1}^{2}}{2{{m}_{1}}}=\frac{p_{2}^{2}}{2{{m}_{2}}}\] or \[\frac{{{p}_{1}}}{{{p}_{2}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\] or \[\frac{{{p}_{1}}}{{{p}_{2}}}=\sqrt{\frac{3}{1}}\] or \[\frac{{{p}_{1}}}{{{p}_{2}}}=\frac{1}{\sqrt{3}}\]
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