A) \[\frac{2.4}{49}m{{s}^{-1}}\]
B) \[\frac{2.4\times 48}{49}m{{s}^{-1}}\]
C) \[\frac{{{10}^{3}}}{7}cm{{s}^{-1}}\]
D) \[\frac{{{10}^{6}}}{7}cm{{s}^{-1}}\]
Correct Answer: C
Solution :
\[\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}=2.4\] or \[{{m}_{1}}v_{1}^{2}+{{m}_{2}}v_{2}^{2}=4.8\] ... (i) now, \[{{m}_{1}}{{v}_{1}}={{m}_{2}}{{v}_{2}}\] or \[{{v}_{1}}=48\,{{v}_{2}}\] Using Eq. (i) \[\frac{1}{1000}\,{{(48{{v}_{2}})}^{2}}+\frac{48}{1000}v_{2}^{2}=4.8\] or \[\frac{48}{1000}(49\,v_{2}^{2})=4.8\] \[\therefore \] \[{{v}_{2}}=\frac{10}{7}m/s=\frac{{{10}^{2}}}{7}cm/s\]
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