A) m
B) 2m
C) 3m
D) 1m
E) 5m
Correct Answer: C
Solution :
\[\frac{1}{2}{{m}_{1}}u_{1}^{2}-\frac{1}{2}{{m}_{1}}v_{1}^{2}=\frac{75}{100}\times \frac{1}{2}{{m}_{1}}u_{1}^{2}\] \[\Rightarrow \] \[u_{1}^{2}-v_{1}^{2}=\frac{3}{4}u_{1}^{2}\] \[\Rightarrow \] \[{{v}_{1}}=\frac{1}{2}{{u}_{1}}\] ?.. (1) Now \[{{v}_{1}}=\frac{({{m}_{2}}-{{m}_{1}}){{u}_{1}}}{({{m}_{1}}+{{m}_{2}})}\] Thus, \[\frac{1}{2}{{u}_{1}}=\frac{({{m}_{2}}-{{m}_{1}}){{u}_{1}}}{({{m}_{1}}+{{m}_{2}})}\] \[\Rightarrow \] \[{{m}_{2}}=3{{m}_{1}}=3m\]
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