A) \[-2\sqrt{2}\]
B) \[+2\sqrt{2}\]
C) \[\sqrt{2}\]
D) 2
E) \[1/\sqrt{2}\]
Correct Answer: B
Solution :
Let\[{{v}_{M}}\]is velocity of man,\[{{v}_{B}}\]of boy, then kinetic energy \[K=\frac{1}{2}M{{v}^{2}}M=\frac{1}{2}.\frac{M}{2}.v_{B}^{2}\] \[\Rightarrow \] \[v_{M}^{2}=\frac{v_{B}^{2}}{2}\] ...(1) \[\Rightarrow \] \[\sqrt{2}{{v}_{M}}={{v}_{B}}\] When man speeds up\[2\text{ }m{{s}^{-1}}\]and boy changes his speed by\[x\text{ }m{{s}^{-1}}\]. Then, \[\frac{1}{2}M{{({{v}_{M}}+2)}^{2}}=\frac{1}{2}.\frac{M}{2}.{{({{v}_{B}}+x)}^{2}}\] \[\Rightarrow \] \[{{({{v}_{M}}+2)}^{2}}=\frac{{{({{v}_{B}}+x)}^{2}}}{2}\] ?. (2) From Eq. (1), we have \[2{{({{v}_{M}}+2)}^{2}}={{(\sqrt{2}{{v}_{M}}+x)}^{2}}\] \[\Rightarrow \] \[\sqrt{2}({{v}_{M}}+2)=\sqrt{2}{{v}_{M}}+x\] \[\Rightarrow \] \[+2\sqrt{2}=x\]
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