A) \[\sqrt{2}\,\,m/s\]
B) \[(\sqrt{2}-1)\,\,m/s\]
C) \[\frac{1}{(\sqrt{2}-1)}\,m/s\]
D) \[\frac{1}{\sqrt{2}}\,m/s\]
Correct Answer: C
Solution :
Let m = mass of the boy, M = mass of the man, v = velocity of the boy and V= velocity of the man Initial kinetic energy of man \[=\,\,\,\frac{1}{2}M{{V}^{2}}\,=\,\,\frac{1}{2}\left[ \frac{1}{3}m{{v}^{2}} \right]\,=\,\frac{1}{2}\,\left[ \frac{1}{2}\left( \frac{M}{2} \right){{v}^{2}} \right]\] \[\left[ As\,\,m=\frac{M}{2}\,\,\,given \right]\] \[\Rightarrow \,\,\,{{V}^{2}}=\frac{{{v}^{2}}}{4}\Rightarrow V=\frac{v}{2}\] ?. (i) When the man speeds up by 1 m/s, \[\frac{1}{2}M{{(V+1)}^{2}}\,=\,\,\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\,\left( \frac{M}{2} \right){{v}^{2}}\] \[\Rightarrow \,\,{{(V+1)}^{2}}\,=\,\,\frac{{{v}^{2}}}{2}\] \[\Rightarrow \,\,\,\,V+1\,=\,\,\frac{{{v}^{2}}}{\sqrt{2}}\] ?. (ii) From (i) and (ii) we get speed of the man \[V=\frac{v}{\sqrt{2}-1}\,m/s\]
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