A) \[(\sqrt{2}-1)m/s\]
B) \[\sqrt{2}m/s\]
C) \[\frac{1}{(\sqrt{2}-1)m/s}\]
D) \[\frac{1}{\sqrt{2}}m/s\]
Correct Answer: C
Solution :
The kinetic energy of moving body is equal to half the product of the mass\[(m)\]of the body and the square of its speedy\[({{v}^{2}})\]. kinetic energy\[=\frac{1}{2}\times mass\times {{(speed)}^{2}}\] i.e., \[=\frac{1}{2}m{{v}^{2}}\] Let mass of man is\[Mkg\]and speed is\[v\]and speed of body is\[{{v}_{1}}\], then \[\frac{1}{2}M{{v}^{2}}=\frac{1}{2}\left\{ \frac{1}{2}\frac{M}{2}v_{1}^{2} \right\}\] ? (i) when man speed up by\[1m/s\], than \[v=v+1\] \[\therefore \] \[\frac{1}{2}M{{(v+1)}^{2}}=\frac{1}{2}\left\{ \frac{M}{2} \right\}\cdot v_{1}^{2}\] ... (ii) Dividing Eqs. (i) by (ii), we get \[\frac{{{v}^{2}}}{{{(v+1)}^{2}}}=\frac{1}{2}\] \[\sqrt{2}v=v+1\] \[v=\frac{1}{\sqrt{2}-1}m/s\]
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