A) \[\left( \sqrt{2}-1 \right)m/s\]
B) \[\sqrt{2}\,m/s\]
C) \[\frac{1}{\sqrt{2}-1}\,m/s\]
D) \[\frac{1}{\sqrt{2}}\,m/s\]
Correct Answer: C
Solution :
Key Idea: The kinetic energy of a moving body is equal to half the product of the mass (m) of the body and the square of its speed\[{{\left( v \right)}^{2}}\]. Kinetic \[enrgy=\frac{1}{2}\times mass\times {{\left( speed \right)}^{2}}\] \[i.e.\], \[=\frac{1}{2}m{{v}^{2}}\] Let mass of man is \[M\] kg and speed is \[v\] and speed of boy is \[{{v}_{1}}\], then \[\frac{1}{2}M{{v}^{2}}=\frac{1}{2}\left( \frac{1}{2}\frac{M}{2}{{v}_{1}}^{2} \right)\] ?\[\left( 1 \right)\] When man speeds up by \[1\,m/s\], then \[v=v+1\] \[\therefore \] \[\frac{1}{2}M{{\left( v+1 \right)}^{2}}=\frac{1}{2}\left( \frac{M}{2} \right).{{v}_{1}}^{2}\] ?\[\left( 2 \right)\] Diving \[Eq.\]\[\left( 1 \right)\] by \[\left( 2 \right)\], we get \[\frac{{{v}^{2}}}{{{\left( v+1 \right)}^{2}}}=\frac{1}{2}\] \[\Rightarrow \] \[\frac{v}{v+1}=\frac{1}{\sqrt{2}}\] \[\Rightarrow \] \[\sqrt{2\,}v=v+1\] \[\Rightarrow \] \[v=\frac{1}{\sqrt{2}-1}m/s\] Note: In the formula for kinetic energy speed occurs in the second power and so has a larger effect compared to mass.
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