A) \[2(\sqrt{2}-1)\]m/s
B) \[2(\sqrt{2}+1)\] m/s
C) 4.5 m/s
D) none of these
Correct Answer: B
Solution :
\[KE=K=\frac{1}{2}m{{v}^{2}}\] Given: \[{{v}_{2}}={{v}_{1}}+2\] \[\therefore \] \[\frac{{{K}_{1}}}{{{K}_{2}}}={{\left( \frac{{{v}_{1}}}{{{v}_{2}}} \right)}^{2}}\] Or \[\frac{{{K}_{1}}}{{{K}_{2}}}={{\left( \frac{{{v}_{1}}}{{{v}_{2}}} \right)}^{2}}\] Or \[\frac{{{K}_{1}}}{2{{K}_{2}}}=\frac{v_{1}^{2}}{{{({{v}_{1}}+2)}^{2}}}\] Or \[v_{1}^{2}+4{{v}_{1}}+4=2v_{1}^{2}\] Or \[v_{1}^{2}-4{{v}_{1}}-4=0\] \[{{v}_{1}}=\frac{4\pm \sqrt{16+6}}{2}\] \[=\frac{4\pm \sqrt{32}}{2}\] \[=2(\sqrt{2}+1)m/s\]
You need to login to perform this action.
You will be redirected in
3 sec
