A) \[{{v}_{0}}\sqrt{\frac{2m}{k}}\And \sqrt{\frac{k}{2m}}\]
B) \[{{v}_{0}}\sqrt{\frac{m}{2k}}\And \sqrt{\frac{2m}{k}}\]
C) \[{{v}_{0}}\sqrt{\frac{m}{2k}}\And \sqrt{\frac{k}{2m}}\]
D) \[{{v}_{0}}\sqrt{\frac{m}{k}}\And \,\sqrt{\frac{k}{m}}\]
Correct Answer: C
Solution :
By conservation of momentum, \[m{{v}_{0}}=(m+m)V\] \[V=\frac{{{v}_{0}}}{2}\to \] Velocity after collision Also, \[\omega =\sqrt{\frac{k}{m+m}}=\sqrt{\frac{k}{2m}}\] \[{{v}_{\max }}=A\omega =\frac{{{v}_{0}}}{2}\] \[\therefore \] \[A={{v}_{0}}\sqrt{\frac{m}{2k}}\]
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