A) \[h=\frac{V_{0}^{2}}{8g}\]
B) \[\sqrt{{{V}_{0}}g}\]
C) \[2\sqrt{\frac{{{V}_{0}}}{g}}\]
D) \[\frac{V_{0}^{2}}{4g}\]
Correct Answer: A
Solution :
Initial momentum of particle = \[m{{V}_{0}}\] Final momentum of system (particle + pendulum) = 2mv By the law of conservation of momentum Þ \[m{{V}_{0}}=2mv\]Þ Initial velocity of system v = \[\frac{{{V}_{0}}}{2}\] \ Initial K.E. of the system = \[\frac{1}{2}(2m){{v}^{2}}\]=\[\frac{1}{2}(2m){{\left( \frac{{{V}_{0}}}{2} \right)}^{2}}\] If the system rises up to height h then P.E. = \[2mgh\] By the law of conservation of energy \[\frac{1}{2}(2m){{\left( \frac{{{V}_{0}}}{2} \right)}^{2}}=2mgh\] Þ \[h=\frac{V_{0}^{2}}{8g}\]You need to login to perform this action.
You will be redirected in
3 sec