question_answer

A particle of mass m moving with velocity \[{{V}_{0}}\]strikes a simple pendulum of mass \[m\]and sticks to it. The maximum height attained by the pendulum will be          [RPET 2002]                

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 \[h=\frac{V_{0}^{2}}{8g}\]