Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति
JEE Main & Advanced
Physics
Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति
Question Bank
Conservation of Energy and Momentum
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 \[h=\frac{V_{0}^{2}}{8g}\]