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
Two identical blocks A and B, each of mass 'm' resting on smooth floor are connected by a light spring of natural length L and spring constant K, with the spring at its natural length. A third identical block 'C' (mass m) moving with a speed v along the line joining A and B collides with A. the maximum compression in the spring is [EAMCET 2003]
A) \[v\sqrt{\frac{m}{2k}}\]
B) \[m\sqrt{\frac{v}{2k}}\]
C) \[\sqrt{\frac{mv}{k}}\]
D) \[\frac{mv}{2k}\]
Correct Answer:
A
Solution :
Initial momentum of the system (block C) = mv After striking with A, the block C comes to rest and now both block A and B moves with velocity V, when compression in spring is maximum. By the law of conservation of linear momentum mv = (m + m) V Þ \[V=\frac{v}{2}\] By the law of conservation of energy K.E. of block C = K.E. of system + P.E. of system \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}(2m)\,{{V}^{2}}+\frac{1}{2}k{{x}^{2}}\] Þ \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}(2m)\ {{\left( \frac{v}{2} \right)}^{2}}+\frac{1}{2}k{{x}^{2}}\] Þ \[k{{x}^{2}}=\frac{1}{2}m{{v}^{2}}\] Þ \[x=v\sqrt{\frac{m}{2k}}\]