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  • Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant K. A third block C of mass m moving with a speed v along the line joining A and B collides with A

Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant K. A third block C of mass m moving with a speed v along the line joining A and B collides with A. The maximum compression in the spring is
 
Option: 1 \sqrt{\frac{m}{2K}}
Option: 2 v\sqrt{\frac{m}{2K}}
Option: 3 \sqrt{\frac{mv}{K}}  
Option: 4 \sqrt{\frac{mv}{2K}}

Answers (1)

best_answer

After collision between C and A, C stops while A moves with speed of C i.e. v [in head on elastic collision, two equal masses exchange their velocities]. At maximum compression, A and B will move with same speed v/2 (From conservation of linear momentum).

Let x be the maximum compression in this position.

∴ KE of A-B system at maximum compression
\begin{aligned} &\text { or }\\ &\begin{array}{l} =\frac{1}{2}(2 m)\left(\frac{v}{2}\right)^{2} \\ \\ =K_{\max }=m v^{2} / 4 \end{array} \end{aligned}

From conservation of mechanical energy in two positions shown in above figure
\begin{aligned} &\text { or }\\ &=\frac{1}{2} m v^{2}=\frac{1}{4} m v^{2}+\frac{1}{2} k x^{2}\\ &=\frac{1}{2} k x^{2}=\frac{1}{4} m v^{2} \\ \\ \Rightarrow \therefore x=v\sqrt{\frac{m}{2 k}} \end{aligned}

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Deependra Verma

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