JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 16)
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
_17th_March_Evening_Shift_en_16_1.png)
$$\sqrt {{{mv} \over K}} $$
$$\sqrt {{m \over {2K}}} $$
$$\sqrt {{{mv} \over {2K}}} $$
$$v\sqrt {{m \over {2K}}} $$
Explanation
_17th_March_Evening_Shift_en_16_4.png)
From conservation of momentum
mv = mv1 + mv1
$${V_1} = {V \over 2}$$
_17th_March_Evening_Shift_en_16_3.png)
And from energy conservation
$${1 \over 2}m{v^2} = \left( {{1 \over 2} \times m{{\left( {{V \over 2}} \right)}^2}} \right) \times 2 + {1 \over 2}K{x^2}$$
$${{m{v^2}} \over 2} - {{m{v^2}} \over 4} = {1 \over 2}k{x^2}$$
$${{m{v^2}} \over 4} = {1 \over 2}k{x^2}$$
Then the maximum compression in the spring is
$$x = \sqrt {{{m{v^2}} \over {2K}}} $$
$$ \Rightarrow $$ x = $$v\sqrt {{m \over {2K}}} $$
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