JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 15)
As per the given figure, two blocks each of mass $$250 \mathrm{~g}$$ are connected to a spring of spring constant $$2 \,\mathrm{Nm}^{-1}$$. If both are given velocity $$v$$ in opposite directions, then maximum elongation of the spring is :
_26th_July_Morning_Shift_en_15_1.png)
$$\frac{v}{2 \sqrt{2}}$$
$$\frac{v}{2}$$
$$\frac{v}{4}$$
$$
\frac{v}{\sqrt{2}}
$$
Explanation
$$\because$$ Loss in KE = Gain in spring energy
$$ \Rightarrow {1 \over 2}m{v^2} \times 2 = {1 \over 2}kx_m^2$$
$$ \Rightarrow 2 \times {1 \over 4} \times {v^2} = 2 \times x_m^2$$
$$ \Rightarrow {x_m} = \sqrt {{{{v^2}} \over 4}} = {v \over 2}$$
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