QJEEM22Phbba3: JEE Main 2022 Physics - Class 11 JEEM22Phbba3 Solution: As per the given figure, two blocks each of mass 250   g…

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QJEEM22Phbba3: JEE Main 2022 Physics - Class 11 JEEM22Phbba3 Solution: As per the given figure, two blocks each of mass 250   g…

As per the given figure, two blocks each of mass $250 g$ are connected to a spring of spring constant $2 {Nm}^{- 1}$ . If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is :

JEE Main 2022 (Online) 26th July Morning Shift Physics - Work Power & Energy Question 58 English

Multiple Choice Options

A v22
B v2
C v4
D v2

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Accepted Answer

$∵$ Loss in KE = Gain in spring energy

$\Rightarrow \frac{1}{2} m v^{2} \times 2 = \frac{1}{2} k x_{m}^{2}$

$\Rightarrow 2 \times \frac{1}{4} \times v^{2} = 2 \times x_{m}^{2}$

$\Rightarrow x_{m} = \sqrt{\frac{v^{2}}{4}} = \frac{v}{2}$

QJEEM22Phbba3: JEE Main 2022 Physics - Class 11 JEEM22Phbba3 Solution: As per the given figure, two blocks each of mass 250 &nbsp; g…

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QJEEM22Phbba3: JEE Main 2022 Physics - Class 11 JEEM22Phbba3 Solution: As per the given figure, two blocks each of mass 250 g…