As per the given figure, two blocks each of mass 250 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 :

(A) $\frac{\mathrm{V}}{2 \sqrt{2}}$

(D) $\frac{\mathrm{v}}{\sqrt{2}}$

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Work, Power & Energy

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 :

(A) $\frac{V}{2 2}$
(B) $\frac{V}{2}$ ✓ Correct
(C) $\frac{V}{4}$
(D) $\frac{v}{2}$

Answer: (B)

Solution

Using energy conservation, the total kinetic energy of both blocks equals the elastic potential energy at maximum elongation:

2 \times \frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}

m v^{2} = k x^{2}

0.25 \times v^{2} = 2 \times x^{2}

x = \frac{v}{2}

The answer is (B).

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option

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