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) (v)/(2√(2)) (b) v/2 (c) v/4 (d) v/√(2)

Answer: (b). Momenta are equal and opposite, so at maximum elongation both blocks are momentarily at rest. 2 × 1/2mv² = 1/2kx² ⇒ x² = (2mv²)/(k) x² = (2(0.25)v²)/(2) = v²/4 So the answer is (b) v/2.

As per the given figure, two blocks each of mass 250 g are connected… | JEE Main 2022 (26 Jul, Shift 1)

Q11JEE Main 2022 · 26 JUL · SHIFT 1Medium

Spring & Elastic Energy

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

Show answer and solution

Answer: (b)

Momenta are equal and opposite, so at maximum elongation both blocks are momentarily at rest.
$2 \times \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \Rightarrow x^2 = \frac{2mv^2}{k}$
$x^2 = \frac{2(0.25)v^2}{2} = \frac{v^2}{4}$
So the answer is (b) $\frac{v}{2}$ .

Report a mistake

Spring & Elastic Energy question from JEE Main 2022 (26 Jul, Shift 1)

More questions on Spring & Elastic Energy