JEE Main 2022 · 26 Jul · Shift 1 · Physics Q47

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PhysicsSingle correct+4 / -1Question 47 of 90

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 :

Two 250 g blocks joined by a horizontal spring.

\dfrac v{2\sqrt2}

\dfrac v2

\dfrac v4

\dfrac v{\sqrt2}

Answer and worked solution

Official answer: B

Centre of mass at rest. All kinetic energy goes into the spring: $2\cdot\frac12(0.25)v^2=\frac12(2)x^2$ , $x=\frac v2$ .

(B)

Topics:Spring-mass systems Conservation of linear momentum

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JEE Main 2022 · 26 Jul · Shift 1, Physics question 47