JEE Main 2015 · 10 Apr · Shift 1 · Physics Q4

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PhysicsSingle correct+4 / -1Question 4 of 90

A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance $x$ from its equilibrium position and released from rest. After approaching half the distance $\left(\dfrac x2\right)$ from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity 3 ms $^{-1}$ . The total initial energy of the spring is :

1.5 J

0.6 J

0.3 J

0.8 J

Answer and worked solution

Official answer: B

The block's speed at $\frac x2$ is passed on in the collision (it stops; the other block, presumably of equal mass, takes 3 m/s). So the block had $\frac12(0.1)(9)=0.45$ J there.

$\frac12kx^2=\frac12k\frac{x^2}4+0.45$ , so $\frac34E=0.45$ and $E=0.6$ J.

Topics:Spring potential energy Elastic and inelastic collisions

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 4