Work Energy Power

JEE Main level, single correct, medium

126 of 183

Question 126 of 183

A block of mass m=0.1kgm=0.1kg is connected to a spring of unknown spring constant k. It is compressed to a distance xx from its equilibrium position and released from rest. After approaching half the distance (x2)\left(\dfrac{x}{2}\right) from the equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with velocity 3ms−13m{s}^{-1}. The total initial energy of the spring is:

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Answer: (A) 0.6J0.6J

By mechanical energy conservation between compression positions xx and x2\dfrac{x}{2}

12kx2=12k(x2)2+12mv2\dfrac{1}{2}k{x}^{2}=\dfrac{1}{2}k{\left(\dfrac{x}{2}\right)}^{2}+\dfrac{1}{2}m{v}^{2}

12kx2−12kx24=12mv2\dfrac{1}{2}k{x}^{2}-\dfrac{1}{2}k\dfrac{{x}^{2}}{4}=\dfrac{1}{2}m{v}^{2}

12kx2(34)=12mv2\dfrac{1}{2}k{x}^{2}\left(\dfrac{3}{4}\right)=\dfrac{1}{2}m{v}^{2}

v=3kx24m=3kmx2v=\sqrt{\dfrac{3k{x}^{2}}{4m}}=\sqrt{\dfrac{3k}{m}}\dfrac{x}{2}

On collision with a block at rest

∵∵ Velocities are exchanged ⇒\Rightarrow elastic collision between identical masses.

∴v=3=3kmx2∴v=3=\sqrt{\dfrac{3k}{m}}\dfrac{x}{2}

⇒6=3kmx\Rightarrow 6=\sqrt{\dfrac{3k}{m}}x

⇒x=6m3k\Rightarrow x=6\sqrt{\dfrac{m}{3k}}

∴∴ The initial energy of the spring is

U=12kx2=12k×36m3k=6mU=\dfrac{1}{2}k{x}^{2}=\dfrac{1}{2}k\times 36\dfrac{m}{3k}=6m

U=6×0.1=0.6JU=6\times 0.1=0.6J

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