A block of mass m=0.1 kg is… - JEE Main 2015 Physics PYQ

JEE Main 2015 — Physics Mechanics

2015mcqhard

A block of mass $m=0.1 \mathrm{kg}$ 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 $(\frac{x}{2})$ from the equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with velocity $3 m{s}^{-1}$ . The total initial energy of the spring is:

Official previous-year question

Held on 10 Apr 2015 · Verified 6 Jul 2026.

Options

A
$0.6 J$
B
$0.8 J$
C
$1.5 J$
D
$0.3 J$

Did you get this right?

Solution

By mechanical energy conservation between compression positions $x$ and $\frac{x}{2}$

$\frac{1}{2}k{x}^{2}=\frac{1}{2}k{(\frac{x}{2})}^{2}+\frac{1}{2}m{v}^{2}$

$\frac{1}{2}k{x}^{2}-\frac{1}{2}k\frac{{x}^{2}}{4}=\frac{1}{2}m{v}^{2}$

$\frac{1}{2}k{x}^{2}(\frac{3}{4})=\frac{1}{2}m{v}^{2}$

$v=\sqrt{\frac{3k{x}^{2}}{4m}}=\sqrt{\frac{3k}{m}}\frac{x}{2}$

On collision with a block at rest

$\because$ Velocities are exchanged $\Rightarrow$ elastic collision between identical masses.

$\therefore v=3=\sqrt{\frac{3k}{m}}\frac{x}{2}$

$\Rightarrow 6=\sqrt{\frac{3k}{m} } x$

$\Rightarrow x=6\sqrt{\frac{m}{3k}}$

$\therefore$ The initial energy of the spring is

$U=\frac{1}{2}k {x}^{2}=\frac{1}{2}k\times 36\frac{m}{3k}=6m$

$U=6\times 0.1=0.6 J$

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Solution

By mechanical energy conservation between compression positions $x$ and $\frac{x}{2}$

$\frac{1}{2}k{x}^{2}=\frac{1}{2}k{(\frac{x}{2})}^{2}+\frac{1}{2}m{v}^{2}$

$\frac{1}{2}k{x}^{2}-\frac{1}{2}k\frac{{x}^{2}}{4}=\frac{1}{2}m{v}^{2}$

$\frac{1}{2}k{x}^{2}(\frac{3}{4})=\frac{1}{2}m{v}^{2}$

$v=\sqrt{\frac{3k{x}^{2}}{4m}}=\sqrt{\frac{3k}{m}}\frac{x}{2}$

On collision with a block at rest

$\because$ Velocities are exchanged $\Rightarrow$ elastic collision between identical masses.

$\therefore v=3=\sqrt{\frac{3k}{m}}\frac{x}{2}$

$\Rightarrow 6=\sqrt{\frac{3k}{m} } x$

$\Rightarrow x=6\sqrt{\frac{m}{3k}}$

$\therefore$ The initial energy of the spring is

$U=\frac{1}{2}k {x}^{2}=\frac{1}{2}k\times 36\frac{m}{3k}=6m$

$U=6\times 0.1=0.6 J$

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics