A particle of mass m moving in… - JEE Main 2015 Physics PYQ

JEE Main 2015 — Physics Mechanics

2015mcqmedium

A particle of mass $\text{m}$ moving in the $x$ direction with speed $2v$ is hit by another particle of mass $2m$ moving in the $y$ direction with speed $v.$ If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

Official previous-year question

Held on 4 Apr 2015 · Verified 6 Jul 2026.

Options

A
$62%$
B
$44%$
C
$50%$
D
$56%$

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Solution

The initial momentum of system is $\vec{{P}_{i}}=m(2V)\hat{i}+(2m)V \hat{j}$

According to question as

JEE Main 2015 Physics, Center of Mass Momentum and Collision — solution figure

On perfectly inelastic collision the particles stick to each other.

$\vec{{P}_{f}}=3m \vec{{V}_{f}}$

By conservation of linear momentum

$\vec{{P}_{f}}=\vec{{P}_{i}} \Rightarrow 3m \vec{{V}_{f}}=m2V \hat{i}+2mV\hat{j}$

$\Rightarrow \vec{{V}_{f}}=\frac{2V}{3}(\hat{i}+\hat{j})\Rightarrow | {V}_{f} | =\frac{2\sqrt{2}}{3}V$

$\therefore$ loss in KE. of system $={K}_{initial}-{K}_{final}$

$=\frac{1}{2}m{(2V)}^{2}+\frac{1}{2}(2m){V}^{2}-\frac{1}{2}(3m){ ( \frac{ 2 \sqrt{ 2 } V }{ 3 } ) }^{2}$

$=2m{V}^{2}+m{V}^{2}-\frac{4}{3}m{V}^{2}=3m{V}^{2}-\frac{4m{V}^{2}}{3}=\frac{5}{3}m{V}^{2}$

% Loss in KE $=100\times \frac{\Delta K}{{K}_{i}}=\frac{\frac{5}{3}m{v}^{2}}{3m{V}^{2}}=\frac{5}{9}\times 100=56%$

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Solution

The initial momentum of system is $\vec{{P}_{i}}=m(2V)\hat{i}+(2m)V \hat{j}$

According to question as

JEE Main 2015 Physics, Center of Mass Momentum and Collision — solution figure

On perfectly inelastic collision the particles stick to each other.

$\vec{{P}_{f}}=3m \vec{{V}_{f}}$

By conservation of linear momentum

$\vec{{P}_{f}}=\vec{{P}_{i}} \Rightarrow 3m \vec{{V}_{f}}=m2V \hat{i}+2mV\hat{j}$

$\Rightarrow \vec{{V}_{f}}=\frac{2V}{3}(\hat{i}+\hat{j})\Rightarrow | {V}_{f} | =\frac{2\sqrt{2}}{3}V$

$\therefore$ loss in KE. of system $={K}_{initial}-{K}_{final}$

$=\frac{1}{2}m{(2V)}^{2}+\frac{1}{2}(2m){V}^{2}-\frac{1}{2}(3m){ ( \frac{ 2 \sqrt{ 2 } V }{ 3 } ) }^{2}$

$=2m{V}^{2}+m{V}^{2}-\frac{4}{3}m{V}^{2}=3m{V}^{2}-\frac{4m{V}^{2}}{3}=\frac{5}{3}m{V}^{2}$

% Loss in KE $=100\times \frac{\Delta K}{{K}_{i}}=\frac{\frac{5}{3}m{v}^{2}}{3m{V}^{2}}=\frac{5}{9}\times 100=56%$

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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