Two particles of masses $M$ and $2M are$moving with speeds of $10 m{s}^{-1}$ and $5 m{s}^{-1}$, as shown in the figure. They collide at the origin and after that they move along the indicated directions with speeds ${v}_{1}$ and ${v}_{2}$, respectively. The values of ${v}_{1}$ and ${v}_{2}$ are, nearly ![JEE Main 2019 Physics, Center of Mass Momentum and Collision — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Center_of_Mass_Momentum_and_Collision/648b5a66417cc3fb48d665ea/question_1__q_648b5a66417cc3fb48d665ea__cdn-question-pool.getmarks.app__c9f6b946-3b16-4bbf-8a45-e0be90fb396b-image9__6e63d77d3a_final_ppt_sync.png)

Two particles of masses M and… - JEE Main 2019 Physics PYQ

JEE Main 2019 — Physics Mechanics

2019mcqhard

Two particles of masses $M$ and $2M are$ moving with speeds of $10 m{s}^{-1}$ and $5 m{s}^{-1}$ , as shown in the figure. They collide at the origin and after that they move along the indicated directions with speeds ${v}_{1}$ and ${v}_{2}$ , respectively. The values of ${v}_{1}$ and ${v}_{2}$ are, nearly

JEE Main 2019 Physics, Center of Mass Momentum and Collision — question figure

Official previous-year question

Held on 10 Apr 2019 · Verified 6 Jul 2026.

Options

A
$6.5 m{s}^{-1}$ and $3.2 m{s}^{-1}$
B
$3.2 m{s}^{-1}$ and $12.6 m{s}^{-1}$
C
$13.02 m{s}^{-1}$ and $19.7m{s}^{-1}$
D
$3.2 m{s}^{-1}$ and $6.3 m{s}^{-1}$

Did you get this right?

Solution

JEE Main 2019 Physics, Center of Mass Momentum and Collision — solution figure

From conservation of linear momentum,

Along $x$ -axis:

${\text{10Mcos30}}^{0}+5(\text{2M}cos{45}^{0})=\text{M}{v}_{1}cos{30}^{0}+2\text{M}{v}_{2}cos{45}^{0}$

$10 M \frac{\sqrt{3}}{2}+ 10M\frac{1}{\sqrt{2}}=M\frac{\sqrt{3}}{2} {v}_{1}+2M{v}_{2}\frac{1}{\sqrt{2}}$

$\frac{\sqrt{3}}{2} {v}_{1}+\sqrt{2}{v}_{2}=5(\sqrt{3}+\sqrt{2})$ … (1)

Along $y$ -axis:

$-10M\frac{1}{2}+10M\frac{1}{\sqrt{2}}=M{v}_{1}\frac{1}{2}-2M{v}_{2}\frac{1}{\sqrt{2}}$

$\frac{ {v}_{1}}{2}-\sqrt{2}{v}_{2}=5(\sqrt{2}-1)$ … (2)

From equations (1) and (2)

${v}_{1}(\frac{\sqrt{3}}{2}+\frac{1}{2})=5(\sqrt{3}+\sqrt{2})+5(\sqrt{2}-1)$

${v}_{1}=13.02 m/s$

${v}_{2}=\frac{5\sqrt{3}+5\sqrt{2}-13.02\times \frac{\sqrt{3}}{2}}{\sqrt{2}}$

$=19.72 m{s}^{-1}$

Your note

JEE Main Mechanics in other years

Mechanics 2026 Mechanics 2025 Mechanics 2024 Mechanics 2023 Mechanics 2022 Mechanics 2021 Mechanics 2020 Mechanics 2018 Mechanics 2017 Mechanics 2016 Mechanics 2015 Mechanics 2014

Keep practicing Mechanics

Work through every JEE Main Mechanics PYQ, year by year.

Browse all Physics topics

Solution

JEE Main 2019 Physics, Center of Mass Momentum and Collision — solution figure

From conservation of linear momentum,

Along $x$ -axis:

${\text{10Mcos30}}^{0}+5(\text{2M}cos{45}^{0})=\text{M}{v}_{1}cos{30}^{0}+2\text{M}{v}_{2}cos{45}^{0}$

$10 M \frac{\sqrt{3}}{2}+ 10M\frac{1}{\sqrt{2}}=M\frac{\sqrt{3}}{2} {v}_{1}+2M{v}_{2}\frac{1}{\sqrt{2}}$

$\frac{\sqrt{3}}{2} {v}_{1}+\sqrt{2}{v}_{2}=5(\sqrt{3}+\sqrt{2})$ … (1)

Along $y$ -axis:

$-10M\frac{1}{2}+10M\frac{1}{\sqrt{2}}=M{v}_{1}\frac{1}{2}-2M{v}_{2}\frac{1}{\sqrt{2}}$

$\frac{ {v}_{1}}{2}-\sqrt{2}{v}_{2}=5(\sqrt{2}-1)$ … (2)

From equations (1) and (2)

${v}_{1}(\frac{\sqrt{3}}{2}+\frac{1}{2})=5(\sqrt{3}+\sqrt{2})+5(\sqrt{2}-1)$

${v}_{1}=13.02 m/s$

${v}_{2}=\frac{5\sqrt{3}+5\sqrt{2}-13.02\times \frac{\sqrt{3}}{2}}{\sqrt{2}}$

$=19.72 m{s}^{-1}$

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

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