An alpha- particle of mass m… - JEE Main 2019 Physics PYQ

JEE Main 2019 — Physics Mechanics

2019mcqhard

An alpha- particle of mass $m$ suffers $1-$ dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing $64%$ of its initial kinetic energy. The mass of the nucleus is

Official previous-year question

Held on 12 Jan 2019 · Verified 6 Jul 2026.

Options

A
$1.5m$
B
$4m$
C
$3.5m$
D
$5m$

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Solution

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

From momentum conservation,

$mv+0=-m{v}_{1}+M{v}_{2}...(1)$

Using kinetic energy conservation,

$\frac{1}{2}m{v}^{2}=\frac{1}{2}m{v}_{1}^{2}+\frac{1}{2}M{v}_{2}^{2} \Rightarrow \frac{1}{2}m{v}^{2}=0.36\times \frac{1}{2}m{v}^{2}+\frac{1}{2}M{v}_{2}^{2} \Rightarrow 0.64m{v}^{2}=M{v}_{2}^{2} \Rightarrow {v}_{2}=0.8v\sqrt{\frac{m}{M}}$

As it retain $36%$ , $\Rightarrow 0.36\times (\frac{1}{2}m{v}^{2})=\frac{1}{2}m{v}_{1}^{2}\Rightarrow 0.36{v}^{2}={v}_{1}^{2}$ Hence, ${v}_{1}=0.6v$

From $(1)$ , $mv=-m(0.6v)+M(0.8v\sqrt{\frac{m}{M}})$

$\Rightarrow 1=-0.6+0.8\sqrt{\frac{M}{m}}$

On solving, we get $M=4m$

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Solution

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

From momentum conservation,

$mv+0=-m{v}_{1}+M{v}_{2}...(1)$

Using kinetic energy conservation,

As it retain $36%$ , $\Rightarrow 0.36\times (\frac{1}{2}m{v}^{2})=\frac{1}{2}m{v}_{1}^{2}\Rightarrow 0.36{v}^{2}={v}_{1}^{2}$ Hence, ${v}_{1}=0.6v$

From $(1)$ , $mv=-m(0.6v)+M(0.8v\sqrt{\frac{m}{M}})$

$\Rightarrow 1=-0.6+0.8\sqrt{\frac{M}{m}}$

On solving, we get $M=4m$

Options

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Solution

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JEE Main Mechanics in other years

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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