Topic: Conservation of linear momentum and applications· Laws of Motion

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is…

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed \(v\), the speed of the third (lighter) fragment is :

1. \(3\sqrt{2}\,v\)

2. \(v\)

3. \(\sqrt{2}\,v\)

4. \(2\sqrt{2}\,v\)

Doubtly Asked 30 Oct 2026 NEET UG 2022 Practice test neet Save WhatsApp Report

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Answer: 4. \(2\sqrt{2}\,v\)

The shell is at rest, so total momentum after explosion is zero. Masses are \(2m/5,\ 2m/5\) and \(m/5\).

The two equal fragments move at right angles with speed \(v\): resultant momentum \(=\sqrt2\cdot\dfrac{2m}{5}v\).

The third fragment must cancel it: \(\dfrac{m}{5}V=\dfrac{2\sqrt2\,m v}{5}\Rightarrow V=2\sqrt2\,v\).

Option (4).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is…

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