1. 1/9 2. 8/9 3. 4/9 4. 5/9

Answer: 2. 8/9 Use the formula for the final speed of a body in a head-on elastic collision with a body at rest. v_A=m₁-m₂/m₁+m₂u=4m-2m/4m+2mu=u/3 . Kinetic energy is proportional to v² , so the final energy of A is (1/3)²=1/9 of the initial. Fraction lost =1-1/9=8/9 .

Topic: Elastic and inelastic collisions in one and two dimensions· Work, Energy and Power

Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is :

1. \(\frac{1}{9}\)

2. \(\frac{8}{9}\)

3. \(\frac{4}{9}\)

4. \(\frac{5}{9}\)

Doubtly Asked 30 Oct 2026 NEET UG 2019 Practice test neet Save WhatsApp Report

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Answer: 2. \(\frac{8}{9}\)

Use the formula for the final speed of a body in a head-on elastic collision with a body at rest.

\(v_A=\dfrac{m_1-m_2}{m_1+m_2}u=\dfrac{4m-2m}{4m+2m}u=\dfrac u3\).

Kinetic energy is proportional to \(v^2\), so the final energy of A is \(\left(\tfrac13\right)^2=\tfrac19\) of the initial.

Fraction lost \(=1-\tfrac19=\tfrac89\).

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

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Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is :

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