JEE Main 2018PhysicsCenter of Mass Momentum and CollisionHardMCQ

JEE Main 2018 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2018 (08 Apr)

Question

It is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss of its energy is Pd, while for its similar collision with a carbon nucleus at rest, the fractional loss of energy is Pc. The values of Pd and Pc are respectively

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Show full solutionCorrect option: B
Correct answer
B0.89, 0.28

Step-by-step explanation


Since, the linear momentum is conserved, u=V1+2V2.

e=1=V2-V1u (where e⇒the coefficient of restitution), u=V2-V1

V2=2u3 ;V1=-u3. Initial energy, =12u2. Final energy, =12×1×u29=u218. Fractional change =u22-u218u22=0.88



u=V1+12V2⇒1=V2-V1u⇒u=V2-V1
⇒V2=2u13⇒V1=-11u13.
Change in energy,=u2-11u132×100=0.294

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About this question

This is a previous-year question from JEE Main 2018, covering the Center of Mass Momentum and Collision chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.