JEE Main 2022PhysicsCenter of Mass Momentum and CollisionHardNumerical

JEE Main 2022 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2022 (25 Jul Shift 2)

Question

Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be x m. The value of x is

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Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

The diagram represents the locations of the masses as mentioned in the question.

For point 1, r1→=0i^+0j^ for point 2, r2→=3i^+0j^ and for point 3, r3→=0i^+3j^.

The formula for centre of mass is

r→com=m1r1→+m2r2→+m3r3→m1+m2+m3

⇒r→com=M0i^+0j^+M3i^+M3j^3M

⇒r→com=i^+j^

r→com=12+12=2=x

⇒x=2

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About this question

This is a previous-year question from JEE Main 2022, 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.