JEE Main 2024PhysicsCenter of Mass Momentum and CollisionMediumNumerical

JEE Main 2024 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2024 (01 Feb Shift 1)

Question

The identical spheres each of mass 2M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is 42x, where the value of x is ________.

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Show full solutionCorrect answer: 3
Correct answer
3

Step-by-step explanation

Position vector r→COM=m1r→1+m2r→2+m3r→3 m1+m2+m3

r→COM =2M×0+2M×4i^+2M×4j^6M

⇒r→=43i^+43j^

|r→|=432+432=423

Therefore, x=3.

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About this question

This is a previous-year question from JEE Main 2024, 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.