JEE Main 2019PhysicsCenter of Mass Momentum and CollisionMediumMCQ

JEE Main 2019 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2019 (12 Jan Shift 1)

Question

The position vector of the center of mass r→cm  of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

  

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Show full solutionCorrect option: A
Correct answer
Ar→cm=138Lx^+58Ly^

Step-by-step explanation

The position vector of center of mass r→cm is given as 

r→cm=Xcmx^+Ycm y^

Where, Xcm=x-coordinate of center of mass  and Ycm=y-coordinate of center of mass

r→cm=m1x1+m2x2+m3x3m1+m2+m3 x^ +m1y1+m2y2+m3y3m1+m2+m3 y^

r→cm=2m×L+m×2L+m5L22m+m+mx^+2m×L+m×L2+m×02m+m+my^

r→cm=13L8x^+5L8y^

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About this question

This is a previous-year question from JEE Main 2019, 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.