Physics-
General
Easy

Question

Two balls each of mass m are placed on the vertices A and B of an equilateral triangle A B C of side 1m. A ball of mass 2m is placed at vertex C. The centre of mass of this system from vertex A (located at origin) is

  1. open parentheses fraction numerator 1 over denominator 2 end fraction m comma fraction numerator 1 over denominator 2 end fraction m close parentheses    
  2. open parentheses fraction numerator 1 over denominator 2 end fraction m comma square root of 3 blank m close parentheses    
  3. open parentheses fraction numerator 1 over denominator 2 end fraction m comma fraction numerator square root of 3 over denominator 4 end fraction m close parentheses    
  4. open parentheses fraction numerator square root of 3 over denominator 4 end fraction m comma fraction numerator square root of 3 over denominator 4 end fraction m close parentheses    

The correct answer is: open parentheses fraction numerator 1 over denominator 2 end fraction m comma fraction numerator square root of 3 over denominator 4 end fraction m close parentheses


    The centre of mass is given by

    stack x with minus on top equals fraction numerator m subscript 1 end subscript x subscript 1 end subscript plus m subscript 2 end subscript x subscript 2 end subscript plus m subscript 3 end subscript x subscript 3 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript plus m subscript 3 end subscript end fraction
    stack x with minus on top equals fraction numerator m cross times 0 plus m cross times 1 plus 2 m cross times open parentheses fraction numerator 1 over denominator 2 end fraction close parentheses over denominator m plus m plus 2 m end fraction
    stack x with minus on top equals fraction numerator 2 m over denominator 4 m end fraction equals fraction numerator 1 over denominator 2 end fraction m
    stack y with minus on top equals fraction numerator m subscript 1 end subscript y subscript 1 end subscript plus m subscript 2 end subscript y subscript 2 end subscript plus m subscript 3 end subscript y subscript 3 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript plus m subscript 3 end subscript end fraction
    stack y with minus on top equals fraction numerator m cross times 0 plus m cross times 0 plus 2 m cross times square root of 3 divided by 2 over denominator m plus m plus 2 m end fraction
    equals fraction numerator square root of 3 over denominator 4 end fractionm
    therefore Centre of mass is open parentheses fraction numerator 1 over denominator 2 end fraction m comma fraction numerator square root of 3 over denominator 4 end fraction m close parentheses.

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