Physics-
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Question

A particle of mass m moving with a velocity u makes an elastic one dimensional collision with a stationary particle of mass m establishing a contact with it for extremely small time T. Their force of contact increases from zero to F subscript 0 end subscript linearly in time T divided by 4, remains constant for a further time T divided by 2 and decreases linearly from F subscript 0 end subscript to zero in further time T divided by 4 as shown. The magnitude possessed by F subscript 0 end subscript is

  1. fraction numerator m u over denominator T end fraction    
  2. fraction numerator 2 m u over denominator T end fraction    
  3. fraction numerator 4 m u over denominator 3 T end fraction    
  4. fraction numerator 3 m u over denominator 4 T end fraction    

The correct answer is: fraction numerator 4 m u over denominator 3 T end fraction


    Change in momentum = Impulse
    = Area under force-time graph
    therefore m v equals Area of trapezium
    rightwards double arrow m v equals fraction numerator 1 over denominator 2 end fraction open parentheses T plus fraction numerator T over denominator 2 end fraction close parentheses F subscript 0 end subscript rightwards double arrow m v equals fraction numerator 3 T over denominator 4 end fraction F subscript 0 end subscript rightwards double arrow F subscript 0 end subscript equals fraction numerator 4 m u over denominator 3 T end fraction

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