Physics-
General
Easy

Question

Two blocks A and B each of mass m are connected by a massless spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in figure. A third identical block C also of mass m moves on the floor with a speed v along the line joining A and B and collides with A. Then

  1. The kinetic energy of the A-B system at maximum compression of the spring is zero    
  2. The kinetic energy of the A-B system at maximum compression of the spring is m v to the power of 2 end exponent divided by 4    
  3. The maximum compression of the spring is v square root of m divided by K end root    
  4. The maximum compression of the spring is v square root of fraction numerator m over denominator 2 K end fraction end root    

The correct answer is: The kinetic energy of the A-B system at maximum compression of the spring is m v to the power of 2 end exponent divided by 4


    Let the velocity acquired by A and B be V, then
    m v equals m V plus m V rightwards double arrow V equals fraction numerator v over denominator 2 end fraction
    Also fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction m V to the power of 2 end exponent plus fraction numerator 1 over denominator 2 end fraction m V to the power of 2 end exponent plus fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    Where x is the maximum compression of the spring. On solving the above equations, we get x equals v open parentheses fraction numerator m over denominator 2 k end fraction close parentheses to the power of 1 divided by 2 end exponent
    At maximum compression, kinetic energy of the
    A – B system equals fraction numerator 1 over denominator 2 end fraction m V to the power of 2 end exponent plus fraction numerator 1 over denominator 2 end fraction m V to the power of 2 end exponent equals m V to the power of 2 end exponent equals fraction numerator m v to the power of 2 end exponent over denominator 4 end fraction

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