Physics-
General
Easy

Question

Two rectangular blocks A blankand B blankof masses 2kg and 3 kg respectively are connected by spring of spring constant 10.8 N m to the power of negative 1 end exponentand are placed on a frictionless horizontal surface. The block A blankwas given an initial velocity of 0.15 m s to the power of negative 1 end exponent in the direction shown in the figure. The maximum compression of the spring during the motion is

  1. 0.01 m    
  2. 0.02 m    
  3. 0.05 m    
  4. 0.03 m    

The correct answer is: 0.05 m


    As the block A moves with velocity with velocity 0.15 m s to the power of negative 1 end exponent, it compresses the spring Which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A, i.e. 0.15 m s to the power of negative 1 end exponent. Let this velocity be v. Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage.

    According to the law of conservation of linear momentum, we get
    m subscript A end subscript u equals left parenthesis m subscript A end subscript plus m subscript B end subscript right parenthesis v
    Or v equals fraction numerator m subscript A end subscript u over denominator m subscript A end subscript plus m subscript B end subscript end fraction
    fraction numerator 2 cross times 0.15 over denominator 2 plus 3 end fraction equals 0.06 m s to the power of negative 1 end exponent
    According to the law of conservation of energy
    fraction numerator 1 over denominator 2 end fraction m subscript A end subscript u to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction open parentheses m subscript A end subscript plus m subscript B end subscript close parentheses 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
    fraction numerator 1 over denominator 2 end fraction m subscript A end subscript u to the power of 2 end exponent minus fraction numerator 1 over denominator 2 end fraction open parentheses m subscript A end subscript plus m subscript B end subscript close parentheses v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    fraction numerator 1 over denominator 2 end fraction cross times 2 cross times open parentheses 0.15 close parentheses to the power of 2 end exponent minus fraction numerator 1 over denominator 2 end fraction open parentheses 2 plus 3 close parentheses open parentheses 0.06 close parentheses to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    0.0225 minus 0.009 equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    o r blank 0.0135 equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    Or x equals square root of fraction numerator 0.0027 over denominator k end fraction end root equals square root of fraction numerator 0.0027 over denominator 10.8 end fraction end root blank equals 0.05 m

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