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Question

Two blocks of equal mass m are connected by an Un stretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force F is applied on the first block pulling it away from the other as shown in figure. If the extension of the spring is X subscript 0 end subscript at time t, then the displacement of the first block at this instant is :

  1. fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator F t to the power of 2 end exponent over denominator 2 m end fraction plus x subscript 0 end subscript close parentheses    
  2. negative fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator F t to the power of 2 end exponent over denominator 2 m end fraction plus x subscript 0 end subscript close parentheses    
  3. fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator F t to the power of 2 end exponent over denominator 2 m end fraction minus x subscript 0 end subscript close parentheses    
  4. open parentheses fraction numerator F t to the power of 2 end exponent over denominator 2 m end fraction plus x subscript 0 end subscript close parentheses    

The correct answer is: fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator F t to the power of 2 end exponent over denominator 2 m end fraction minus x subscript 0 end subscript close parentheses

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Two blocks of equal mass m are connected by an unstretched spring, and the system is kept at rest on a frictionless horizontal surface. A constant force F is applied on the first block pulling it away from the other as shown in figure Then the displacement of the centre of mass at time t is:-

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physics-General
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If net force on a system in a particular direction is zero (say in horizontal direction) we can apply: sum m subscript R end subscript x subscript R end subscript equals sum m subscript L end subscript x subscript L end subscript comma sum m subscript R end subscript v subscript R end subscript equals sum m subscript L end subscript v subscript L end subscript text  and  end text sum m subscript R end subscript a subscript R end subscript equals sum m subscript L end subscript a subscript L end subscript Here R stands for the masses which are moving towards right and L for the masses towards left, x is displacement, v is velocity and a the acceleration (all with respect to ground). A small block of mass m = 1 kg is placed over a wedge of mass M = 4 kg as shown in figure. Mass m is released from rest. All surfaces are smooth. Origin O is as shown. Final velocity of the wedge is m/s :–

If net force on a system in a particular direction is zero (say in horizontal direction) we can apply: sum m subscript R end subscript x subscript R end subscript equals sum m subscript L end subscript x subscript L end subscript comma sum m subscript R end subscript v subscript R end subscript equals sum m subscript L end subscript v subscript L end subscript text  and  end text sum m subscript R end subscript a subscript R end subscript equals sum m subscript L end subscript a subscript L end subscript Here R stands for the masses which are moving towards right and L for the masses towards left, x is displacement, v is velocity and a the acceleration (all with respect to ground). A small block of mass m = 1 kg is placed over a wedge of mass M = 4 kg as shown in figure. Mass m is released from rest. All surfaces are smooth. Origin O is as shown. Final velocity of the wedge is m/s :–

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