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rajan jha Grade: 12
        

A tunnel is dug in the earth across one of its diameter. two masses m and 2m are dropped from the two ends of the tunnel. The masses collide and stick to each other. they perform SHM, the amplitude of which is ----------------- 

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

gravitational potential at center  = -3/2 (GMe/Re)


gravitational potential at surface = -GMe/Re


accleration of each particle = -GMx/Re3 , this is independent of mass so both of these particle will meet at center of earth...


applying energy conservation for mass m


 mV2/2 - 3/2(GMem/Re) = -GMem/Re 


here in the above eq mass cancels out so velocity is independent of mass of particle


 Vm = V2m = V = (GMe/Re)1/2                    ................1


now applying momentam conservation at center of earth


-mV + 2mV = (m+2m)V1                           (due to opposite direction one of the velocity is -ve)


  3mV1 = mV


      V1 = V/3        ...............2   


now this combined mass will perform SHM because force acting on this mass is -GMe3mx/Re3


    F=3m(a)


   a (accleration)= -GMex/Re3


     -W2 x = -GMex/Re3                                  (a = -w2x)


      W = (GMe/Re3)1/2                .........3


now maximum velocity of particle is at mean position which is given by eq 2 ,


 Vmax = V/3


 Vmax = AW                             (from eq of shm)


    AW = V/3


    A = V/3W                           (A is amplitude of SHM)


from eq3 & eq1


  A = Re/3


thus ampitude will be 1/3rd times radius of earth


 

6 years ago
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