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kanika kaul Grade: 12
        

 


A small ball of mass m is released at ht R above the surface of earth.If the maximum depth of the ball to which it goes is R/2 inside the earth through a marrow groove before coming to rest momentarily. The groovet, contains an ideal spring of spring constant k and natural length R find the value of  k if R is the radius of earth and M mass of earth?


 


                         


                                        


Ans: 7GMm/( R cube)


 

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

gravitational potential inside the earth is given by -3GM/2R [1-r2/3R2]


potential energy of ball when it was at a height of R above earth is-GMm/2R


kinetic energy at this instant is 0...


total energy = -GM/2R    .................1


finally ball agian comes to rest  , now ball is at a depth of R/2 from earth so


its gravitational energy is -3GMm/2R[1-r2/3R2]


                                  =-3GMm/2R[11/12]                             (r=R/2)


                                  =-11GMm/8R


potential energy stored in spring is KX2/2 = KR2/8                   (compression is R/2)


total energy now is sum of potential energy of spring + gravitational potential energy


                       TE = -11GMm/8R + KR2/8 ................2


total energy remains same so eq1 = eq 2


          KR2/8 -11GMm/8R = -GMm/2R


          KR2 /8 = 7GMm/8R      or


           K = 7GMm/R3      


approve my ans if u like

6 years ago
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