a block of mass m1 fastened to a spring and d block of mass m2is placed against it on an inclined plane.the blocks are pushed to a dist of (2/k)(m1+m2)gsin@ against d spring and releasd . . find the position wher the 2 blocks separate?????

Dear Sneha,

ANS:-Let the velocity of the masses when the spring is again totally relaxesd is v, Then from the energy eq we get

1/2kd²=1/2(M1+M2)v²+(M1+M2)gdsinθ

Putting d=2/k(M1+M2)gsinθ & Solving we get v=0.

Hence the velocity is o, So, it will again compress the spring and let the comp is x then from energy eq we get,

(M1+M2)gsinθ x=1/2kx²

So again x=2(M1+M2)gsinθ/k

So, the two mass will execute Periodic motion with amplitude (M1+M2)gsinθ/k

THUS THEY WILL NEVER SEPARATE OUT

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