Please explain this interesting question in the figure...The answer is (3).
Q. Starting from rest on her swing at initial height h 0 above the ground, Saina swings forward. At the lowest point of her motion, she grabs her bag that lies on the ground. Saina continues swinging forward to reach maximum height h 1 . She then swings backward and when reaching the lowest point of motion again, she simple lets go off the bag, which falls freely. Saina's backward swing then reaches maximum height h 2 . Neglecting air resistance, how are the three heights related ?


(1) h 0 > h 1 > h 2
(2) h 0 = h 1 = h 2
(3) h 0 > h 1 = h 2
(4) h 0 = h 2 > h 1

Dear Student,


let the mass of the Saina be M and that of bag be mat every point the system energy remain constant E=Mgh0whenpicks the bag the system mass increasesMgh0=M+mgh1h1=Mh0M+mclearly from above eqn h1<h0now when she return back she releases bag at lowest point .w system energy becomesE'=Mgh0-mgh1this energy takes upto height h2Mgh0-mgh1=Mgh2M+mh1-mh1=Mh2h2=Mh1M=h1hence h0>h2=h1Regards

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