A fig. shows a frictionless hemispherical ball of radius 'R'.A ball of mass 'm' is pushed down from A it rises to pt. B.Find the speed with which ball is pushed down the wall.
the answer is √(2g(h-R)) but i am getting √(2g(R-h)). can you explain how?
In the fig. point B should be above point A.
hey tanvi its been given in the question that it has fallen down from A and has then gone to pt B.
and at the highest point only potential energy is there ,not the kinetic energy but at pt B both kinetic and potential energies are present.
shouldnt it be: 1/2mv^2+mgh=mgR?
If m is pushed down from A and there is no friction it will rise to a higher height. This means point B should be above point A.
mgR + 1/2*mu2 = mgh
u^2 = 2g( h-R )
u = [2g(h-R)]1/2
here , u = initial speed
this is done by energy conservation
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