Question

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.0 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

Answer 1

V = 6m/s

a.) From conservation of energy K = U

Taking basketball as a hollow shower

=) 1/2 Iw^2 + 1/2 mV^2 = mgh

=) 1/2 (2/3mr^2)w^2 + 1/2mv^2 = mgh

=) 5/6 mv^2 = mgh

=) h = 5V^2/6g = 5×36/6×9.8 = 3.06m

b.) Since final speed depends only upon height of inclination h

Hence speed at bottom for frozen juice is also same aa basketball .

Hence V = 6m/s