Question

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 7.5 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? (a) Number Units (b) Number Units

Answer 1

Given

   basket ball of mass m , rolling from top of a hill to the bottom

   as it rolls down it would have both rotational and translational motion

   given that at the bottom the speed of the ball is v = 7.5 m/s

as there is no frictionaal loseese then the gravitiational potential energy at top will be completely converts into kinetic energy at bottom

that is mgh = 0.5 mv^2 ==> h = 0.5*v^2 = 0.5*7.5^2 = 28.125 m

b) if the ball was released from rest from the top of the hill ,the can of frozen juice is also having same velocity

   if there is any friction oo the surface, and the speed does not depend on mass of the body ,
and there is no much information about the radius of the ball and can so they can be treated as point mass then both have same speed at the bottom