Question

A ball-bearing rolls down a smooth ramp, continues rolling on a flat surface, and then goes up a second ramp, identical to the first one. The ball-bearing starts at height h. On the horizontal surface it moves at speed v. Assume there is no significant friction between the ball-bearing and the surfaces.

At what height on the second ramp is the speed of the ball bearing going to be equal to v/3?

Answer 1

Initial height of the ball-bearing = h

Speed of the ball-bearing at the flat surface = v

Suppose mass of the ball bearing = m

So, from conservation of energy -

loss in potential energy = gain in kinetic energy

=> m*g*h = (1/2)*m*v^2

=> v^2 = 2*g*h

=> v = sqrt(2*g*h)

Now, suppose, at height h', speed of ball bearing is v/3.

So, again apply conservation of energy -

Total energy at the flat surface = total energy at a height h'

=> (1/2)*m*v^2 = (1/2)*m*(v/3)^2 + m*g*h'

=> 0.5*m*v^2 - 0.056*m*v^2 = m*g*h'

=> 0.444*v^2 = g*h'

put the value of v^2 -

=> 0.444*2*g*h = g*h'

=> h' = 0.888*h

Therefore, at a height of 0.888h on the second ramp, speed of the ball bearing will be v/3. (Answer)