if it starts 2 feet high is kinetic energy at the bottom will be m g h.
it will go that same height up the other ramp before it stops.
Now you did not say if it was a solid ball or whatever
If I assume it is solid the its moment of inertia is (2/5) m R^2
and its rotational kinetic energy is (1/2)I omega^2
since omega = v/R
the total kinetic energy = (1/2) m v^2 + (1/2) (2/5)m v^2
= (7/10) m v^2
so
m g h = (7/10) m v^2
v = sqrt (10 g h/7)
If ball is rolling from 2 ft down an inclined ramp at an angle of 45deg, what is the speed of the ball when it reaches the ground and how far will it go up a second ramp that is inclined at 30deg?
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