Problem — A ball descending three tracks

Problem

Descent of the ball. A ball is released along three different tracks, all starting from the same height $h$ and ending at the same low level: (A) a frictionless incline; (B) a curved “roller-coaster” track; (C) a free vertical fall. In which case does the ball reach the ground with the greatest speed? Explain based solely on conservation of mechanical energy.

Solution

In all three cases there is no friction, so mechanical energy is conserved. The ball starts from rest at the same height $h$ and reaches the same low level, so the change in potential energy is identical: $\Delta E_p = m\,g\,h \quad\text{(the same for A, B, C)}$ This energy is converted entirely into kinetic energy at the bottom: $m\,g\,h = \frac{1}{2}\,m\,v^2 \quad\Rightarrow\quad v = \sqrt{2gh}$ The result does not depend on the shape of the track nor on the path followed: it depends only on the height drop $h$. So the ball reaches the ground with the same speed in all three cases (A, B and C).

Note: it is the magnitude of the speed that is equal, not the time taken nor the direction of arrival. The vertical fall (C) is the fastest, but the final speed’s magnitude is identical. $\ev{v_A = v_B = v_C = \sqrt{2gh}}$

Links

Topics: Work and energy Concepts: Conservation of mechanical energy Skills: Analysis of limiting cases and thought experiments Objects: Inclined plane