---
title: "A small cart of mass \\(m\\) is released from rest at a variable height \\(h\\) above the lowest point of a track that forms a vertical circular loop of radius \\(R\\), as shown. Friction and air resistance are negligible, and the track is designed such that it can only exert an inward normal force on the cart (\\(F_N \\ge 0\\)). In the limiting case as the release height approaches the theoretical minimum value \\(h_{\\min}\\) required for the cart to navigate the loop without losing contact at the apex, which of the following correctly identifies \\(h_{\\min}\\) and the kinetic energy \\(K_{\\text{apex}}\\) of the cart at the apex, along with the correct physical justification?"
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url: "https://nerd-notes.com/ubq/124200/"
date_modified: "2026-09-28T13:31:16+00:00"
---

# A small cart of mass \(m\) is released from rest at a variable height \(h\) above the lowest point of a track that forms a vertical circular loop of radius \(R\), as shown. Friction and air resistance are negligible, and the track is designed such that it can only exert an inward normal force on the cart (\(F_N \ge 0\)). In the limiting case as the release height approaches the theoretical minimum value \(h_{\min}\) required for the cart to navigate the loop without losing contact at the apex, which of the following correctly identifies \(h_{\min}\) and the kinetic energy \(K_{\text{apex}}\) of the cart at the apex, along with the correct physical justification?

A small cart of mass \(m\) is released from rest at a variable height \(h\) above the lowest point of a track that forms a vertical circular loop of radius \(R\), as shown. Friction and air resistance are negligible, and the track is designed such that it can only exert an inward normal force on the cart (\(F_N \ge 0\)). In the limiting case as the release height approaches the theoretical minimum value \(h_{\min}\) required for the cart to navigate the loop without losing contact at the apex, which of the following correctly identifies \(h_{\min}\) and the kinetic energy \(K_{\text{apex}}\) of the cart at the apex, along with the correct physical justification?

![A schematic diagram of a frictionless track viewed from the side. On the left, a curved ramp descends from an elevated release point to the ground level. At the top of the ramp, a small solid rectangular cart labeled \(m\) rests at a vertical height indicated by a vertical double-headed dimension line labeled \(h\). The ramp connects smoothly to a flat horizontal section at ground level, which then transitions into a vertical circular loop of radius \(R\). A dashed vertical line extends from the geometric center of the circular loop to the highest point of the loop, labeled \(R\). A dashed horizontal line indicates the ground level beneath the circular loop. The entire track is drawn as a smooth solid curve. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790602276-zxhH5O.jpg)

- **A.** \(h_{\min} = 2R\) and \(K_{\text{apex}} \to 0\), because the cart only requires sufficient initial mechanical energy to reach the gravitational potential energy of the apex, where its velocity can momentarily reach zero without losing contact.
- **B.** \(h_{\min} = \dfrac{5}{2}R\) and \(K_{\text{apex}} \to \dfrac{1}{2}mgR\), because the inward normal force approaching zero requires gravity alone to supply the downward centripetal acceleration of magnitude \(g\).
- **C.** \(h_{\min} = \dfrac{5}{2}R\) and \(K_{\text{apex}} \to \dfrac{1}{2}mgR\), because the normal force approaching zero requires the net force on the cart to vanish so that it is in translational equilibrium at the apex.
- **D.** \(h_{\min} = 3R\) and \(K_{\text{apex}} \to mgR\), because the track must exert an inward normal force equal to the cart's weight to keep the cart moving along the circular path.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/124200/*