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    $\begingroup$ "An object in free fall is inertial and weightless" Careful. In Newtonian mechanics which is the best way to model this scenario, an object in freefall is not on an inertial path. It is accelerating under the force of gravity. In General Relativity you would say the object is on a geodesic path, but the surface of the Earth is not. It's a wholly different paradigm of mechanics that's totally unnecessary here. $\endgroup$ Commented 2 days ago
  • $\begingroup$ You're using language that throws a lot of ambiguity into the mix. You're referring to a ball in freefall, but then say that the hill contacts the ball. You're not mentioning that the hill is frictionless. Therefore, the touch must impart some kind of friction, which in turn means the ball is not in freefall as the imparted friction infringes on the "free" fall that the ball is supposedly in. Secondly, are you assuming that the hill is an unmalleable solid? $\endgroup$ Commented 2 days ago
  • $\begingroup$ In the hill's frame, the ball accelerates down but is also moving horizonally; I think that's where the intuition is mistaken. $\endgroup$ Commented yesterday
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    $\begingroup$ @Flater There's no theoretical problem with having the ball "just touch" the surface with zero normal or frictional force, whether the surface has a nonzero coefficient of friction or not. "Contact" is simply a matter of position and doesn't require a reaction force. Freefall just implies there is no force acting other than gravity, which there isn't for an object falling over a hill that perfectly matches its freefall trajectory. If there was another force acting, the trajectory by definition wouldn't be the freefall trajectory. It's clear gravity is the only force acting. $\endgroup$ Commented yesterday
  • $\begingroup$ "such that it contacts the ball" <-- The contact force is zero though because the hill doesn't accelerate the ball by making contact, rather the earth accelerates the ball with gravity and the path is unchanged by the hill because it "precisely matches the ball's path". I assume that when you say precisely matching the ball's path it implies that the hill exerts no additional force on the ball to change its path from what it would otherwise be due to gravity alone. Related question about contact force $\endgroup$ Commented yesterday