The actual force experienced by the ball and earth on impact depends on the nature of the impact (elastic or inelastic} as well as the stopping distance/time of the ball. The acceleration of the earth is so small due to its large mass that only the acceleration of the ball is observed. To determine the effect of the equal and opposite force on each object you need to apply Newton’s second law $F=ma$ to each object individually. Yes the ball and the floor (Earth) exert equal and opposite forces on one another per Newton’s third law but they don't "cancel" each other. The ball should be net ZERO ( $X + (-X) = 0$). Minus indicating opposite direction), then the total force acting on If the ball (which has a constant force when it hits the ground ( $X$))Įxperiences the same constant force in the opposite direction ( $-X$, The force of the floor on the ball is electromagnetic in nature and results in elastic or plastic deformation, and must be larger than X for some time interval unless the ball crashes through the floor. You don't observe that, so you can conclude it isn't true. If it was X, the ball would keep moving at constant velocity. The force which the floor exerts upward on the ball cannot be X. I believe that is the big mistake you are making. While the interaction of the ball with Earth manifests in a constant force on the ball, that force has very little to do with the force magnitude between the floor and ball, beyond the velocity which results from the acceleration of $g$. A force results from an interaction of two things (ball/Earth or ball/floor) and acts on an object. Objects do not have or possess or carry force. If the ball (which has a constant force when it hits the ground (X)) experiences the same constant force in the opposite direction (−X, minus indicating opposite direction), then the total force acting on the ball should be net ZERO ( X+(−X)=0). Second, Newton's 3rd Law (N3L) is always in effect the ball is pulling on the Earth while the ball falls. Now when the ball makes contact with the ground, Newtons 3'rd Law takes effectįirst, $X=mg$, force of Earth's mass pulling on the ball. Moves to the ground with a constant force of say (X). Shouldn't the ball just stay on the ground? Bouncing back means a force greater than (-X) was applied to the ball giving it upward motion. So there are no net forces acting on the ball, so why does it BOUNCE BACK? What am I missing? If the ball (which has a constant force when it hits the ground ( $X$)) experiences the same constant force in the opposite direction ( $-X$, minus indicating opposite direction), then the total force acting on the ball should be net ZERO ( $X + (-X) = 0$). If object A (the ball) exerts a force on object B (the floor), then object B will exert an equal force on object A in the opposite direction. Now when the ball makes contact with the ground, Newtons 3'rd Law takes effect (no air resistance): Gravity acts and this ball as it has mass and then the ball now moves to the ground with a constant force of say ( $X$). Take a ball that's been dropped to the ground.
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