When a ball is thrown onto a wall, the ball exerts a force onto the wall. According to Newton's Third Law, the wall will exert an equal and opposite force to the ball. Thus, how would the ball be able to bounce back? Shouldn't it just slide down the wall?
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$\begingroup$ Consider that generally, the wall is a fixed object, and the ball is not. So the reaction force from the wall pushing on the ball has far less resistance, hence bounces. $\endgroup$– user60063Commented Oct 2, 2014 at 0:21
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$\begingroup$ Related: physics.stackexchange.com/questions/132605 $\endgroup$– Kyle KanosCommented Oct 2, 2014 at 0:38
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$\begingroup$ That entirely depends on the elasticity of the "ball". If it's perfectly elastic, it will bounce back, if it's perfectly inelastic, it will stick to the wall and slide down. That's pretty much the physics behind "egging". $\endgroup$– CuriousOneCommented Oct 2, 2014 at 2:59
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$\begingroup$ Think of it as an elastic collision between the ball and the wall(which is connected to Earth). Since the mass of the latter is so great, the ball would make the wall move by less than an atom's width (but still non zero, else it would violate the conservation of momentum). $\endgroup$– t.cCommented Oct 2, 2014 at 7:46
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$\begingroup$ Possible duplicate of Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground? $\endgroup$– knzhouCommented Apr 22, 2019 at 12:36
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3 Answers
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You have to consider this as a dynamic process. To make sense of it you also have to think of the ball as NOT a point particle, so it has a centre of gravity, probably in its actual centre. As the surface of the ball impacts the wall the centre of gravity slows down, which means its momentum drops. (It also means that the wall or the ball or both must deform during the impact) Newton's second law requires that momentum can only change if there is an external force applied and this comes from the wall. This force increases from zero as the impact proceeds, it is exerted in the opposite direction of the balls motion. At some point the momentum will reach zero, but the force is still there and still directed outwards so now it begins to accelerate the ball's centre of gravity back in the opposite direction. As it moves out the balls momentum increases and the force drops but still points in the same direction. So the ball accelerates until the surfaces separate at which point the force is back to zero.
Now it is true that Newton3 requires that throughout this process the ball is exerting an equal and opposite force on the wall but the point here is that these two forces are exerted on DIFFERENT objects. So the force of the ball on the wall acts to try and increase the momentum of the wall (if it was somehow moveable it would accelerate away from the impact). For a fixed wall the force goes to make some deformation of the wall (bending it / making a dent in it etc.)
If both objects are perfectly elastic, meaning they return to their original shape once forces are removed, then the ball will rebound with exactly the same but opposite momentum it had before the impact. If either object is permanently deformed in the impact, the ball will rebound with less momentum.
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$\begingroup$ ...but the force is still there...: but why is the force still there if the ball has no more momentum? because of the wall's (or ball's) deformation rebounding? Please explain. $\endgroup$ Commented Nov 20, 2018 at 18:17
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2$\begingroup$ The force is still there because the ball ( and the wall ) are elastically deformed. The atoms within both objects have been displaced from their equilibrium positions so the inter-molecular forces do not cancel out at that point and both will operate to return the objects to equilibrium. $\endgroup$ Commented Dec 6, 2018 at 19:07
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$\begingroup$ thanks for your response (on such an old question!), I think I've cleared up what's been bothering me; see my posted answer. $\endgroup$ Commented Dec 8, 2018 at 18:19
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The forces don't anulate each other because they are not on the same body. The ball creates a force F on the wall. Then, according to the Newton's Third Law, the wall will create a force N with the same module of F but with oposite direction. Pay attention that F is on the wall and N is on the ball, that's why you can't sum them.
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$\begingroup$ Good answer. A diagram would really help! $\endgroup$ Commented Sep 7, 2015 at 22:10
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1$\begingroup$ The opposing force on the ball should be just enough to stop it (counteract it's forward inertia), not to send it backwards... $\endgroup$ Commented Nov 20, 2018 at 17:54
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(The other answers have focused on the fact that the equal and opposite forces of Newton's law are not acting on the same object, rather the ball on the wall and the wall on the ball. However, as I see it, the question already takes that for granted, but wants to know why that opposite force is enough to rebound the ball and not just stop its motion)
The question assumes that because the ball is moving with a momentum of $mv$ (mass-of-ball x velocity), the force it applies to the wall is also $mv$. However, this is incorrect. The actual force given is $2mv$, so the 'equal and opposite force' on the ball will be $-2mv$ - just enough to stop the ball and send it back at the same speed.
Keep in mind that when we say $2mv$ is applied to the wall, the $m$ is the mass of the ball, so relative to the mass of the wall (and the earth it's attached to) the force $2mv$ is tiny and will only move the wall/earth an immeasurable amount.
To be really accurate, the force is not exactly $2mv$/$-2mv$, because that would be a violation of conservation of energy - if the ball rebounds at the same speed, how could any energy be transferred to the wall? Instead, it's a bit less than $2mv$; The exact value has to be calculated to obey both the rule of conservation of momentum and conservation of energy. (This point has been discussed further here).
The fact the the force given is not the same as the momentum of the ball may be surprising, but because energy and momentum are conserved no laws of physics are violated. [Here's a hint: from the perspective (or: rest frame) of the ball, the wall is rushing towards it with a tremendous momentum of $2Mv$ ($M$ being the mass of the wall/earth) so it would expect to fly off at tremendous speed, upon collision. Obviously the force does not necessarily correspond to the momentum.]
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$\begingroup$ Why is it $F=2mv$ and not $F \Delta t=m \Delta v = \Delta p$? On the physics hypertext book it defines Impulse as $F \Delta t$. I had originally conceived of Impulse as $F _{tot}$, which I derived from $\int_{0}^{f}Fdt = \int_{0}^{f}madt = m\int_{0}^{f}adt = m(v(f)-v(0)) = m \Delta v$ where 0 and f are measurements of time, m is constant, and F, v, and a are actually functions of time. $\endgroup$– phlaxyrCommented Oct 6, 2020 at 4:16
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1$\begingroup$ @phlaxyr The force in Newton's 2nd law refers to that applied to the 2nd object i.e. $F=MA$ not $F=ma$; if we use the law of elastic collisions for a stationary $M$ (see en.wikipedia.org/wiki/Elastic_collision) the velocity gained by $M$ is ${2m \over {m+M}}v$, and we can substitute it for $A$ in $F=MA$. When $m$ is small and $M$ very large then $M$ and $m+M$ almost cancel out, leaving approximately $F=2mv$. Does this clarify things? $\endgroup$ Commented Oct 12, 2020 at 8:45
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$\begingroup$ Oh, so if I'm understanding correctly, $F=2mv$ is for elastic collisions? $\endgroup$– phlaxyrCommented Oct 13, 2020 at 2:24
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$\begingroup$ @phlaxyr - [*BTW, I intended Newton's 3rd law, not 2nd]. $F = 2mv$ for *totally elastic collisions (in reality there are no totally rigid objects, but a massive object like a wall anchored to earth relative to a ball is is pretty close). You need to work it out for other cases. If a ball of mass $m$ hits a ball of equal mass, then $F = mv$ and the first ball will just stop (by Newton's 3rd law), and the second ball will continue with velocity $v$ as demonstrated by Newton's cradle - en.wikipedia.org/wiki/Newton%27s_cradle. $\endgroup$ Commented Oct 13, 2020 at 7:04