According to Newton's third law, to each action there is an equal and opposite reaction, which means that no matter how heavy is the object, if I give it X amount of force, it will give that force back to me. However, if I were standing on top of a skateboard and I throw a bowling ball with X amount of force, it will push me to the back more than if I threw something really light like a golf ball with that same force. Why is that?

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    $\begingroup$ Because the bowling ball is pushing back on you with the force that you push on it, and the golf ball is pushing back on you with the force that you push of it. $\endgroup$ – Chet Miller Feb 28 '16 at 21:36
  • $\begingroup$ Yes but lets say i am applying the same force on both balls. As a result, the golf ball will have more acceleration than the bowling ball $\endgroup$ – user3929076 Feb 28 '16 at 21:46
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    $\begingroup$ So? But you will have the same acceleration in the opposite direction if you apply the same force to each. $\endgroup$ – Chet Miller Feb 28 '16 at 22:16
  • $\begingroup$ @user3929076 Yes, and force and acceleration are related by the mass... nothing new there. For the same force, a higher mass means a lower acceleration and vice versa; or for the same acceleration, a higher mass requires a higher force and vice versa, etc. $\endgroup$ – Asher Feb 29 '16 at 4:25
  • $\begingroup$ What's the status of this question. There are several posted answers. Are you looking for additional clarification? $\endgroup$ – DanielSank Feb 29 '16 at 17:55

Suppose we have an object of mass $m$. You stand on a skateboard and throw that object as hard as you can. Suppose your arm can put out a maximum force of $F_\text{max}$.

When you push the object with your maximum force, its acceleration is $a = F_\text{max}/m$. The position of the object as a function of time during this acceleration is the usual

$$x = \frac{1}{2} a t^2 \, .$$

Your arm has only a certain length $L$ so you can only apply this force and get the object to experience that acceleration over a distance $L$. Therefore, the maximum time over which you can push the object is

$$t = \sqrt{2L / a} \, .$$

The final momentum of the bowling ball is therefore

$$p = F_\text{max} t = F_\text{max} \sqrt{\frac{2L}{a}} = \sqrt{2 L F_\text{max} m} \, . $$

So you see, the amount of momentum you can impart to an object goes proportional with the square root of that object's mass, the force you can exert, and the length of your arm.

Newton's action-reaction law says that the force exerted by you on the ball is equal to the force exerted by the ball on you. Note, of course, that these forces are exerted over equal amounts of time, so the thing that's really equal in the end is the momentum imparted onto you and the ball (in opposite directions). Since we see that the momentum you can impart on the ball increases with increasing ball mass, then the momentum exerted on you also increases with increasing ball mass.

Intuitively, this is all just saying that when you throw a golf ball, it's so light that it leaves your hand before it's had a chance to push back on you very much.

  • $\begingroup$ Very nice explanation sir. But this drove another question to my head. What if instead of throwing the balls, we assume that the balls are free falling from some distance right in front of you, and when they reach the same level as your face, you punch them with your maximum force $F_{max}$. The bowling ball will most likely crush your knuckles because, i imagine, it gives you that very same force, which will be to great for your bones to handle, while if you punch the golf ball you will send it flying. Why is that? Shouldn't the golf ball give you back the same force? $\endgroup$ – user3929076 Feb 29 '16 at 4:35
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    $\begingroup$ @user3929076 I can answer this but I hesitate to do so here. It's not a good idea for us to go through extensions of the original question here in the comments. We could go on and on where you bring up new questions and I try to answer them. This site works best when each post contains a single question and answers designed to address that one question. If you ask this specific question in another post, please put a link here and I can write an answer. $\endgroup$ – DanielSank Feb 29 '16 at 5:26

Short intuitive answer: because they have more inertia, and therefore will less easily modify their trajectory under interaction. So the interactor (if lighter) will have endorse most of the trajectory or deformation change, which translates by "strong reaction".


What a great Fermi problem! Consider a bowling ball of 16 pounds mass. If you toss it directly away from you, horizontally, at a height of 1.2 m, such that it lands 2 m from the release point. If you exert same force profile (impulse) to a golf ball at the same initial height, the golf ball would take the same time to fall, but would have a horizontal velocity (and consequently, distance) that scales like the square root of the ratio of masses (see @DanielSank s answer). The standard golf ball has a mass of about 1.6 ounces, (16 ounces per pound), the that scale factor is $\sqrt{160}= 12.6$. So you would have to throw the golf ball over 25 meters, not a trivial throw.

I believe you are underestimating what it takes to exert an equal force.

  • $\begingroup$ But assuming that the same force is applied on both objects, will the push back distance be the same? $\endgroup$ – user3929076 Feb 29 '16 at 4:28
  • $\begingroup$ If the impulses applied are identical, the push back following the imppulses will be the same. $\endgroup$ – Bill N Feb 29 '16 at 4:31

Newton's third law of motion, the action-reaction law, does not directly depend upon the mass of an object; it depends upon the force.

By Newton's second law of motion we have $\text{Force} = \delta \text{momentum}/\delta \text{time}$, and momentum is the product of the mass and velocity; thus a measure of force is the change in $mass \times velocity$ per unit of time; this idealizes as $F=dp/dt$.

If the mass is unchanged by the action, then $F=\text{mass} \ \times \ \text{acceleration}=ma$, so mass enters into the reaction force. You can obtain the same force with different masses: if the mass is doubled, only 1/2 the acceleration is required, but if the mass is halved, you must double the acceleration.

By including the distance over which the force is applied, you bring up the question of work performed, which is a separate question.

  • $\begingroup$ I understand, but lets say i gave both the golf and the bowling balls the exact same force. Thus giving the golf ball higher acceleration due to its small mass $\endgroup$ – user3929076 Feb 28 '16 at 21:48
  • $\begingroup$ There's more to this. It's in some sense the limited length of your arm that determines which object pushes you faster. $\endgroup$ – DanielSank Feb 28 '16 at 21:48

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