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If a ball is moving towards me, I can kick it further than if I were to kick it if it was stationary. But surely if the ball is moving in the opposite direction, it should take more force to kick it the same distance as I am accounting for the initial movement.

Does this have a theoretical explanation, or is it that I can perform better technique if it's moving?

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  • $\begingroup$ I deleted some comments that looked like they were attempting to answer the question, and their responses. $\endgroup$
    – David Z
    Jun 27, 2018 at 18:57

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The first comment above hits the answer right on the head, but I'll give some more details.

The interaction between your foot and the soccer ball is inelastic - which shouldn't be surprising because all real-life interactions are essentially inelastic, but it's more elastic then a collision between say, a concrete block and your foot, because of the construction of the ball. The ball is essentially made to store elastic energy.

To study the energy of this system, let's call the work you can do on the ball (no matter if it's moving or not) $W$. So, if the ball is stationary, you give the ball energy $W$, some energy is stored in the ball elastically when it deforms, but when it returns to shape the energy is converted to kinetic. So the kinetic energy of the ball is $W$.

If the ball is moving towards you, with kinetic energy $K$, and then you kick it, what happens? Well first, you have to make sure you kick it hard enough for it to turn around, so roughly $K>W$ (you should actually do momentum conservation for this, but let's assume the collision is essentially elastic). The instant the ball is stopped, the elastic energy of the ball is $U=K+W$ - all of the energy has gone into deforming the ball. But when the ball returns to shape, all of the elastic energy goes into kinetic, and so the final kinetic energy of the ball is

$$K_f=U=K+W$$

that is, more then the energy you could give it when you kicked it from a stationary position.

Compare this to kicking a concrete block, which does not have the ability to store energy elastically. If you stop the block, all the energy $K+W$ goes into heating/destroying the block - or actually, destroying your foot!

So there are certainly some effects I am ignoring, like the fact that the interaction doesn't happen instantaneously and there are some sources of energy which I am not keeping track of, but in essence this is what happens.

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    $\begingroup$ Energy is always conserved. It's mechanical energy that's not 100% conserved here. $\endgroup$
    – probably_someone
    Jun 27, 2018 at 17:34
  • $\begingroup$ Haha ok yes of course. I didn't want to bring that up because the work done by your foot includes some of that non-mechanical energy. But I'll edit! $\endgroup$
    – levitopher
    Jun 27, 2018 at 17:37
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I will give a simpler explanation. A ball hitting a wall, reverses direction and if friction is small, it is elastically scattered, i.e. keeps its speed. Momentum is conserved because the mass of the wall is very large . It is more complicated in real situations, see the answer here for details.

So, if your leg, where the ball hits, were a wall you need not move and the ball would be reflected with its incoming speed. The extra energy you provide with the kick increases its speed. Of course you are not a wall, you balance through your contact to the earth,and because you will move some and there will be friction,it will not really be an elastic collision , still your weight is many times larger than the ball's, and you can take advantage of this initially favorable situation to reflect the ball with a higher speed than it came in.

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  • $\begingroup$ friction only works in the direction perpendicular to the surface, and so is not applicable here (unless the ball is spinning). And conservation of momentum has nothing to do with the mass of the wall - momentum would be conserved also if the wall was extremely small, as long as the system was isolated. It's the structure of the ball which creates this effect, not the wall or the leg. $\endgroup$
    – levitopher
    Jun 29, 2018 at 14:11
  • $\begingroup$ @levitopher conservation of momentum is necessary to explain the reflex extra momentum, which is balanced by the very very small velocity given to the large mass of the wall. Ball hitting a wall is a classic conservation of momentum exercise. The answer here gives some numbers quora.com/… $\endgroup$
    – anna v
    Jun 29, 2018 at 16:23
  • $\begingroup$ It is true that deformation and elasticity will enter the final balance, but even with a solid billiard ball this would happen ( it would be hard on the leg :). $\endgroup$
    – anna v
    Jun 29, 2018 at 16:25
  • $\begingroup$ You said, "Momentum is conserved because the mass of the wall is very large", which is false. The wall does not move because the mass is very large. $\endgroup$
    – levitopher
    Jun 29, 2018 at 18:21
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    $\begingroup$ @levitopher well, maybe I should clarify by: when the mass is very large, momentum is conserved by an infinitesimally small velocity needed to conserve it. Theoretically the earth should move an infinitesimal amount if the wall is rigid on the earth . Otherwise, as in earthquakes, it would vibrate, dissipating the momentum in friction. $\endgroup$
    – anna v
    Jun 30, 2018 at 3:24
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If would certainly take more force and time (impulse force $F\Delta t$) to reverse the speed of the ball than to just accelerate it to the same final speed, since $m(V_{init}+V_{final}) \gt mV_{final}$.

And you can feel the difference in force (pain), if you kick a ball moving at you at high speed, provided that you kick it "the same way", or more precisely with the same speed, as you would kick a resting ball.

Whether the reversed ball moves further or not, depends on whether you are willing to tolerate the pain and whether can maintain the speed of your foot over the duration of the kick.

If you can, the reversed ball should move further. To see it, you can model your foot as a moving wall. If the wall was stationary, the ball would have bounce at about the same speed as it hit the wall (assuming low losses). If the wall is moving, the speed of the moving wall should be added to the final speed of the ball (consider the reference frame associated with the moving wall).

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  • $\begingroup$ This answer is essentially correct, but I think talking about impulse here is more confusing then talking about energy. What exactly the elastic force is from the ball on the foot and the reaction on the ball is complicated, which is why I choose to describe the ball as simply a device that stores elastic energy. $\endgroup$
    – levitopher
    Jun 27, 2018 at 17:05
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A short answer to this: I will treat the event of the ball being kicked as an elastic collision between two bodies (the foot (with the body attached) moving with initial velocity $u_1$, and the ball approaching with negative velocity $u_2$. The problem of elastic collisions of 2 particles is a problem from various physics courses, I won't derive it here, but I'll use the result for $v2$, the velocity of the football after the collision occured, given in Wikipedia:

\begin{align} v_{2}={\frac {u_{2}(m_{2}-m_{1})+2m_{1}u_{1}}{m_{1}+m_{2}}} \end{align}

The mass of the foot is bigger than the mass of the football, which makes the quantity $(m_2 - m_1)$ negative. As $u_2$ is negative as well, the velocity $v_2$ will increase for bigger absolute values of $u_1$.

Of course this argument is just an approximation to reality, where the foot is connected to the leg (and the leg to the body and so on). An easy way to model this would be to increase the mass of the foot, taking into account that it is attached to a leg and to a body.

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If a ball is moving toward you it will exert a force on you.If you kick it in opposite direction of its its inital direction it will require more force for same distance in case it was moving away from you.You hit it.Now the ball will cover same distance with less force than first case as by now there is already a force acting on ball in that particula direction with the force of your kick added.

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