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I remember the first time I saw this formula, $F = ma$, and it seemed perfectly correct and natural to me (in the sense that it is very straight-forward and logical, not that it is easy by any means). But then, I wondered, what if an object is moving with a constant speed? Maybe a car. If it travels with constant speed and hits me, it still hurts. So it's force must not be $0$.

One more thing related to this problem. If the same car is pulling another car behind it, if the car goes with constant speed, it implies that the first car doesn't exert any force to the second car at that moment, since it is not technically pulling it anymore (espacially if we don't take into consideration the friction).

So my question would be the following: How do these two phenomenons describe the same law of motion?

PS: If gravity wasn't behaving as some kind of acceleration pulling me constantly to the Earth's core, then if I jumped upwards, would I start slowly getting away from Earth?

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  • $\begingroup$ Acceleration is a change of velocity. When the car hits you, does your velocity change? Does its? $\endgroup$ – J... Dec 22 '19 at 14:20
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You are confused about what the variables mean in Newton's second law.

Newton's second law is $$\mathbf F_{net}=m\mathbf a$$

What this means is that a net force (the sum of all forces acting on an object) produces an acceleration. This acceleration is determined by how strong the force is as well as the mass of the object.

What Newton's second law does not mean is that if an object has an acceleration $\mathbf a$ then it will produce a force consistent with $\mathbf F=m\mathbf a$ onto another object. All we know of an object with an acceleration $\mathbf a$ is that the net force is acting on this object consistent with $\mathbf F=m\mathbf a$.

Certainly, if a car hits you it will exert a large force on you, which will then cause you to have a large acceleration.

In your second example, $\mathbf F_\text{net}=0$ does not imply there are no forces present at all. If the front car is pulling another car at a constant velocity, this just means that the total forces acting on the system is $0$. But $\sum_ i0$ is not the only way to add up numbers to get to $0$

This is a great lesson in using equations in physics. You have to know what each variable in the equation means before you start trying to understand how to apply that equation to physical systems. Mathematically there is nothing wrong with saying "more acceleration means more force with constant mass". But what force and what acceleration are we talking about here? Without knowing the physical meaning of the equation, you could essentially say anything.

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I wondered, what if an object is moving with a constant speed? Maybe a car. If it travels with constant speed and hits me, it still hurts. So it's force must not be 0

Just to be clear, prior to impacting you the car travels with constant speed because the net force acting on the car is zero, not because there are no forces acting on the car. The force that the drive train applies to the drive wheels of the car to propel it forward equals the sum of all the dissipative forces (air drag, rolling resistance, mechanical friction of the moving parts, etc.) acting on the car in the opposite direction, for a net force of zero.

Once the car hits you, it is no longer traveling at constant speed. There will be a change in its velocity, $\Delta v_{M}$ and momentum, $M\Delta v_{M}$, where $M$ is the mass of the car and $\Delta v_{M}$ is the change in its velocity. In order for this to happen you had to exert a force on the car to decelerate it and it exerted an equal and opposite force on you per Newton's third law to accelerate you.

The change in momentum is called impulse which in turn equals the average impact force applied to the car times the duration of the impact, $\Delta t$, or

$$F_{average}\Delta t = M\Delta v_{M}$$

$$F_{average}=M\frac{\Delta v_{M}}{\Delta t}$$

Per Newton's third law, the object applies an equal and opposite impact force on you, resulting in a change in your momentum and accelerating you as well,

$$F_{average}=m\frac{\Delta v_{m}}{\Delta t}$$

where $m$ is your mass and $\Delta v_{m}$ is the change in your velocity.

One more thing related to this problem. If the same car is pulling another car behind it, if the car goes with constant speed, it implies that the first car doesn't exert any force to the second car at that moment, since it is not technically pulling it anymore (espacially if we don't take into consideration the friction).

If the car being pulled is going at constant speed it simply means that the net force acting on the car being pulled is zero. That does not mean the first car doesn't exert a force on the second car. It simply means that the force exerted by the first car equals the sum of all the dissipative forces acting on the second car in the opposite direction. It is the same reason the velocity of the first car was constant before the collision.

So my question would be the following: How do these two phenomenons describe the same law of motion?

They describe the same law of motion when Newton's second law is stated in terms of momentum. That is, the net external force equals the change in momentum of a system divided by the time over which the momentum changes, or

$$F_{net}=\frac{\Delta p}{\Delta t}$$

$$p=mv$$

In the case of the car initially traveling at constant velocity, and the one car towing the other with both having constant velocity, the change in momentum is zero (no change in velocity) and so the net force acting on each, and both as a system, is zero.

In the case of the car impacting you, there is a net force acting on each of you (the car on you and you on the car) resulting in a change in momentum of each of you.

PS: If gravity wasn't behaving as some kind of acceleration pulling me constantly to the Earth's core, then if I jumped upwards, would I start slowly getting away from Earth?

Yes.

If there were no gravitational force and you jumped upward you would exert a downward force $F$ on the Earth and the Earth would exert an equal and opposite upward force $F$ on you per Newton's third law. The force you exert on the Earth of mass $M$ will give it an acceleration of

$$a_{M}=\frac{F}{M}$$

with respect to the center of mass of the combination of you and the Earth.

And the equal and opposite force the Earth exerts on you of mass $m$ will give you an upward acceleration with respect to the center of mass of

$$a_{m}=\frac{F}{m}$$

So you and the Earth will separate. However, the position of the center of mass of the combination of you and the Earth will remain the same in order to have conservation of momentum of the system.

Hope this helps.

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The force acting on the car is zero when it is travelling but non-zero when it hits you. If a car hits you, its speed changes as the collision causes it to slow down. This means an impulse acts on you and the car exerts a force on you.

If the two cars travel at constant speed then the resultant force on each is zero. If there is no friction, this means they dont exert any force on each other (this can be seen by drawing a free body diagram of each car). If there is friction, the tension must cancel the frictional force.

Both situations are described by $F=ma$ because in the first the car decelerates as a force acts to oppose its motion, and in the second neither car accelerates and the resultant force on each is zero.

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Whenever you see a mathematical formula describing a physical law, try to see which of the variables describe causes and which describe effects. It may be helpful to rewrite the formula as $\mathrm{effect} = \mathrm{cause}$. In this case, forces cause acceleration, so $\mathbf{a} = \frac{\mathbf{F}_\mathrm{net}}m$. Similarly, having a velocity causes a body to travel a certain distance, connecting a power source to a circuit causes electric current to flow etc.

What if an object is moving with a constant speed?

It means the forces acting on it are such as to cause it to move with zero acceleration, and Newton’s second law says this means they add up to exactly zero.

If it travels with constant speed and hits me, it still hurts.

The car traveled with constant speed before it encountered a stationary pedestrian. Then they touched, and forces arose (specifically electromagnetic forces between atoms of the two bodies now that they are in close proximity; just about everything you can see is caused by electromagnetic interactions, the only exceptions are subatomic things and gravity). Those forces caused acceleration of both bodies. Newton’s third law says both the car and the pedestrian experience forces equal in magnitude and opposite in direction, but the second law says the accelerations are going to be anything but equal because $\mathbf a=\frac{\mathbf {F}}m$. The car, being way more massive than the pedestrian, is going to experience a slight deceleration, while the pedestrian will feel a significant acceleration. What’s worse is that the acceleration will be different for different parts of the pedestrian’s body (e. g. two ends of the same bone) depending on which were struck first, causing those to move relative to each other in unnatural ways, inflicting injuries.

So no, you can’t say that the car traveling at constant speed hit a pedestrian. During the impact its speed was not constant. However, you can’t say that the change of speed, i. e. acceleration, of the car was what hurt the pedestrian either; instead, forces caused acceleration in both bodies.

On the other hand, if the car is truly traveling with constant speed, it can never hurt anyone. For example, the driver’s seat is constantly interacting with the driver but it doesn’t hurt them. Instead, all the parts of the driver’s body are comfortably moving at the same speed.

So it's force must not be 0.

You can’t say that an object has a force. Objects have masses, energies, temperatures etc., but not forces. What you can say is that right now, these two objects exert forces on each other (to which Newton’s third law applies).

Another way of looking at it is from the standpoint of energy. Objects do have energies assigned to them. A raised hammer has more energy than one lying on a table. A car moving fast has more energy than one moving slow. To say forces are exerted is equivalent to saying something loses energy while something else gains it. So when a car hits a pedestrian, it slows down and imparts some of its energy to the pedestrian in a way they don’t like. But a steady-moving car does not transfer energy to its equally steady-moving driver, so nothing happens.

If the same car is pulling another car behind it, if the car goes with constant speed, it implies that the first car doesn't exert any force to the second car at that moment, since it is not technically pulling it anymore (espacially if we don't take into consideration the friction).

That’s very true, and it can be easily observed by watching the tow rope. Part of the time it will be taut, and part of the time it will sag. Whenever you see it sag, it means no force is exerted.

If gravity wasn't behaving as some kind of acceleration pulling me constantly to the Earth's core, then if I jumped upwards, would I start slowly getting away from Earth?

Consider the Pioneer and Voyager probes. Effectively they made a strong enough jump and then turned their engines off. They left the solar system (not very slowly, the Sun’s escape velocity is 42 km/s), and aren’t going to return.

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