Force is explained as a push or pull, feels quite intuitive at first.

My question is if an object accelerates over distance of $100$ meter hits a car, and another object of same mass at same acceleration over a distance of $1000$ meters hits the same car, then the force of these two objects is the same, since mass and acceleration of the two objects are the same. Yet, the "push" felt would be very different, because surely the second object will "push" the car much "harder" and "farther" than the first.

Applying another physic equation, work, forced over distance, may explain this, but I am left not truly understanding what/where is force in everyday situations - how can force be understood as a "push" if the push felt is different from force of the same magnitude?

My second question is the basic math for Newton unit. My understanding of multiplication is a way of counting: unit per group multiplied by number of groups yielding total number of units. How/what's the best way to understand the math of physics composite units, such as force: the product of $\text{kg}$ and $\text{m/s$^2$}$ yielding a composite unit, $\text{N}$? Within the context/definition of multiplication, What am I actually counting with two different mixed units?

  • $\begingroup$ Think about the total energy of the objects due to the applied force. I think force should be intuitively understood as 'instantaneous change in momentum' since, in fact, $F=ma=dp/dt$. $\endgroup$
    – Danu
    Jun 29, 2014 at 14:20

3 Answers 3


You have misunderstood the concept of force and conflated it with momentum.

Force is not something objects possess, it is something that is exerted on them externally, or something that they exert. Newton's second law stating $\vec{F} = m\vec{a}$ means just that the acceleration of a body is proportional (with the proportionality constant the mass) to the force exerted on it, this does not make the force a property of the body.

What you seem to intuitively think of a force is momentum, which Newton called quantity of motion. Momentum is (speed times mass) $\vec{p} = m\vec{v}$, and it measures how strong the impact of any object colliding with any other will be, since, if you plug the formulae into each other, you will find $\vec{F} = \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$, so if any object exerts a force on any other, the force exerted will be determined by the amount of momentum the object loses in that process, this amount is sometimes called impulse.

Now, to go to your specific question, if two objects accelerate with the same $\vec{a}$ over different distances $d_1,d_2$, then their momenta at the end of that acceleration will depend on the times $t_1,t_2$ needed to traverse that distance, which are (at constant acceleration $a$), given by $t_i = \sqrt{\frac{2d_i}{a}}$ (since $s(t) = \frac{1}{2}at^2$). Therefore, the momenta will be $\vec{p}_1 = mt_1 \vec{a}$ and $\vec{p}_2 = mt_2 \vec{a}$. Obviously, if $d_2 > d_1$, then $t_2 > t_1$ and so $|\vec{p}_2| > |\vec{p}_1|$, so the longer the distance the object has to accelerate, the larger its momentum will be (of course - the longer it accelerates, the faster it will be!). If they now hit a target that brings them to a complete stop, the change in momentum - the impulse - will be the entirety of their momentum, so the larger the momentum of the objects was (equivalently, the faster they were), the larger will be the force they exert on the target.

  • 1
    $\begingroup$ "Force is not something objects possess, it is something that is exerted on them externally" I see this particular confusion of vocabulary a lot in my intro-physics for non-majors courses. Once they get the words squared away, they often turn out to have a pretty good mental model of what is going on. $\endgroup$ Jun 29, 2014 at 15:06
  • $\begingroup$ There's something weird going on with your units. How is $p = da$? The latter has units of velocity squared, not momentum. $\endgroup$
    – Javier
    Jun 29, 2014 at 15:12
  • $\begingroup$ @Javier Radia: Argh, you are right...I'll fix this right away. $\endgroup$
    – ACuriousMind
    Jun 29, 2014 at 15:13
  • $\begingroup$ And sorry for misspelling your name. Not my best day, it seems :) $\endgroup$
    – ACuriousMind
    Jun 29, 2014 at 15:20

It's dangerous to endorse someone's intuition because it may be wrong in some cases and different people may use different intuition – sometimes correct, sometimes incorrect.

The force is a quantity $F$ that tells us how strong is a cause of the mechanical change of some objects or their motion. How the objects actually change – or how their motion changes – when the force is applied is a different question that requires another calculation or reasoning on top of the knowledge of the value of the force.

When a large car collides with a small car, they are exerting the same forces of the opposite direction (Newton's third law) on one another. But that doesn't mean that both cars will be "equally affected". If the effect of the force is to change the speed of the cars, the lighter car will be affected more by a given force because $F=ma$.

Similarly, a force is something that may tear a thread to pieces. But the same force $F$ may be enough to tear a thin thread but not a thick one. Each thread has its own value of $F$ at which it (probably) tears apart.

You may visualize every force as being "equivalent" to the force obtained by hanging a weight of mass $m$ somewhere. This creates the force $F=mg$ in the gravitational field $g$. If one is vague enough, one could confuse force with "the distance by which a car gets kicked" or "the momentum" or "the energy" or dozens of other related quantities. All these quantities are completely different and it is indeed a terrible beginner's mistake to confuse any pair of these. But unless one uses some precision and mathematical expressions – and it seems that the OP has avoided them – it is very likely that one will confuse pretty much everything with everything else.

Composite units are no different from the units of areas. The area "one squared meter" is "one meter" times "one meter" and one may literally visualize "one squared meter" as the are of the square with both sides equal to "one meter". The volume is totally analogously the product of three distances and the "cubic meter" may be defined as the volume of the cube with the "one meter" sides in all three independent directions. All these things are rooted in the fact that the area of a rectangle is $A=L_x L_y$, the product of two distances (or three, in the case of volume).

The other composite (product) units are analogous in the mathematical sense. One Newton is one kilogram times meter over squared second (mass times acceleration). If one multiplies the mass $m$ and acceleration $a$, one simply has to get units that are products of the units used for $m$ times the unit used for $a$. That's the case for the very same reason why the squared meters are units of areas.

Imagine that we decide that the distance we know as "one meter" will be called "two meters" because we redefine meters. Because the areas are $A=L_x L_y$, this means that "one squared meter" will be called "four squared meters" because this square may be divided to $2\times 2 = 4$ "new squared meters". So it's important to remember the powers of the base units because different quantities would be affected differently if the meaning of the base units changed. For areas and volumes, these things are particularly self-evident and may be honestly visualized. For all other composite units, the visualization is a bit abstract but the mathematical reason why the units must be multiplied correctly is always the same.


Force exerted by the car accelerated over large distance is greater than the force exerted by other car accelerated to the same extent over a shorter distance. In your context, the acceleration $'a'$ of force equation refers to the extent to which the object accelerated and not the acceleration of the car. This might clarify your conjecture.

Reference needed from reliable sources.


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