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?
Best Answer
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.