Velocity in Newton's second law

I have a basic conceptual issue when thinking about the following equation describing Newton's second law: $$F(t)=ma(t)$$ To me, the functional form as well as its specification seems counterintuitive. Specifically, I wonder why this does not include an additional term, velocity at time $$t.$$ Specifically, imagine two objects hurtling towards a wall. Both are identical in their masses and acceleration, but the velocity of one is twice that of the other. How come the force exerted on the wall upon impact exhibits independence of this difference in velocity?

• Where you put your “additional term” and what form would it have? Commented Sep 9, 2023 at 14:51
• F=m*a gives the force to accelerate the mass, if a mass with velocity v hits a wall, you have to know the deceleration so for example if the impact is elastic, the velocity v may be -v after the impact, to know the force you have to know how long it took from v to -v. and so can find the force of impact. It has nothing to do with the force F which initially gave the mass its velocity. Commented Sep 9, 2023 at 14:58

How come the force exerted on the wall upon impact exhibits independence of this difference in velocity?

It doesn't. The impact force of the object on the wall is due to the change in momentum of the impacting object which, in turn, depends on the velocity of the object just prior to impact, plus the force giving it its acceleration prior to impact, assuming it continues to be applied.

Prior to impact the acceleration of the object is based only on the net force acting on it prior to impact and is independent of the velocity of the object.

Hope this helps.

happens

How come the force exerted on the wall upon impact exhibits independence of this difference in velocity ?

It doesn't. If the balls are more or less the same size, and so are in contact with the wall for about the same amount of time, then the force exerted on the wall by the faster ball is approximately twice that of the slower ball, because the faster ball has twice as much momentum.

Another way to think of this is that the average acceleration of the faster ball during the collision with the wall must be twice that of the slower ball, because the change in velocity of the faster ball is twice that of the slower ball in the same amount of time.

There's no any "hidden rule" in Physics that some variables in any Physics equation must be constant (Unless it's not a variable, but Physics constant instead like $$\text{G, c, k, h, e}$$ and etc). In practice, they usually aren't due to many reasons. Physics equation relates dependent variables to the output variables.

In fact not only acceleration can vary in Newton law with time, but body mass as well in so called variable-mass systems. One example of that is rocket,- due to leaving exhaust, rocket engine constant force produces greater and greater acceleration, hence Newton second law for varied mass system is re-defined as : $$F = ma + v_e \frac {dm(t)}{dt} \tag 1$$ Here $$a=\text{const}$$ or can be $$a=a(t)$$ as well. Note, here $$m=m(t)$$, because otherwise you will get usual Newton law instead.

(1) equation can be derived from more general Newton law form $$F = \dot p \tag 2$$ using multiplication rule of derivatives. So better get used to (2) expression than to simple $$F=ma$$.

What you are doing here is essentially trying to treat $$F=ma$$ as a model for the force on the object. That is, you are demanding the accelerations of the two objects to be equal and trying to deduce the force from that, and wondering why other effects aren't being accounted for.

But we don't get to choose the acceleration, and $$F=ma$$ is not a model for the force. The acceleration is determined by the force, which itself is determined by the physics of the wall, which must be specified separately. This force will depend on the velocity of the object colliding with it, as your intuition suggests.

For example, we could consider a simple model where the wall/object acts like a spring: $$F=-kx$$, where $$k$$ is a number quantifying the stiffness of the wall/object and $$x$$ is the length of the deformation of the wall/object during the collision. Then, what $$F=ma$$ says is that $$-kx(t)=ma(t)=m\frac{{\rm d}^2x(t)}{{\rm d}t^2},$$ which can be solved for $$x(t)$$.

The dependence on velocity is easy to see in this model. Suppose the object retains a fraction $$\varepsilon$$ of its kinetic energy during the collision. Then, by conservation of energy, the potential energy stored in the maximally deformed wall/object is $$\frac{1}{2}kx_{\rm max}^2=\varepsilon\frac{1}{2}mv^2,$$ so we find that $$x_{\rm max}=\sqrt{\varepsilon m/k}\,v$$. So the maximum force felt by the object is dependent on its incoming velocity, $$F_{\rm max}=kx_{\rm max}=\sqrt{\varepsilon km}\,v$$, as expected.