I am a first year math student taking physics $0$ right now and I got this problem on an exam. Suppose the mass of the ball is $6.0 \mathrm{kg}$ and it is accelerating at $1.0 \mathrm{\frac{m}{s^2}}$. You might wonder, how is the ball accelerating? Well the teacher didn't say, but I imagined it was gong down a hill or something. The question is , what's the force that the wall exerts over the ball when they contact? The intended solution was to use Newton's second law, multiply mass times acceleration and that would give you the force. Since the problem also states that the direction the ball was going in was the positive one, then the direction of the force the wall excerts over the ball would be negative, then the right answer was $-6.0 \mathrm{N}$

I did this in the exam, and got it "right" but I had my questions, like if the ball had just started accelerating from rest it would hit the wall with much less force than if it had accumulated bigger speed, so then knowing the acceleration and mass couldn't be enough information. I asked another teacher about this, he told me the right answer would be to look at the change in momentum, which we could calculate if we knew the velocity an instant before the collision and an instant after the collision. This change in momentum would therefore be the impulse or force times time, we could then get the force by dividing the impulse by the time.

My first question is, would the change in momentum always be twice the momentum before impact? Assuming of course the velocity just changes direction and keeps the same magnitude. If not, would we need to measure velocity twice? Once before impact and another time after impact in order to figure out the change in momentum? I'm guessing it depends on the material of the ball and the wall.

My second question is about the time, we're supposed to divide the impulse over time but how do we compute that time? In the physics experiment that we did in class, we just assumed it was something small like $0.2 \mathrm{s}$ and didn't measure it at all. If we wanted to be precise, would we need to count the amount of time that the ball and the wall are in contact and divide the impulse by that?

Lastly, it is my understanding that impulse is only equal to force times time when the force is constant, if the force is not constant impulse would be the area below the force over time curve. In this particular case, is it reasonable to assume the force is indeed constant and that we can find it by simply dividing impulse over time?

thanks and sorry if some of this is obvious, I basically didn't learn any physics in high school and I just started learning it in college this semester.


1 Answer 1


I would claim depending on, how accurate you want your result to be, all of your doubts are justified.

Your take on the exam problem seems correct to me. The presented intended solution does only work if the ball hits the wall at rest (lul), which means you look at a static problem where the ball is pressing against the wall, which would bring the formulation of it accelerating with $1.0\frac{m}{s}$ ad absurdum.

The approach through the momentum conservation would work for calculating the average force during the collision, if you knew the collision time.

The absolute value of momentum change would be twice the absolute value of the momentum before the impact only if the wall's mass is infinite or in good approximation if its way larger then the ball's and you have no energy dissipation. Otherwise you would have to solve a collision problem, or measure the momentum twice, to get the momentum change. If you assume an elastic collision, which means kinetic energy is preserved during the collision and not turned into heat, this calculation should be independent of the used materials. If you want to consider energy dissipation in the materials, that would depend on the materials.

As for the time, if you wanted to calculate the average force during the collision, yes measuring the time, the ball is in contact with the wall would give you the right time.

Now, as you appear to already have figured, depending on what you want to do, focusing on the average force might not be ideal. For example the ball might be exposed to significantly larger peak forces, not accurately desribed by looking at the average. Assuming a ball acting like an ideal spring i.e. the force changing linearly with the distance the ball travels after hitting the wall, if I am not mistaken, your equation of motion during the collision would take the form of a harmonic oscillator equation. That would mean that the force over time being proportional to the displacement would take the form of a sine (just one fraction of a period around one extremum ofc). Assuming such a time dependence would already significantly increase your accuracy when trying to calculate the maximum force acting on the ball, compared to just taking the average force during the collision. Using this equation of motion one also could calculate the collision time from it, which one then could compare to the measruement as a self consistency check for example. Of course there is still be allot of room for improvement through use of more complicated models of the collision.

Overall this appears to be a good example of how a seemingly simple problem can get arbitrarily complicated depending on how close you want to look at it.

Edit to answer comment: The ball acting as a spring is just a simple approximation, you can think about it as cutting of a Taylor-expansions after the linear term. Now how well that approximation works will generally depend on your materiel and on the shape, where the dependence on the shape would again depend on the material I think. But also for all materials, this approximation should be valid in the limit of small deformations. For larger deformations one should perhaps use some nonlinear model. Generally I intuitively believe this approximation should be sufficient for an estimate on the maximal forces acting on an object for most everyday cases, but I also have no practical experience to really make any reliable statement on how good that kind of modelling really works.

  • $\begingroup$ Thanks, I read this comment multiple times and it really helped me understand it all fully. I do have one question, when you talk about the ball acting like a spring, does that depend on the shape of the ball or the material or both? $\endgroup$
    – zlaaemi
    Commented Apr 6 at 22:08
  • $\begingroup$ @zlaaemi I made an edit to adress your comment. $\endgroup$
    – Zaph
    Commented Apr 7 at 7:42

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