# How long will the ball and the spring be in contact?

• B

## Main Question or Discussion Point

This is my complete question.
I've been wondering on this topic for quite a while.
I really can't visualize what happens at the border between the spring and the ball when we release the potential energy stored in the compressed spring.
I would like to know how long would the contact beììtween the spring and the ball last. How long will the restoring force of the spring be exerted on the ball?
I imagine a fricitonless floor and an elastic collision (push).

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## Answers and Replies

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Dale
Mentor
Assume that the ball is rigid (or at least much stiffer than the spring) and that the spring is massless (or at least much lighter than the ball). Then draw the free-body diagram for the ball, write the equations, and solve.

russ_watters
Mentor
Where exactly is your issue? Do you know the length of the compression? Mass of the ball? The equation for spring force vs distance compressed? Newton't laws of motion? Please try to set something up and we can nudge you from there.

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Well I didn't take any physis course, I was just wondering about that scenario. I usually think of the time of application of the force and then calculate it's acceleration. I would have liked to have an intuitive answer which was not dependent on the particular string ( even though I would intuitively assume that its mass is negligible respect to the ball's one and that the ball it's rigid for the sake of simplicity I guess). I would intuitively guess that in this situatuion given that the spring will accelerate faster if there was no ball, the whole siste (spring +ball ) will accelerate with an a=(spring constant)*(distance form equilbrium)/(mass of the ball), so the ball will be in contact with the spring till the equilibrium point ( then ball wil keep its speed while spring will decrese so there will be a detachment). Anyway this is what I imagined ideally, but I couldn't be sure it happens in reality too. I had some doubts about if the spring and the objects will really stick together for the whole time or maybe the spring will accelerate the ball infinetly times for an infinite small time (which is quite the same, but miscroscopically mmaybe different). Thank everyone for the partecipation.

Dale
Mentor
Well I didn't take any physis course, I was just wondering about that scenario.
Ah, ok, I have changed the level from I to B then to indicate that the answer should be at a basic level. I had mistakenly assumed that you would have already seen free body diagrams, etc.

I would have liked to have an intuitive answer which was not dependent on the particular string ( even though I would intuitively assume that its mass is negligible respect to the ball's one and that the ball it's rigid for the sake of simplicity I guess).
Unfortunately, it will depend on the mass of the ball and the spring constant.

I would intuitively guess that in this situatuion given that the spring will accelerate faster if there was no ball, the whole siste (spring +ball ) will accelerate with an a=(spring constant)*(distance form equilbrium)/(mass of the ball), so the ball will be in contact with the spring till the equilibrium point ( then ball wil keep its speed while spring will decrese so there will be a detachment).
This is a very good start for not having taken physics before. I am impressed! (you are a natural). I will work it out in detail shortly.

Anyway this is what I imagined ideally, but I couldn't be sure it happens in reality too. I had some doubts about if the spring and the objects will really stick together for the whole time or maybe the spring will accelerate the ball infinetly times for an infinite small time (which is quite the same, but miscroscopically mmaybe different). Thank everyone for the partecipation.
The considerations for reality differing from the ideal you described are as follows. If the ball is not rigid then you would have to consider the spring constant of the ball also which would typically lengthen the duration of contact. If the spring is not massless then it will not stop immediately upon reaching the equilibrium position, but will gradually decelerate resulting in a longer contact time. Also, although we didn't explicitly discuss it, if there is any surface adhesion or any resistance then the contact time will also be lengthened. So the ideal contact time is a minimum time and everything else should make it slightly longer.

jbriggs444
Homework Helper
2019 Award
Start with Newton's second law, F=ma. If you apply a force F to a mass m, the resulting acceleration will be a.

In this case, F is the force of the spring, m is the mass of the ball on top of the spring and a is the resulting rate of acceleration.

Proceed with Hookes law. F=kx. If you compress a spring with spring constant k by distance x, the resulting force will be F.

In this case, x is how far you compress the spring from its equilibrium position, k is the spring constant and F is the resulting force.

The problem you face is that as the ball undergoes acceleration, its position is not fixed. So the force on the ball is not fixed. So the ball's acceleration is not fixed. You need to use calculus to determine how the position of the ball varies over time. Or, if you do not have the appropriate calculus background, you can get an approximate result by simulating the motion of the ball, over tiny steps of time and assuming that the acceleration is close to constant for each small time step.

@Dale Thank you Dale! I'm realy happy with your insights and I think I understood partially the reasons behind them. So, just to be completely sure, the ball will stick to the spring at least till the equilibrium point (will not detach even microcospically, given the considerations above)?!

Dale
Mentor
So, with the ideal assumptions we have $f=ma$ gives $-k x = m x''$ where $x$ is the distance from equilibrium. This is the same as what you said earlier $a=x''=-kx/m$ except that you missed the minus sign. Solving that with initial conditions $x(0)=x_0$ and $v(0)=x'(0)=0$ we get $x(t)=x_0 \cos(t \sqrt{\frac{k}{m}})$.

The spring detaches from the ball when $x(t)=0$ which can be solved for t to obtain $t_{\text{detach}}=\frac{\pi}{2}\sqrt{\frac{m}{k}}$