# What is the relationship between force and momentum in collisions?

I know that $\Sigma F = \Delta mv/\Delta t$. But if we had a marble that moves in a straight line at a constant velocity and colloids with another marble. Because of the law of conservation of momentum, the second marble now had the velocity of the first marble. But what is the force that the first marble applied one the second marble? The collision is almost instantaneous. Wouldn't that make the force in $\Sigma F = \Delta mv/\Delta t$ insanely large because $\Delta t$ is so small?

• This simply shows that the collision isn't instantaneous. Try (if you have the time) filming such a collision with a high-speed camera and looking at it in slow-motion. A) It should be cool and (more importantly) B) You'll see that the collision isn't instantaneous - it's simply very brief. – HDE 226868 Oct 22 '14 at 1:47

The force can be surprisingly large, but $\Delta t$ is not zero, and the force is not infinite.

Make some estimates: the duration of the collision is so short that our eyes and brain cannot perceive it. Make an estimate for an upper limit for the duration. (There's no right answer, but a lot of wrong answers. For example, I would think that a duration of 0.1 s would be perceivable, and "wrong". My upper limit should be smaller.)

From this you can get a lower limit on the force.

You can improve your estimate for $\Delta t$. You know the speed of the marble. You can make a guess at the size of the deformation of the ball that occurs during the collision. It's certainly less than one tenth of the radius. Probably less that 1/100 ... (Make your own estimate.) From there calculate a $\Delta t$.

Try this: make a crude model of the force generated when two satellites collided in 2009. The relative speed of the collision is known. The masses and sizes can be found or estimated. Calculate the duration of the collision and the force generated. (They did not collide head-on, so divide by 2 to crudely account for this. :) They also disintegrated. So there the analysis will have flaws. But it's an interesting exercise as long as you keep in mind that it is unrealistic. )

In this type of collision where you have what amounts to a very quick change in velocity, the force is called an impulse force and it is best to think of the equation a little differently. For example, instead of: $$\sum F = \frac{\Delta mv}{\Delta t}$$

Think of $\int F \mathrm{d}t$ being equal to the change in momentum, that is: $$\Delta mv = \int F\,\mathrm{d}t \qquad\text{impulse force}$$ The change in momentum itself is equal to this integral where the actual functional form of F may not be known at all.

• plus the impact forces (impulse forces) which act on a very small time-frame are usualy ignored in calculating inital-final momentums etc... If one would take into account such impulse forces. 1. They are unknown, 2. diificult to put into analytic form. In any case in the integral inital-final conditions, they smooth out – Nikos M. Oct 22 '14 at 10:21
• Mathematically, nothing new was added. But, looking at the problem a little differently, mathematically rearranging things, often simplifies the problem of understanding. I think the OP problem was dealing with infinite or very large force but actually the change is momentum is due to the combination of force over time as I showed by the integral above. – K7PEH Oct 23 '14 at 14:31

what is the force that the first marble applied one the second marble? The collision is almost instantaneous. Wouldn't that make the force in ΣF=Δmv/Δt insanely large because Δt is so small?

Suppose two steel balls A, B of equal mass (m = 0.1 r = 0.03 m) collide and B is at rest:

Ball A will exert on b the Impulse of a Force $J$ and its velocity, momentum and KE will increase: $$J = [F . t] = \Delta p$$ If you know exactly of what material the balls are made you can calculate the time of the collision and consequently the impulse of the force, almost exactly: kinetic energy travels in steel at the speed of sound $\approx 6000m/s$ (5800 -6100), let's say 6000.

Therefore the collision will last 0.06/ 6000 = 0.00001 sec, if A has v = 20m/s, its $$KE = \frac{0.1*20^2}{2} = 20J, \Delta p = 2 Kg*m/s$$ (or any value you like). You can find out the force exerted $$F * \frac{1}{10^5} = [\Delta p] = 2 \rightarrow F = 200,000N$$

• That's interesting. But is "exactly" too optomistic? The $\Delta t$ you calculate it the length of time it takes the influence to get across the ball. It's not clear to me how that is related to the duration of the collision. Is it possible that the balls would have disengaged before the disturbance traversed the ball, making $\Delta t$ even shorter? Or that they stay engaged longer than that? – garyp Oct 22 '14 at 14:06
• @garyp , ". Is it possible that the balls would have disengaged before the disturbance traversed the ball, making Δt even shorter? Or that they stay engaged longer than that? " . No, the second ball doesn't move before KE reaches the other end it's like a locomotive getting to the head of the train. Take 'exactly' with a pinch of salt :) – bobie Oct 23 '14 at 7:45

As stated in other answers, $\Delta t$, although small than times you're used to perceiving, is not 0. Because the marbles do not deform very much, the collision does not appear to take much time. However, in the grand scheme of things, the change in momentum for each marble during the collision is also small. Although answers vary, a quick google search suggests that mass of 10-20 grams might be reasonable. That means that even if the marble is traveling at highway speeds when it hits the other marble,$\approx 30m/s$, which seems very unlikely, the total momentum of the marble is less than $1 kgm/s$. So even for small time intervals (under the upper-limits that garyp suggests in his answer), I can convince myself that the force will certainly be on an order of magnitude of 100N-1000 N, which may seem like lots, but not more than the weight of a typical human being.

If you don't find the back-of-the-envelope calculations convincing, you could always try looking at collisions with other types of balls. Although the collision will not be perfectly elastic, ping pong balls, tennis balls, or base balls should still all separate after the collision as well, and the deformation, and time in contact, might be just enough that you can convince yourself that all collisions take place over a non-zero interval of time.

As mentioned in the comment here, the impact forces that are active during the time-frame of the actual impact are

1) unknown

2) difficult to put in analytic form

That is why results like the conservation of momentum theorem are used. One can do estimations or approximations of these impact forces but would have to use more information by employing other equipment (like high-speed cameras and detectors, high-sensitivity heat detectors etc).

If you assume linear force displacement relationship (small displacements) with stiffness $k$ then $F(t) = k\, x(t)$ where $x(t)$ is the separation of the two objects. Also initially the impact speed is $V = \dot{x}(0)$.

The impact time is characterized by the natural frequency of the system $$\omega = \sqrt{\frac{k}{m}}$$ where $m = \left( \frac{1}{m_1} + \frac{1}{m_2} \right)^{-1}$ is the reduced mass of the system. If we assume a elastic harmonic response to the impact with displacement $x(t) = X \sin (\omega t)$ lasting from $t=0\ldots \frac{\pi}{\omega}$ then the contact force is $$F(t) = m \omega^2 x(t) =m \omega^2 X \sin(\omega t)$$.

From the conservation of impact momentum $J=(\epsilon +1) m V$ (see this post for more details on the treatment of collisions) with $\epsilon$ the coefficient of restitution. In addition, the impulse is defined as $J=\int _0^\frac{\pi}{\omega} F(t) \,{\rm d} t = 2 X m \omega$. So the displacement amplitude is estimated at $$X = \frac{(\epsilon+1) V}{2 \omega}$$ with peak force $$F_{max} = F(t=\frac{\pi}{2\omega}) = m \omega^2 X = \omega \frac{(\epsilon+1) m V}{2}$$

This can be stated as the fraction of the peak force to impact momentum

$$\boxed{ \dfrac{F_{max}}{m V} = \omega \dfrac{(\epsilon+1)}{2} }$$

For a purely elastic impact $\epsilon=1$ then $$F_{max} = V \sqrt{k m}$$

Appendix

According to Hertzian contact theory two identical spheres or radius $r$, and of the same material with elastic modulus $E$ and Poisson ratio $\nu$ the force displacement relation ship is

$$F^2 =r \frac{2}{9} \left( \frac{E}{1-\nu} \right)^2 \delta^3$$

which is used to estimate the stiffness $k = \frac{F}{\delta}$ of the contact at the impact time.