If an object of mass $m_1$ moving with velocity $v_1$ hits an object of mass $m_2$ at rest,then is it possible to calculate the force with which mass $m_1$ hits $m_2$ given that $m_1,\,v_1,\,m_2$ are only given parameters? Does this thing exist? What are other parameters needed to make proper calculation here?

  • $\begingroup$ If you knew the velocity of $m_1$ after the collision and how long it was in contact with $m_2$, you could use $\mathbf F\simeq m_1\Delta \mathbf v/\Delta t$ (i.e., the impulse), otherwise you don't have enough information. $\endgroup$
    – Kyle Kanos
    Dec 31, 2016 at 13:45
  • $\begingroup$ @KyleKanos Suppose that m1 is a solid iron ball and m2 is a solid beam and when the ball collides with the beam,it stops instantaneously,now can we calculate? $\endgroup$ Dec 31, 2016 at 13:50
  • $\begingroup$ Well you could use the equation I give: $F=m(v_{1,f}-v_{1,i})/(t_{f}-t_{i})$. But saying "instantaneously" typically means $t_f=t_i$, which means you have a "divide by zero" error here. $\endgroup$
    – Kyle Kanos
    Dec 31, 2016 at 13:52
  • $\begingroup$ @SurazBasnet It's much more complicated than that. The answer depends on the material of the beam, the way the beam is fixed, what position the ball collides with the beam, whether the beam material is permanently deformed (i.e. it ends up bent of dented), etc, etc. Doing this "for real" involves computer simulations, which can take literally days to run to simulate a single impact. The only thing you can calculate easily is the "Impulse" or change of momentum, as in the answers. $\endgroup$
    – alephzero
    Dec 31, 2016 at 13:55

2 Answers 2


We need to know how long they are in contact. then we are able to calculate impulse (not force).

$$\mathbf{I}=\int_{t{\text{ impact}}}^{t{\text{ impact }+ \tau}} \mathbf{F} dt=\Delta \mathbf{p}$$

Impulse is a measure of blow exerted to mass.

Force varies with time. But If we assume the force is constant during the impact, we can find it using

$$\bar{\mathbf{F}} \tau=\Delta \mathbf{p} \ \implies \ \bar{\mathbf{F}}=\frac{\Delta \mathbf{p}}{\tau}$$

I found this useful picture:

impact force

The force during the impact varies with time. Without knowing the function of that curve, we cannot calculate it as a function of time. One thing we can do is to assume it's constant so that we know the function. Now we can integrate.

  • $\begingroup$ What is a force really?If a body is subjected to some force in space,will it accelerate forever?As f=m.a .If it doesn't(which is certainly true),till what time will the force of 1N produce an acceleration in space? $\endgroup$ Dec 31, 2016 at 13:55
  • 1
    $\begingroup$ @SurazBasnet Physics doesn't try to answer the question of what anything is "really." In classical mechanics, a force is just a way to measure the amount of momentum (mass x velocity) transferred from one object to another - in other words, it is defined by what Newton's second law says about it. A force will produce an acceleration for any length of time - though of course if the velocity of the bodies becomes very large, you have to replace Newtonian mechanics with special or general relativity! $\endgroup$
    – alephzero
    Dec 31, 2016 at 14:03
  • $\begingroup$ @alephzero so what would be the velocity pattern in this space case I just talked above? $\endgroup$ Dec 31, 2016 at 14:04
  • $\begingroup$ @SurazBasnet If you are sitting on the rocket measuring acceleration with an accelerometer, then if there is a 1 N force acting forever, your accelerometer will register an acceleration of $a=F/m$ forever. If you are viewing the rocket from an inertial frame, the rate of velocity increase will get smaller and smaller with time. When the velocity is low, the acceleration might be so close to $F/m$ as to be indistinguishable from it, but in fact there is always Special Relativity which tells us that the acceleration is never exactly $F/m$ $\endgroup$
    – garyp
    Dec 31, 2016 at 20:30

Work put the final velocities from conservation of momentum, you don't need to determine the forces. $$m_1v_1=m_1v'_1+m_2v'_2$$ momentum is shared equally in the collision $$m_1v'_1=m_2v'_2$$ (these are the proper parameters)

  • $\begingroup$ I think momentum is a vector. And we have no clue that the masses are limited to one dimension. Even though, conservation of momentum alone can't solve the unknowns, we will need conservation of kinetic energy, too! And if the motion is in 3D, these will not be sufficient, we will need the specification of the shape of objects and their orientation when they hit. Forgetting all those above, I think the OP wanted to know the interaction force between the masses. Not final velocities. $\endgroup$
    – AHB
    Jan 1, 2017 at 7:48
  • $\begingroup$ Why would the momentum will be equally shared after collision? Where did you get this condition from? $\endgroup$
    – nasu
    Oct 18, 2023 at 3:09

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