What would be an intuitive way to damage objects in a physics game using impulses? Since momentum is conserved, so is impulse (the change in momentum for any two time periods) in a closed system.

So if I have the impulse of a collision (that is, the change in momentum during a time interval for one of the two bodies) what would be the intuitive way to damage both the objects. Should each object receive the same damage based on the impulse of the overall system, or should the damage incurred be split unevenly based on the relative mass of each object? My intuition is that objects with larger mass would have higher hit-points, and so would be less affected if the same damage was incurred to both the objects in the collision.


2 Answers 2


Objects are not damaged by momentum $\vec{p}$ they are damaged by force $\vec{F}$. When two objects collide their momenta changes because of forces they apply each other while being in contact. According to 2nd Newton's law the force can be calculated as follows: $$ \vec{F} = \frac{d\vec{p}}{dt} $$

So the force is determined by the time of interaction. When two objects contact their surfaces are flexed. The bigger is the flex the higher is the force which changes the momentum. At some moment of time the relative normal momentum (and velocity) becomes zero and the objects start move backward.

At this moment the force and the flex is maximal. If the surface (armor) is not strong enough for this flex the object is damaged. This can happen long before the relative normal velocity become zero. In that case the interaction between internal parts of the objects started.


Hence damage is not determined only by momentum. It is determined by the force and the ability of the object resist it. The force is not constant during the collision and its maximal value depends on momentum and the time of interaction of the surfaces. The time of interaction depends on both flexibility and strength.

When a basketball hits the floor its momentum changes from $\vec{p}$ to almost $-\vec{p}$ so the total change is almost $2p$. When a glass hits the floor its momentum becomes zero so the change is $p$. The ball has higher flexibility and strength and is not destroyed even though the momentum change is two times higher.

Another example is a bullet that hits a door. It makes hole before the door opens. The force is huge but the momentum of the door is almost not changed because the time necessary to reach critical flex is too short. When one pushes the door with his finger the force is small and does not destroy the surface before the flex stop the finger. The door gets enough momentum to start move.


When the surface is flexed in irreversible way or damaged some part of kinetic energy of the objects turns into kinetic energy of their parts, heat, sound, light etc. This is called inelastic collision. Elastic collision means no damage. If you need a good model this should be taken into account.

  • $\begingroup$ A bullet will not open a door? This can't be correct. The momentum of a bullet is enough to recoil on your shoulder with a good shove, sufficient to swing open a door. $\endgroup$
    – Ron Maimon
    Commented Dec 18, 2011 at 10:24
  • $\begingroup$ Hmm. I think what he meant was that the time taken to distribute the force has different results. From the formulas I think momentum leads to the force, since a force is essentially the first difference of the momentum. I'm not sure if I'll be able to go into such detail as surface forces and kinetic energy, but I think the impulse (change in momentum) should suffice as a measure of the impact of a collision, no? $\endgroup$ Commented Dec 18, 2011 at 11:13
  • $\begingroup$ @RonMaimon, probably you are right. I have read this example in a schoolbook long ago but have never seen any experiments. I will correct the post and may be will find a less doubtful illustration. $\endgroup$ Commented Dec 18, 2011 at 11:56
  • $\begingroup$ Actually the introduction in this post about momentum is somewhat misleading as the OP is not talking about momentum in the question, he is talking about the same thing you are (impulse is force during a timestep). But the bottom line, that it is complicated to calculate fractures, is correct of course. $\endgroup$
    – BjornW
    Commented Dec 18, 2011 at 12:08
  • $\begingroup$ @AramKocharyan, momentum is not a sufficient measure of damage. I will change the example with the bullet to more clear one proving this. At least two more magnitudes is needed: rigidity to calculate time and force, and strength to recalculate force to damage. $\endgroup$ Commented Dec 18, 2011 at 12:24

There is a whole field of interest in breaking things realistically in computer games and movie CG effects.

Basically the impulse of the collision has to, like you say, be propagated internally in the objects that collide. This is a very complex problem if you want to do it realistically, essentially you need to solve the stress diagram of any number of suitably small pieces of the object and fracture them as the propagated impulse breaks their connections.

Some games get around this by letting the artist create "pre-fractured" models with known sub-parts that can be split upon an impulse over a certain threshold at a certain place. This is more flexible than it sounds because the artist can create the pre-fracturing using mechanical engineering programs, so the fracturing itself is realistic, but the exact impulse -> fracture shape will be mostly inexact.

In either case, simply using the mass of the two colliding objects to determine which should break, is overly simplistic (overall mass is not an accurate predictor of "breakability"), but a computer game is a work of art and not an accurate simulation in most cases, so if you implement it and it feels right, go for it :)

Here is a discussion from the Bullet Physics forum about fracturing collision-shapes depending on the impulse (somewhat technical but you might get some inspiration). Bullet is a well-known open-source 3D physics library used in hundreds of computer-games and Hollywood movies.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.