# Conservation of linear mometum neglects friction?

1. Oct 5, 2007

### Rocket254

Lets suppose a vehicle (A) stopped at a red light is rear ended by another vehicle(B). Both of the cars slid with locked wheels until frictional forces bring them to a stop. When trying to find the velocity of the second car just before impact and assuming that linear momentum is conserved in the collision I would use :

Vb=(MaVa+MbVb)/Mb

Now my question is, what possible ways could this assumption be invalid? I know that the conservation of lin. momentum during the impact depends on the fact that the only significant force is the force of contact between the cars. Am I correct in saying that is in turn ignores the force of friction exerted by the road on the cars during the impact and this would cause error in calculation?

2. Oct 5, 2007

### Meir Achuz

Good question.
The impulse due to friction is f\Delta t, where \Delta t is the time of impact.
For a collision with \Delta t very short, the frilction impulse is negligible compared to the
impact impulse on each object.

3. Oct 5, 2007

### arildno

The magnitude of the frictional force can be regarded as roughly constant.
Thus, over a tiny period of time, it contributes an ignorable impulse compared to those generated by the enormous collision forces between the cars.

Similarly, air resistance is still working on the cars during the collision, but you wouldn't include the impulse from that one as well?

4. Oct 5, 2007

### Loren Booda

There might be two types of friction - that where the tires' contact, initially between solids, skids across the road; and that until near stop with the tires effectively melted at contact.

5. Oct 5, 2007

### Rocket254

Could you also say that another error is the assumption that the transfer of momentum occurs at only one location and the cars never slide during Delta T?

6. Oct 6, 2007

### TVP45

A good way to think about your question is to imagine what the difference would be if both drivers took their foot off the brakes at an instant before the crash, thus almost eliminating friction (and whatever friction drags along).

7. Oct 6, 2007

### rcgldr

Don't forget there's a significant amount of energy consumed by deformation of the car bodies, unless these are "bumper" cars.

8. Oct 6, 2007

### Rocket254

Ah, very good point.

I had not thought of that.

Last edited: Oct 6, 2007
9. Oct 6, 2007

### TVP45

The crumple zone is a good place to look for missing energy, not missing momentum.
Go back to your text and reread the distinctions between elastic (no permanent deformation) collisions and inelastic (permanent deformation) ones. Pay particular attention to what is conserved and what is meant by a "system".
Tom

10. Oct 7, 2007

### Claude Bile

I agree with Tom, conservation of momentum is only valid if there are no external forces acting on the system. If your system is just the two colliding bodies, then friction would be considered an external force and conservation of momentum would not apply.

Thus instead of Sum(P_before) - Sum(P_after) = 0, you should use;

Sum(P_before) - Sum(P_after) = F.dt

Where F.dt is the Impulse supplied by the external force (friction). As has been pointed out, if F.dt is negligible over the time interval of interest, the second equation reverts back to the first one.

The beauty of linear momentum conservation is that details of the collision itself don't affect the end result, though it will only tell you about how the cars translate not rotate, to get the "full picture" as it were, one needs to apply conservation of angular momentum as well.

Claude.