Momentum Conservation: Large vs Small Car

[toesockshoe]

The question is: When a large car collides with a small car, which one undergoes the greater change in momentum: the large one or the small one? Or is it the same for both?

I wanted to plug in dumb numbers:
m_tv_{ti} + m_cv_{ci} = m_tv_{tf} + m_cv_{cf}
lets assume the truck has a mass of 10000kg and the car has a mass of 10kg (lets not care about realisticness)
also let's assume that the initial velocity of the truck is 10 m/s and the cars is -2m/s (assuming the truck is going in the positive direction). Let the final velocity of the truck be 5m/s. Plugging all of these numbers into the top equation, we get that the final velocity of the car HAS to be: 4998 m/s. Again the numbers arent realistic, but that isn't really important. The change in momentum of the truck is: m_tv_{ti} - m_tv_{tf} = 50000 Ns and the change in the momentum of the car is -50000. So are the changes different? One is NEGATIVE 50000 while the other is positive... I would assume they are different...

Also, the 2nd part of the question asks why passengers in the small car are more likely to get hurt... I have no idea how to answer this question... any hints? thanks!

 

[Svein]

Well, there is more to a collision than just conservation of momentum. You also need to know (or specify) whether the collision is elastic, non-elastic or something in between. Also, you cannot specify the speed after the collision, just the momenta.

So: In this case I would assume a non-elastic collision (both having the same speed after the collision). This gives v_{tf}=v_{cf}. Your equation will then be m_{t}v_{ti}+m_{c}v_{ci}=(m_{t}+m_{c})v_{tf}. Plug in your numbers and you get v_{tf}=\frac{10000\cdot 10+10\cdot(-2)}{10000+10}\frac{m}{s}\approx 9.99 \frac{m}{s}.

 

[PeroK]

As momentum is conserved the change in momentum is equal and opposite for two objects in collision.

As for your question about passengers getting injured. What can you say about how the change in velocity of the two objects is related to their mass?

 

[sophiecentaur]

which one undergoes the greater change in momentum

This is potentially a very complicated problem. People often post questions about their recent traffic accident / insurance claims. Basic Physics cannot help them, I'm afraid but it's an interesting thing to study.
The momentum change has to be the same for both cars (different signs, of course). However, what counts is the change in velocity and, importantly, the way that the dissipated Kinetic Energy is shared between them. (KE is not conserved.) More of the energy of the collision will be dissipated on the smaller vehicle (KE = mv²/2 and the velocity is squared so faster means much more KE).
The fact that the truck and car may apply brakes afterwards can cloud the issue about what actually happens. You have to ignore this for a simple treatment.
As for what happens to the occupants, you have, effectively, a sequence of collisions. Car to Truck then car to occupant. Each collision will take a finite time. What happens to the occupants will depend upon the time taken to come to a halt. Your truck may (probably will) push the small car backwards, making the change in velocity for the occupants even worse.
Change in momentum is called Impulse. Impulse is Force times the time it's applied for. Being brought to a halt slowly (crumple zone and resilient seat belt) will bring you to rest over a longer time and involves a smaller Force.
Sitting down with a pencil and paper and drawing some diagrams can help in getting an idea of what's "really" happening. If you apply Newton's Laws of motion rigidly, the situation can get much clearer.
And you will often notice that intuition can take you in the wrong direction.

 

[toesockshoe]

As momentum is conserved the change in momentum is equal and opposite for two objects in collision.

As for your question about passengers getting injured. What can you say about how the change in velocity of the two objects is related to their mass?

right, although the changes in momenti are the same, one object has a much smaller mass meaning it would have a much greater velocity than the other mass. Does this mean that the person in the small car would hit the airbag or steering wheel with a much greater momentum (separate from the cars momentum)?

 

[sophiecentaur]

the person in the small car would hit the airbag or steering wheel with a much greater momentum

Yes. The steering wheel could actually be coming towards him, pushed by the truck. Two velocities would add together.

 

[toesockshoe]

Yes. The steering wheel could actually be coming towards him, pushed by the truck. Two velocities would add together.

im still not too sure why the smaller car passengers feel it more... if the velocity changes faster does this mean acceleration is greater and the force at which he hits the steering wheel will also be greater?

 

[sophiecentaur]

The massive truck may only slow down a bit. The small car can end up being carried backwards fast. (This is just after the impact.) so the change In its velocity can be much more than for the truck.briefly, the occupant could still be moving in the original direction and meet the steering wheel coming back towards him.
This is yet another situation where the Maths can give a good explanation when you start with a couple of basic formulae.
Momentum is conserved ( for a start). KE will not be. KE is 'shared out' unequally and the smaller mass cops the worst of it.
"Force of impact" is not a meaningful term; Force X distance (= Work) and Force X Time (impulse) are the relevant quantities.