Some basic questions about collision force

[12thString]

1. a small car and a big truck collided. as Newton's 3rd law states, they should apply the same amount of force with each other. so if the big truck has a mass of (just to make it simple) 5kg and acc of 2m/s to the right and the small car has a mass of 2kg and acc of 3m/s to the left. (Ftruck=+10N, Fcar=-6N) of course the car has a smaller force, but as stated above, upon collision, they should have the same force? how do i determine that force given their initial[?] forces. please clarify me on this. a follow up question would be that if given the initial forces acting on two known masses, how can i determine the force upon their impact?

2. I've read in a magazine that when 2 motorbikes collide, they'll produce roughly 124K N of energy. Given that their initial velocity is the same at 50mph (22.2m/s, one positive and one negative of course since this is just one dimensional collision) and their masses are 80 and 90kg with a collision time of 0.015s, i was wondrin with how they derived this 124K N of force. I can't use conservation of momentum (and F=m\Deltav/\Deltat also) since i can't assume that they have zero final velocity as that would mean that they have the same mass. how should i use the given variables?


please enlighten me guys. ^^

 

[Astronuc]

In inelastic collisions, the total momentum is conserved, so the momentum before is equal to the momentum after. Then determine the change in momentum of each mass and divide by the time over which it occurs. That would given an averate force over that time period (collsion time).

http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html

In inelastic collisions, the kinetic energy is not conserved.

 

[12thString]

does that mean that the information provided in question number 2 is insufficient?

and how about question number 1?

thanks for the reply.

 

[surinderjitsd]

The cars made now are ones that absorb the shock for an instant, giving you time to react or buckle you feet and prepare for impact.