De-acceleration of cars in wreck

[mljoslinak]

Homework Statement

Two automobiles of 540 and 1400 kg collide head-on while moving at 80km/h in opposite directions. After the collision the automobiles remain locked together. The front end of each automobile crumples by 0.60 m during the collision. Find the acceleration (relative to the ground) of the passenger compartment of each automobile; make the assumption that these accelerations are constant during the collision.

v=80 km/h
m1= 540
m2= 1400
displacement =.6 m
a1= ?
a2= ?
velocity after collision=9.9 m/s

Homework Equations

P=mv
KE=(1/2)mv^2
F=ma

The Attempt at a Solution

I know the momentum is constant--(68800 kg*m/s) and the kinetic energy before and after the collision is--(4.8*10^5 J, 9.4*10^4 J). I think that m1*a1=m2*a2, not too sure about that. After that I am not sure how to calculate the de-acceleration without knowing the impact time.

 

[chaoseverlasting]

I haven't checked your calculations, but I'm assuming that when you calculated the total momentum of the system, you took the direction of velocities into account. As they are in opposite directions, essentially it will be (m1-m2)*v.

How did you find the velocity of the automobiles after the collision?

From what is given, each of the automobiles crumples by 0.6m. So, the work done on each automobile is F1*x=W1, etc.

From the impulse momentum theorem you have the integral of F(t)*dt=m*(delta v). This force is W1/x for car 1 and W2/x for car 2.

Also, you know the work done is equal to the change in kinetic energy (as potential energy here is constant). Thus, you can find the time taken, and the change in velocity for each car and hence the acceleration.

 

[mljoslinak]

Thanks for the help. I think I solved it:
large car a= -130 m/s^2
small car a= 850 m/s^2 since the sign of the velocity switches.