How Do You Calculate the Impact Force in a Two-Car Collision?

[get_physical]

Homework Statement

1200kg car moving at 9.72m/s runs into and sticks to a parked 1000kg car. The collision takes 0.15s and after the crash the cars wheels lock and they skid to a stop on u=0.40 road. What is the impact force on both cars?

Homework Equations

J=F*t=m*dv

The Attempt at a Solution

I've found the distance it takes for it to stop (3.68m) and also the speed after the crash (5.3m/s).

I used the momentum to divide by the time (0.15s) to get the impact force, but the answer is wrong.

 

[jackarms]

Did you make sure to use the change in momentum? Otherwise, your calculations for the speed after the crash and distance may be off.

 

[get_physical]

The calculations for the speed after the crash and the distance are correct. But I just can't find the impact force.

 

[NascentOxygen]

The calculations for the speed after the crash and the distance are correct. But I just can't find the impact force.

Show your arithmetic. Maybe you are making a mistake.

 

[lendav_rott]

What does it mean that the collision takes 0.15 seconds? That deformation takes 0.15 seconds after which the cars start skidding with the remaining energy?

Oh Ok, I think I understand. Think about kinetic energies. If you knew the amount of energy and the time it was being applied, you can find out the force affecting the target mass.
Before the collision, the "speeding" car as a kinetic energy of N. After the collision, the 2-car-mass has a kinetic energy of M. M is dissipated by friction over time. What happened to the N-M , though?

 

[BvU]

Dear get,

All this could have been short-circuited if you had shown why F = 1200 kg * (9.72 - 5.3)m/s / 0.15 s did not yield the right answer. Someone would have told you to check your calculation against 1000 kg * 5.3 m/s / 0.15 s

 

[get_physical]

Dear get,

All this could have been short-circuited if you had shown why F = 1200 kg * (9.72 - 5.3)m/s / 0.15 s did not yield the right answer. Someone would have told you to check your calculation against 1000 kg * 5.3 m/s / 0.15 s

Thank you. I used F= 2200 (0 - 5.3m/s) /0.15
Why do we only use 1200kg car and why do we not use 1000kg car?

 

[Doc Al]

Thank you. I used F= 2200 (0 - 5.3m/s) /0.15
Why do we only use 1200kg car and why do we not use 1000kg car?

To find the force on a car, you must analyze the change in momentum of that car. Not both together! You can use either car.