Car Crash and Such: Investigation and Analysis

[Doc Al]

You seem to have the right idea now, but made an error here:

pxf= (934.40kg)(9.29m/s cos (45))= 61,389.09kg*m/s
pyf= (934.40kg)(9.29m/s sin (45)) = 6,1389.09kg*m/s

These should be: 6,138.09 kg-m/s
...

However about the negative ...not sure where to put that...I was going to say that the momentum for the initial and final was negative since the car 1 goes in the y direction initially and finally goes in the y direction as well.

Don't worry about the direction or sign of the impulse (although you could certainly find the direction if you wanted to, just like you'd find the direction of any other vector), just figure out the magnitude.

 

[~christina~]

You seem to have the right idea now, but made an error here:

These should be: 6,138.09 kg-m/s

Oh my... thanks for catching that Doc Al

well editing that...

m1= 934.40kg
viy= 25.06m/s
vix= 0m/s

vfy= 9.29m/s sin 45
vfx= 9.29m/s cos 45

pxi= (934.40kg)(0m/s)= 0kg*m/s
pyi= (934.40kg)(25.06m/s)= 23,416.064kg*m/s

pxf= (934.40kg)(9.29m/s cos (45))= 6,138.09kg*m/s
pyf= (934.40kg)(9.29m/s sin (45)) = 6,138.09kg*m/s

$$\Delta p_x= p_{xf}-p_{xi}= 6,138.09kg*m/s - 0kg*m/s = 6,138.09kg*m/s$$

$$\Deltap_y= p_{yf}-p_{yi}= 6,138.09kg*m/s - 23,416.064kg*m/s= -17,277.974kg*m/s$$

magnitude $$I= \Delta P = \sqrt{(6,138.09kg*m/s)^2 + (-17,277.974kg*m/s)^2}= 18,335.88kg*m/s$$

If it is now correct then I would think that the impulse is the impulse the car 1 delivered to car 2 when they crashed

Thanks

 

[Doc Al]

Looks good.

If it is now correct then I would think that the impulse is the impulse the car 1 delivered to car 2 when they crashed

Since you calculated the change in momentum of car 1, you found the impulse delivered to car 1. (Which was delivered by car 2, of course.)

 

[~christina~]

Looks good.

Since you calculated the change in momentum of car 1, you found the impulse delivered to car 1. (Which was delivered by car 2, of course.)

Oh..yep..


Thanks for all your help Doc Al