Homework Help: Finding the impluse of a stopping car

1. Aug 21, 2010

Ibby

A car with a total mass of 1400kg,travling at 60kmh hits a large tree and stops in 0.080 s .

2. change in momentium: m(v-u)

3. 1400(0-60)

=-84000

2. Aug 21, 2010

rock.freak667

That should be correct, I am not sure if impulses are positive or negative though. Based on it being the change in momentum, it should be negative.

EDIT: You will need to convert 60 km/h to m/s or else your units will be off.

3. Aug 21, 2010

Ibby

at the back of the book its 2.3 x 10 ^4 opposite tothe intial dircetion of the car

4. Aug 21, 2010

Ibby

got it f= ma 1400 x 16.6ms= 2.3 x 10^4 its right but isnt thsi meant to be the force or net force not the impulse?

5. Aug 21, 2010

cepheid

Staff Emeritus
That's not ma. That's m(vfinal - vinitial), which IS the impulse. The velocity change is 16.67 m/s. This is not an acceleration.

6. Aug 22, 2010

Ibby

iam not getting it . how is the velocity chnage 16.67? isnt that its inital speed since the car stoped when it hit the tree .so its final velocity should be 0 and its inital 16.67 ????

7. Aug 22, 2010

cepheid

Staff Emeritus
The change in velocity is the difference between the final velocity and initial velocity. I hope you understand this.

change in velocity = 0 m/s - 16.667 m/s = -16.667 m/s

Yeah, sure, 16.667 m/s is the initial velocity. But since the final velocity is zero, the change in velocity will be equal in magnitude to the initial velocity (and opposite in direction). Since I was only referring casually to the magnitude of the change, I was a bit lax with the negative sign in my previous post (but so were you). Anyway, multiply this change in velocity by the mass of the car to get its change in momentum (which, by definition, is the impulse).

Nitpicking aside, the main point I was trying to make in post #5 (which I hope you understood) was that for some strange reason you were claiming that 16.667 m/s was an acceleration when, in fact, it is not. It doesn't even have the right units to be an acceleration.

Last edited: Aug 22, 2010