Question on momentum, impulse, and collisions

AI Thread Summary
A 2,800 kg car traveling at 22 m/s comes to a stop after colliding with a wall over 0.3 seconds. The impulse experienced by the car can be calculated using the equation FΔt = m(Vf - Vi), resulting in a force of approximately -205,333 N exerted by the wall. This force is also the same magnitude but opposite in direction to the force the car exerts on the wall, according to Newton's third law. To find the acceleration, use F = ma, and for distance, apply kinematic equations. Understanding the definitions of impulse and force is crucial for solving these types of collision problems.
Kalix
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Homework Statement


Question:A 2,800kg car is traveling at a velocity of 22m/s. The car hits the wall and comes to a complete stop. The collision took 0.3 seconds.

a. What is the cars impulse?
b. What force did the wall exert on the car?
c. What force did the car exert on the wall?
d. Calculate the acceleration rate and the distance the car traveled during its acceleration?

m=2,800kg
Vi=22m/s
Vf=0m/s
t=0.3 seconds
F=?

Am I missing any other variables?

Homework Equations


FΔt=mVf-mVi

There must be other equations but I have no idea what they are.


The Attempt at a Solution


So for part a this is what I tried
FΔt=mVf-mVi
F=(2800)(0)-(2800)(22)/.3
F=-205333
This seems wrong to me just because its so big. So what am I doing wrong?

I don't really know what to do for b, c, and d so if someone could get me started on those that would be great. Thanks!
 
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Kalix said:

The Attempt at a Solution


So for part a this is what I tried
FΔt=mVf-mVi
F=(2800)(0)-(2800)(22)/.3
F=-205333
This seems wrong to me just because its so big. So what am I doing wrong?
You're doing nothing wrong. That's the answer to part b. (Be sure to give the correct units.) It seems big because it is. You have a briskly moving heavy vehicle being stopped quickly.

What about part a?

For the rest, apply Newton's laws.
 
oh I thought I was doing part a. Then how do you find impulse? Is there another equation for that?
The for part c how is that any different than part b. Is it just the opposite. So 205333. (not negative)
and finally for part d I understand how to find acceleration (F=ma i think) but how do I find the distance? Do I have to you use kinematics?
 
Kalix said:
oh I thought I was doing part a. Then how do you find impulse? Is there another equation for that?
You don't need another equation. What's the definition of impulse?
The for part c how is that any different than part b. Is it just the opposite. So 205333. (not negative)
Right. What law tells you that?
and finally for part d I understand how to find acceleration (F=ma i think) but how do I find the distance? Do I have to you use kinematics?
Yes, you'll need to use kinematics.
 
In my book it says the definition of impulse is "the product of the force and the time over which the force acts on an object"
The wording is what really confuses me. What does "the product of the force mean"?

Newton's third law?
 
Kalix said:
In my book it says the definition of impulse is "the product of the force and the time over which the force acts on an object"
The wording is what really confuses me. What does "the product of the force mean"?
"Product" just means multiply. The product of A and B is AxB.

With that in mind, look at the equation you already used.
Newton's third law?
Right!
 
Okay so for impulse I think it is FΔt=ΔmV
So I get the same answer for both a and b?
 
Kalix said:
Okay so for impulse I think it is FΔt=ΔmV
Yes: The left side of that equation is the definition of impulse.
So I get the same answer for both a and b?
No. For a you want the impulse; for b you want the force (which is just one factor of the impulse).
 
So what equation do I use for b. if I use FΔt=mVf-mVi I get the same answer as I do in a. I thought about using F=ma but I don't have the acceleration.
 
  • #10
Kalix said:
So what equation do I use for b. if I use FΔt=mVf-mVi I get the same answer as I do in a.
You already solved b! Anyway, it's the same equation, but you're solving for F instead of impulse.
 
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