Question on momentum, impulse, and collisions

So you just need to rearrange the equation to solve for F.F=m(Vf-Vi)/ΔtF=(2800)(0-22)/.3F=-205333Yes, that's correct. Just remember to include units in your final answer. For part d, you'll need to use kinematics. You have the acceleration and time, so you can find the displacement using the equation:d=Vit+1/2at^2d=(22)(0.3)+1/2(-205333)(0.3^2)d=-429.4995In summary, a 2,800kg car traveling at 22m/s collides with
  • #1
Kalix
26
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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?
 
  • #6
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!
 
  • #7
Okay so for impulse I think it is FΔt=ΔmV
So I get the same answer for both a and b?
 
  • #8
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).
 
  • #9
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.
 

Related to Question on momentum, impulse, and collisions

What is momentum?

Momentum is a measure of an object's motion, determined by multiplying its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

What is impulse?

Impulse is the change in momentum of an object over a period of time. It is equal to the force applied to the object multiplied by the time it is applied.

What is the law of conservation of momentum?

The law of conservation of momentum states that in a closed system, the total momentum before a collision is equal to the total momentum after the collision. This means that momentum is conserved and remains constant in a closed system.

How is momentum related to collisions?

In a collision, momentum is transferred from one object to another. The total momentum of the system before and after the collision must remain the same, as dictated by the law of conservation of momentum.

How does an object's mass and velocity affect its momentum?

An object with a larger mass or a higher velocity will have a greater momentum. This is because momentum is directly proportional to an object's mass and velocity.

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