Force on the car in a collision

[TT0]

Homework Statement

Car A with mass 1000 kg is accelerating at 1ms^-2 at car B.
Car B with mass 1000 kg is accelerating at 1ms^-2 at car A.
What is the force on car B at the collision?

Homework Equations

F = ma
Newton's third law

The Attempt at a Solution

Car A force:
F = ma
F = 1000 x 1
F = 1000 N

Car B force:
F = ma
F = 1000 x 1
F = 1000 N

Then using Newton's third law, since every action has an equal opposite reaction force, if car B was colliding with a stationary object then the force on car B after colliding is 1000 N. But since car A is traveling at car B with 1000 N of force then that also has to be added into car B's total collision force. So therefore car B has a total collision force of 2000 N.

Is the answer right? Also is my explanation correct?

 

[Scott Redmond]

From Newton's third law, each force involves two objects, e.g. weight is exerted on an _apple_ by the _Earth_. What are the objects involved in each of the two forces you've identified?

 

[TT0]

I'm not really sure what you are asking but the objects involved is the 2 cars, car A and car B. Is this what you are asking for?

 

[Scott Redmond]

Not really, but maybe I'm misinterpreting the problem. If car A is accelerating at 1 m/s² "at car B", what is causing that acceleration? Also, what kind of force is it (e.g. gravitational, normal, friction, tension, ...)?

 

[haruspex]

The question makes no sense.
We are not told the velocities at collision, so cannot assess the momenta. Neither are we told the duration of the collision, so there is no way to determine the forces during impact (and they would not be constant during the impact anyway).
Where does this question come from?

 

[TT0]

It was a question my friend had. Can't you explain this through Newton's third law of motion?

 

[Scott Redmond]

You can, but you need a lot more detail first. That's why I was asking about the objects and types of forces involved, and even that's just the beginning.

 

[TT0]

If it was a collision between 2 cars that are traveling at each other in a straight line and they have a "perfect collision". Is this enough information?

 

[haruspex]

It was a question my friend had. Can't you explain this through Newton's third law of motion?

In your attempt to do so, you used an acceleration before impact and related it to a force after impact. That does not constitute a valid application of the law. It has to be the net force producing the acceleration. The accelerations here are produced, I assume, by the engines.

 

[TT0]

How about if I reworded the question a bit: Two cars are traveling at each other with 1000 N of force each. What will be the impact force of one of the cars?

 

[Scott Redmond]

What does it mean to travel with 1000N of force? Sorry for all the questions, but these are things you need to think through.

 

[TT0]

No problems, its my bad that I did not explain the question properly. What I mean by 1000N of force is that the car has a force of 1000N. Eg. Car A has a force of 1000N, car B has a force of also 1000N but in a opposite direction to car A so they will collide together. Is this enough information?

 

[Scott Redmond]

No problems, its my bad that I did not explain the question properly. What I mean by 1000N of force is that the car has a force of 1000N. Eg. Car A has a force of 1000N, car B has a force of also 1000N but in a opposite direction to car A so they will collide together. Is this enough information?

Okay, but what does "a force of 1000N" mean in this question? Usually we speak about cars having a velocity (speed and direction) when talking about collisions (e.g. when the police take measurements to determine each car's speed before a collision and decide whether to give a ticket for dangerous driving). Sometimes we'll talk about a car's weight in Newtons (e.g. a 9810N car, which would have a mass of 1000 kg) but that's not so common.

It's a fantastic idea to create your own questions to help understand how the physics ideas fit together. You still have a lot to review, but you'll come out of it much farther ahead!

 

[TT0]

Force of 1000N means the car has a force of 1000N applied to it. So the car has a net force of 1000N.

 

[Scott Redmond]

But how does that relate to the second car?
Since 2 objects are involved in every force, what other object is exerting that 1000N force on the car?
What direction is the 1000N force in?
Is that before the collision, during the collision, or after the collision?

 

[TT0]

But how does that relate to the second car?

Car A and car B both have 1000N of force applied to it. The force are in opposite direction.

Since 2 objects are involved in every force, what other object is exerting that 1000N force on the car?

The engine is exerting the force on the car. The force from the engine on the car is greater than 1000N. The 1000N is from the force from the engine subtracted by the friction.

What direction is the 1000N force in?

The 2 cars are traveling in opposite directions.
Eg. Car A has 1000N applied to it to the left. Car B has 1000N applied to it to the right. So the 2 cars are traveling at each other.

Is that before the collision, during the collision, or after the collision?

This is before the collision. Disregard air resistance so that before the collision, right up to it the net force of each car is 1000N. I am trying to find what is the force after the collision.

 

[haruspex]

Force of 1000N means the car has a force of 1000N applied to it. So the car has a net force of 1000N.

This is no different from your original statement, that the cars have mass 1000kg and are being accelerated at 1m/s². The longer that state lasts before collision, the greater the speed at impact, so the greater the forces during the impact.
Suppose the time from rest to impact is t₁, that during that time each car is propelled towards the other with a net force F, and the duration of the impact is time t₂. The average force during impact will be $F\frac{t_1}{t_2}$.

 

[TT0]

How about if the force of each car right before the collision is 1000N? Could you then explain it with Newton's third law? My understanding is that if a car has a force of 1000N right before a collision and it hits a brick wall then the force on the car immediately after the collision would be 1000N because of the third law. The question is to help me understand what happens if the car collided with a object that is moving at it? Will the force on the car immediately after the collision be the force of the object (it collided with) + the reaction force of the car?

I apologise for my questions, I have only started physics recently so I am not completely sure as to how to explain a question.

 

[haruspex]

if the force of each car right before the collision is 1000N?

As Scott has tried to explain to you, this doesn't mean anything.
An object has momentum, an object has KE, but an object does not "have" force. Force is something one object exerts on another.
The way you are using the term 'force' here sounds like what you are intuitively thinking of is momentum. The cars will have momentum just before impact. The units of momentum are kg m/s, or N s (Newton-seconds).

 

[TT0]

Ok I think I get where you are coming from now. The purpose of the question is for me to understand Newton's third law, that every action has an equal opposite reaction.

How I understand is if a car hits a stationary object with 1000N of force it will have 1000N of reaction force. Is this the correct way to think about it?

So if my understanding above is correct then what happens if the object that the car hits was not stationary? So if the car hits the object with 1000N of force and the object hits the car with also 1000N of force, will the reaction force on the car be 2000N? I got this from:
Total force = reaction force of the car + the action force of the object.

The equation is something I though of and is not an equation I learned from school so it might be wrong.

Also thanks for helping me even though my question is incorrect.

 

[haruspex]

if a car hits a stationary object with 1000N of force

That still isn't the right way to think of it. A force is exerted at some instant. A collision happens (in reality) over a period, though it may be very short. The impulse (momentum) is the integral of the force over the time for which the force is applied. So you can say car A hits car B and imparts some quantity of momentum to it, but it is not meaningful to say it hits it with a certain force. There will be an average force over the duration of the impact, and there will be a peak force - but that may be very large and difficult to calculate.

what happens if the object that the car hits was not stationary

The momentum imparted depends on what the two vehicles were doing and how massive they are.
You can apply the action equal and opposite to reaction law at each instant during the collision. The consequence is that they impart equal and opposite momentum to each other, that is, total momentum is conserved:
$\vec{F_{AB}}(t) = -\vec{F_{BA}}(t)$, $\int \vec{F_{AB}}(t).dt = -\int \vec{F_{BA}}(t).dt$.

if the car hits the object with 1000N of force and the object hits the car with also 1000N of force, will the reaction force on the car be 2000N?

Let me rephrase that as: "If object A exerts a force F on object B, the reaction force from B onto A is -F. Does this mean the total force is 2F?"
No. Each exerts a force of magnitude |F| on the other. You can think of the equal and opposite pair of forces as constituting a compression, but its magnitude is still F.
See if https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/ helps at all.

 

[TT0]

Ok thanks, I get the idea now, that you don't use force in this scenario.
If I said I kicked a soccer ball with 10N of force is this the correct way to use force?

Thanks

 

[haruspex]

If I said I kicked a soccer ball with 10N of force is this the correct way to use force?
Thanks

No. You could place your boot in contact with the ball then push it with a force of 10N, but kicking implies your boot made contact at some speed. Suppose you had accelerated the boot with a force of 10N for 0.2s then your boot would have acquired momentum of 2 kgm/s. If contact with the ball then lasts 0.05 s and slows your boot to half its prior speed then you will have imparted 1kg m/s of momentum to the ball at an average force of 20N.

 

[TT0]

Ok so you used the impulse formula I'm guessing?

I = F∆t

So my new (and hopefully correct) question is:
If my foot had a momentum of 2kgm/s prior contact and the contact with the ball lasts 0.05s. Then after the contact the foot momentum changed to 1 kgm/s, then what is the reaction force on the foot from the kick?

For the foot:
I = F∆t
1 = F x 0.05
F = 20 N

Then the foots reaction force would be 20N since the force of the foot on the ball is 20N. Is this correct?

 

[haruspex]

Ok so you used the impulse formula I'm guessing?

I = F∆t

So my new (and hopefully correct) question is:
If my foot had a momentum of 2kgm/s prior contact and the contact with the ball lasts 0.05s. Then after the contact the foot momentum changed to 1 kgm/s, then what is the reaction force on the foot from the kick?

For the foot:
I = F∆t
1 = F x 0.05
F = 20 N

Then the foots reaction force would be 20N since the force of the foot on the ball is 20N. Is this correct?

That would be the average reaction force over the .05s. The actual force will vary over that time. You could treat it as a simple elastic impact, with the force increasing to a max then falling away to zero again as a sine function. That would probably be a decent approximation. See the example graphs at the link I posted.

Correction, try this link https://www.physicsforums.com/insights/frequently-made-errors-mechanics-momentum-impacts/

 

[TT0]

Ok thanks for all your help. I'll think about it. I really appreciate you and scott helping me even though my questions are incorrect, so thanks again.

 

[Nile Anderson]

Now I am new here but not new to Physics , I now this may seem very simplified but if they are coming at each other , can we not assume they are coming from opposite directions and hence when they collide they have the same acceleration and mass, then we can assume that on collision the , net force will be zero

 

[haruspex]

Now I am new here but not new to Physics , I now this may seem very simplified but if they are coming at each other , can we not assume they are coming from opposite directions and hence when they collide they have the same acceleration and mass, then we can assume that on collision the , net force will be zero

Net force on what? If there's a fly in between, yes the net force will be zero on the fly. There will, however, be a compression.
Oh, and it's not having the same acceleration that matters. It's having the same speed at impact.

 

[Nile Anderson]

Net force on what? If there's a fly in between, yes the net force will be zero on the fly. There will, however, be a compression.
Oh, and it's not having the same acceleration that matters. It's having the same speed at impact.

I am sorry but how do we define a force again, I think my error lies there , is it not the change of momentum rather than the momentum , and are we not just concerned with forces here, I understand I may be holding some common misconception though.

 

[Nile Anderson]

I am sorry but how do we define a force again, I think my error lies there , is it not the change of momentum rather than the momentum , and are we not just concerned with forces here, I understand I may be holding some common misconception though.

I am thinking really on a resultant if we were to consider the forces involved .

 

[haruspex]

how do we define a force again, I think my error lies there , is it not the change of momentum rather than the momentum

It's neither.
A force is something one body exerts on another. A consequence of a force acting over a period of time can be a change in momentum.

I am thinking really on a resultant

Sure, but you have to identify the body acted on. If you fix on one of the cars, the other car exerts a force on it during the impact, but it does not exert a force on itself, so the resultant is not zero.
To get a zero resultant here you need to introduce a third object for both cars to act on.

 

[Nile Anderson]

It's neither.
A force is something one body exerts on another. A consequence of a force acting over a period of time can be a change in momentum.

Sure, but you have to identify the body acted on. If you fix on one of the cars, the other car exerts a force on it during the impact, but it does not exert a force on itself, so the resultant is not zero.
To get a zero resultant here you need to introduce a third object for both cars to act on.

Newton's Second Law ?

 

[haruspex]

Newton's Second Law ?

Applied here how, exactly?

 

[Nile Anderson]

F=mv-mu/t=m(v-u)/t=ma

 

[haruspex]

F=mv-mu/t=m(v-u)/t=ma

I asked exactly how you are applying it in the present context. Which body has mass m? What force acting on the body is represented by F? What time period does t stand for?
Anyway, F=m(v-u)/t is not quite right. That will give you the average force over the time interval t. Likewise, your a is the average acceleration.,