# Truck collision with a resting car

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1. Feb 27, 2015

### vizakenjack

So a truck hits a car, which is at rest.

Would there be a moment in time when a car stops exerting a force in equal magnitude but in opposite direction (Newton's third law)?

To put in some arbitrarily numbers to make the example easier to understand.
Truck weighs 1000 kg, and the car weighs 100kg.
The truck when it hits the car has a velocity of 100m/s, it is accelerating at constant rate of 10m/s.

So after hitting the car it keeps accelerating 10 m/s.

So the truck exerts a force on the car, which is 1000kg * 9.8 (gravity constant)?
If so, then the car ALSO exerts a force of the same magnitude (1000 * 9.8=9000 N), right?

That means, the truck is being decelerated at a = F/m = 9000N/1000kg = 9 m/s
and the car is being accelerated (because it was at rest) at a = F/m = 9000N/100kg = 90 m/s
?
If it's true, does it mean that the truck will be accelerating forward with a(of the truck itself) - a(from the force exerted by the car on the truck) = 10m/s - 9m/s = 1 m/s
?

And eventually, would the car become one part with the truck? So these two objects: truck and car
will turn into one heavier object m3? No? The car would keep exerting opposite force on the truck? I mean, the car has wheels... I feel like as soon as the car has the same velocity as the truck, it should stop exerting force on the truck...
no?

2. Feb 27, 2015

### Staff: Mentor

When the car and truck collide they exert forces on each other. Those forces change, but at any instant the forces they exert on each other are always equal and opposite. Is that your question?

(The force starts out at zero, peaks rapidly, then goes back to zero as things settle down. There is no simple way to determine what those forces will be, without more detailed modeling.)

3. Feb 27, 2015

### vizakenjack

Well, is everything above was correct then? The truck will have an acceleration of 1m/s?
But then you said that "Those forces change", which means after some time the car will be exerting less force on the truck and vice versa?
How could truck exert less force on the car, though? Is it because the car will be in motion?

"goes back to zero", what do you mean? The truck would eventually stop exerting force on the truck? But it would keep pushing it forever... why would it stop exerting force on the car, then?
Unless the truck and car start acting as one unanimous object, I don't see how that's possible.

4. Feb 27, 2015

### Staff: Mentor

You stated in the problem that the truck is accelerating at 10m/s, but we don't know any of the details about how that happens. However, if that comes from a constant torque in the truck's tires, then you just apply f=ma to the new combined mass, giving the truck+car an acceleration of 9.1 m/s2. As Doc said, the interaction during the collision is very complicated, but that's what the situation looks like after the collision.
I think Doc was considering a more normal collision scenario, where the truck driver doesn't keep his foot on the gas while plowing through a car.

However, not everything in there was correct. Sorry, but most of it was wrong:
1. Acceleration is m/s2, not m/s.
2. When the collision happens, the truck will decelerate slightly, for a certain time, but how long and at what rate is impossible to know. However, you can if you consider the collision instantaneous, calculate how much speed it loses during the collision.
3. You calculated force wrong. You don't use the gravitational acceleration, you use the acceleration of the truck and car.
4. Much of the force of the collision doesn't cause acceleration at all, it just crushes the car.

Last edited: Feb 27, 2015
5. Mar 1, 2015

### vizakenjack

Well, alright, even if the truck stops accelerating. After the collision, how would the force exerted by the car onto the truck change? I figure, if the truck has no acceleration, then it shouldn't exert any force on the car? (F=ma, a = 0) But that doesn't make sense, since the truck has a velocity and is then decelerating... and it will push the car some distance forward...
No, I mean, if the car isn't on the brake, and can simply roll if you push it, will there be a point when the car would stop exerting force back at the truck?
I really do feel like if the truck is accelerating, and then both the truck and the car gain the same velocity, then the car would stop exerting the force back at the truck, the truck would simply become a much heavier object (truck+car), no?

6. Mar 1, 2015

### sophiecentaur

No.
N3 has nothing to do with equilibrium or deformation of one of the colliding objects. It just says action and reaction are equal and opposite.

7. Mar 1, 2015

### Staff: Mentor

Yes, exactly. I think it would be less complicated to consider an idealized collision where the only horizontal forces involved are due to the collision itself.

You need to clearly distinguish before, during, and after the collision. Before the collision, the force the two vehicles exert on each other is zero, of course. During the collision, that zero force increases rapidly and then dies off. After the collision, assuming a perfectly inelastic collision, the truck + crumpled car will continue moving along as a single body. (Again, an idealized case where we ignore the friction and other forces that will eventually slow the system down.)

Sure. But at any point, the force they exert on each other is always equal and opposite.

During the collision, the truck is slowing down (thus accelerating) and the car is speeding up (thus accelerating). During that time of acceleration they are exerting equal and opposite forces on each other. (The forces are the same, not the accelerations.) After the collision, the "truck + car" ends up moving at some final speed. (At that point, the truck and car are no longer exerting force on each other, but are just moving together. The net force on each will be zero.)

At no point will the truck be pushing the car while the car is not also pushing back on the truck.

8. Mar 1, 2015

### vizakenjack

Truck is decelerating, while the car is accelerating. Also, the values of their acceleration will be decreasing (given that gas pedal on the truck isn't being pressed on), right?

But F=ma, force is related to acceleration, less acceleration = less force.
So I was right? The force which the truck is exerting at the car will be less and less as the car is gaining velocity?
Also, I guess, collision wasn't a good scenario. What if it's not collision, but rather, two identical lightweight cars (100kg), both at rest and are touching each other.
First car then decides to start accelerating very slowly with constant rate, let's say 1 m /s^2.
Would the 2nd car even "push" back on the 1st car? If yes, wouldn't it only for a very brief period of time? Then they both would instantly gain the same velocity, thus, stop exerting force on each other, right?

9. Mar 1, 2015

### Staff: Mentor

OK. (In physics language, the term "acceleration" covers both speeding up and slowing down.)

The acceleration is proportional to the force. As the force settles down at the end of the collision, so will the accelerations.

Even though the force on each is the same, the accelerations are different because the masses are different.

No, you were not right. But yes, the force that the truck is exerting on the car will become less and less as the impact forces settle down.

No, wrong. Since the first car keeps accelerating at the same rate as it pushes against the second car, that means that the second car must now be accelerating as it is moved by the first car. Right? And for it to be accelerating, that means the first car must be exerting a force on that second car. So the two cars will continue to exert equal and opposite forces on each other.

Let's make up some silly numbers. Let the cars have a mass of 1000 kg. So for the first car to accelerate at 1 m/s^2, the net force on it must be 1000 N. That force is provided by the ground pushing on the tires. As it begins to push the second car, if you want them to keep accelerating at the same rate, that means the driving force of friction must now be 2000 N, since both cars must be accelerated. The force that the first car exerts on the second must be 1000 N in order for that second car to have an acceleration of 1 m/s^2. And that means that the second car pushes back on the first car with a 1000 N force. Thus the net force on the first car is 2000 - 1000 N = 1000 N.

Make sense?

10. Mar 2, 2015

### sophiecentaur

It could be useful to make a simple model of this, not involving a truck and a car but two point masses, each with buffer, consisting of a single spring and a friction damper. All the variables on each side would be different. You should also introduce some friction with the ground, for each car. The springs would be obeying Hooke's Law (for simplicity). To represent how steel car bodies behave, the friction dampers would have a high damping coefficient so there would be over-damping. This idea doesn't introduce anything extra into the discussion but it addresses the essence of the problem and helps in avoiding too much intuitive mis-reasoning. An equation of motion could easily be built up from here.

11. Mar 3, 2015

### vizakenjack

No.
By the ground? I thought by the motor of the car... or by its wheels...

What is this "driving force of friction", and why is it 2000N?
There would be only one friction - friction of the ground, that's it. It could be very small, for instance, μk (kinetic coefficient of the friction) could be 0.25, therefore friction when the car is moving = μk * Fn
No?
I don't get this part.
They both will be exerting forces on each other simultaneously. And if the first car has only 1000N pushing it forward, it would be cancelled by the second car's force pushing on it backwards.
No?

12. Mar 4, 2015

### Staff: Mentor

In order for the car to accelerate there must be an external force acting on it. That force is provided by the ground. (Try accelerating on a patch of slippery ice.) Of course, the motor and wheels are essential for creating that friction force.

I'm trying to analyze the example you gave, where the first car maintains an acceleration of 1 m/s^2 while pushing the second car. In order to keep accelerating at that rate when the first car is pushing the second car, a force of 2000 N is required.

No.

For one thing, unless the tires are slipping, the friction required is static friction, not kinetic friction. And if the maximum static friction (between tires and ground) is too low, then you will not be able to produce the required force to maintain the acceleration.

No.

If you want to keep the driving force at 1000 N, then when the first car is pushing the second the resulting acceleration will only be 0.5 m/s^2. (That comes from ΣF = ma, where F = 1000 N and m = 2000 kg.) In that case, the force that the first car exerts on the second car will only be 500 N. (And, of course, the second car will push back on the first car with an equal and opposite force.)

But if you want to keep both cars accelerating at 1 m/s^2, you must increase the force applied to 2000 N. Makes sense, does it not, that you must push harder when you are accelerating both cars?

13. Mar 6, 2015

### vizakenjack

Both cars have the same mass of 1000kg.
When the first car pushes the second car, in order for both of them to have an acceleration of 1m/s^2, the first car must exert 2000N.

That's because if the first car only exerts 1000N, according to F=ma, where m the mass of both cars (since they're touching each other), 1000N=2000kg*acceleration
a = 0.5
Which means, effective exerting force of the 1st car onto the 2nd car is only 500N, despite the fact that the first car is trying to exert 1000N.
Hm... did I get it correctly?
If so, let me see, the 1st has 1000N, right? But the acceleration is 0.5, since it has to also push the 2nd car.
But(!) why is the fact that the 2nd car is pushing back onto the 1st car isn't accounted for?
1st pushes forward with 500N and 0.5 acceleration, but the second car pushes back with 500N too, so why don't both cars simply stop due to acceleration being 0?

In other words, if there was only one car involved in the problem, and its mass was 2000kg, and the applied force 1000N, then I see how it would have had an acceleration of 0.5.

But in our case, there are TWO cars... with the second car pushing back onto the 1st one...

Last edited: Mar 6, 2015
14. Mar 6, 2015

### jbriggs444

By this you should mean that the first car must exert 2000N backwards on the ground and the ground, accordingly will exert 2000N forward on the first car.

15. Mar 6, 2015

### Staff: Mentor

That last phrase is a bit ambiguous, which I think is part of the problem here.

I am assuming that only the first car is providing the driving force. As jbriggs444 notes, that driving force is produced by the tires pushing on the ground and thus the ground pushing back on the first car, driving it forward. (I assume that the 2nd car is just rolling without friction; engine off, in neutral.)

So, I would rephrase your sentences as: When the first car pushes the second car, in order for both of them to have an acceleration of 1m/s^2, the net external force on the cars must be 2000 N. That is the force that the ground exerts on car 1. It is not the force that car 1 exerts on car 2.

Not exactly. If car 1 pushes against the ground with a force of 1000 N, then the net driving force on both cars is 1000 N, which would make the acceleration only 0.5 m/s^2. In that situation, car 1 will push on car 2 with a force of 500 N.

Here's how to analyze it. You can analyze it two ways: (1) by treating the two cars as a single system, or (2) by treating each car separately.

Let's start with (1): Treating the two cars as a single composite system we will ignore the forces they exert on each other since those forces are internal to the system. Applying Newton's 2nd law, if the external driving force from the ground on car 1 is 1000 N, then the acceleration of the two-car system is 0.5 m/s^2.

Now let's do (2), treating them separately: First consider the forces on car 2. Since we know that its acceleration is 0.5 m/s^2, we know that the net force on car 2 must be 500 N. That is the force that car 1 exerts on car 2.

Now consider the forces on car 1. There's the driving force from the ground on car 1, which is 1000 N. Then there's the force that car 2 pushes back on car 1, the 500 N. (From Newton's 3rd law.) So the net force on car 1 is 1000 - 500 = 500 N. Which makes sense, since the acceleration is only 0.5 m/s^2.

Because the ground is exerting a force of 1000 N!

16. Mar 15, 2015

### vizakenjack

To put the final nail in the coffin.
One massive car of mass of 2000kg that has a force forward of 1000N, will have an acceleration of 0.5

Two cars, both have the masses of 1000kg, 1st car has a force forward of 1000N, the second is at rest. But they also both will end up with acceleration of 0.5

right?

17. Mar 15, 2015

### Staff: Mentor

Right. (Assuming no other external forces are involved, as we have been doing.)

18. Mar 15, 2015

### vizakenjack

I just have a small problem with how the 1st car is exerting force onto the 2nd car.

I mean, it exerts 1000N, period. Yes, the 2nd car exerts back 500N, but the 1st car would still be exerting 1000N onto the 2nd car.
The 2nd car exerting back 500N doesn't cancel the applied force of the 1st car, it just means that both cars are exerting forces on each other simultaneously.
No? Alright, then, I guess the resisting force of the 2nd car (500N) needs to be subtracted from 1000N applied force of the 1st car in order to find the net effective force of the 1st car...

19. Mar 15, 2015

### Staff: Mentor

No the ground exerts 1000 N on the 1st car. That is what provides the driving force.

No. The ground exerts 1000 N on the 1st car. The 1st car exerts 500 N on the 2nd car and the 2nd car exerts 500 N back on the 1st car.

That's certainly true.

Yes, to find the net force acting on the 1st car you would subtract 500 N from 1000 N to get a net force of 500 N on the 1st car.

The only external force acting on the cars is the 1000 N force from the ground on the 1st car. When the 1st car pushes the 2nd car, that 1000 N force from the ground must accelerate both cars.