B Greater momentum on impact means greater force?

[dibbsy]

TL;DR Summary

Trying to understand force and momentum. If F=ma, then force is about acceleration, not velocity. But what about greater impact with greater momentum and no acceleration?

Sorry for this beginner's question, but...if F=ma, then force is all about acceleration. But if vehicle A moving at constant velocity V hits a wall, and vehicle B moving at constant velocity greater than V hits the wall, then B hits the wall with greater momentum than A and does greater damage etc., so it must have hit the wall with greater force than A. So nothing about acceleration in this scenario, but force is greater. What am I missing here? Thank you in advance.

 

[Ibix]

What will happen to the velocity of the cars when they hit the wall? If the second car is going faster, what's the relative magnitude of the changes? And what does that say about the forces involved in the two collisions?

 

[dibbsy]

Not sure what these questions mean, sorry. Is there a point here? Apologies if I'm missing it. I'd be grateful for an answer rather than more questions.

 

[Ibix]

The aim is to lead you to the answer.

Try these instead. Is the velocity of the cars the same before and after the collision? If it's changed, what's that called?

 

[dibbsy]

At the moment of collision, the velocity of A is V and the velocity of B is >V.
So at the moment of collision, surely B hits the wall with greater force than A, simply due to greater momentum = mV. So nothing to do with acceleration, as far as I can see. I just don't get what I'm missing.

 

[Ibix]

Try answering my question. Did the car's velocity change? And what is it called when a velocity changes?

 

[dibbsy]

No, no change until after impact. At the moment of impact, the velocities of A and B were the same as they were the moment before impact.

 

[Ibix]

So, are you proposing that the car touches the wall, the wall and the car are instantly damaged while the whole car continues at 30mph (or whatever), and then the car stops for no reason related to the damage?

Or do you think the car might be slowed while it crumples and damages the wall?

 

[dibbsy]

OK so the cars are slowed when they hit the wall. That means they decelerate. The faster the car's velocity, the greater the deceleration needed to slow it down, so the greater the force needed to slow it down. So how does that explain the greater damage done to the wall by the faster car?

 

[berkeman]

So how does that explain the greater damage done to the wall by the faster car?

As you say, since F=ma, the faster car experiences higher crash forces compared to the slower car. You can also think about the higher energy that the faster car has (kinetic energy) -- during the crash that energy goes into deforming the car and wall and some is dissipated as heat from the collision.

 

[dibbsy]

OK, so there are two parts to your reply. (i) Not sure what 'experiences higher crash forces' means exactly. I guess that it experiences greater deceleration from the wall. But I don't see how that explains greater damage to the wall in the first place. You can say that Newton's Third Law implies that corresponding to the force from the wall on the car, there is an equal force from the car on the wall. But since the faster car experiences greater deceleration, it also imparts a greater force on the wall than the slower car. Is that right? But neither car is accelerating. They are decelerating. So once again, the explanation of the greater damage done by the faster car seems to have nothing to do with acceleration.

 

[dibbsy]

(ii) The faster car has greater kinetic energy = mv^2, which occurred to me as well and must be right. And that would be a straight and neat explanation of the greater damage done to the wall by the faster car. You wouldn't even need to appeal to 'crash forces' etc. in your first part, which is hard to understand anyway, as I suggested.

But here's the thing: mv^2 has nothing to do with acceleration, so the F generated by the higher kinetic energy in the faster car is proportional to mv^2, not ma. These are two different phenomena, as demonstrated by the fact that neither car is accelerating!

So where does that leave us? Thank you so much for the time you have taken with a non-physicist like me, who finds physics a bit of a mystery.

 

[Frabjous]

So how does that explain the greater damage done to the wall by the faster car?

For this it is better to think about force per area applied which is called stress. Things fail when a certain stress (called strength) is exceeded. Notice that a larger force creates a larger stress.
If you want to understand why you want to use stress, it explains why a pointy object will go through something while a blunt object will not.

 

[Frabjous]

neither car is accelerating. They are decelerating.

The magnitude of the force is proportional to the magnitude of the change. Acceleration or deceleration merely changes the direction.

 

[hutchphd]

But neither car is accelerating. They are decelerating. So once again, the explanation of the greater damage done by the faster car seems to have nothing to do with acceleration.

You know that this is wrong! The guy who falls from the cliff is not killed by acceleration of gravity but by deceleration from dirt.
Deceleration and acceleration are the same thing (just the sign changes)
As @Frabjous mentions it is actually spatially different accelerations that break things.

 

[dibbsy]

For this it is better to think about force per area applied which is called stress. Things fail when a certain stress (called strength) is exceeded. Notice that a larger force creates a larger stress.
If you want to understand why you want to use stress, it explains why a pointy object will go through something while a blunt object will not.

OK but this is getting off the point, imo. It doesn't help explain the greater force in the first place.

 

[dibbsy]

The magnitude of the force is proportional to the magnitude of the change. Acceleration or deceleration merely changes the direction.

Sure, but this doesn't seem to explain the greater damage done by the car to the wall.

 

[dibbsy]

You know that this is wrong! The guy who falls from the cliff is not killed by acceleration of gravity but by deceleration from dirt.
Deceleration and acceleration are the same thing (just the sign changes)
As @Frabjous mentions it is actually spatially different accelerations that break things.

Sure, and the car is damaged by the deceleration generated by the wall. But this still doesn't explain why the damage done to the wall is greater in the faster car than in the slower car, when neither car is accelerating. In the cliff case, the falling guy is accelerating. So the cases aren't parallel, are they?

 

[hutchphd]

The magnitude of the force depends on the rate of change of velocity over time. Higher velocity implies both more change in v and shorter time (hence the damage goes like v²)

 

[dibbsy]

BTW just realised there are several people speaking to me about this. Sorry, thanks all for your contributions. I don't feel much wiser, mind you. Seems to me the answer is kinetic energy, and in this inertial example F is proportional to mv^2 and so F has nothing to do with acceleration in this sort of case. Bet that'd be marked down in a high school exam, though.

 

[hutchphd]

It's all connected

 

[dibbsy]

The magnitude of the force depends on the rate of change of velocity over time. Higher velocity implies both more change in v and shorter time (hence the damage goes like v²)

Velocity is not changing until impact on my scenario. So no rate of change until deceleration by the way. But yes, v^2 is the key as far as I can see, which is not about acceleration.

 

[dibbsy]

It's all connected

LOL it sure is. I just wish I could see the connections! I hate physics lol. Give me biology any day.

 

[hutchphd]

Newton's great leap (called calculus) was how to do the math. He was crazy smart!

 

[Frabjous]

OK but this is getting off the point, imo. It doesn't help explain the greater force in the first place.

The car experiences a change of velocity Δv in a time Δt. The force is then $F=m {\frac {\Delta v}{\Delta t} }$. Δv is obviously larger. Δt is smaller because things are happening faster. So the force is larger.

 

[PeroK]

Velocity is not changing until impact on my scenario. So no rate of change until deceleration by the way. But yes, v^2 is the key as far as I can see, which is not about acceleration.

You've got some funny ideas, if you don't mind my saying.

There's an old joke that if you jump off a tall building it's not the fall that kills you ... but when you hit the ground.

But, actually, it's not hitting ground that kills you ... it's what happens after you hit the ground.

 

[jbriggs444]

As you say, since F=ma, the faster car experiences higher crash forces compared to the slower car.

I disagree. The truth of $\sum F = ma$ does not entail that faster cars experience greater force then slower cars. There are way too many complications to permit a simplistic analysis.

We need another parameter. For instance, the distance through which the two cars can crumple when they collide with the wall. [The length of the car might prove to be a useful upper bound for sufficiently high impact velocities].

The duration of a collision need not decrease due to greater impact speed.

 

[berkeman]

does not entail that faster cars experience greater force then slower cars. There are way too many complications to permit a simplistic analysis.

Yes, I was making several simplifying assumptions in my comment. But given the level of understanding of the OP, I didn't want to introduce too many other considerations.

 

[hutchphd]

The duration of a collision need not decrease due to greater impact speed.

Pedantically true IMHO.

 

[anorlunda]

In my opinion, visualizing a car crumpling as in the photo can help the OP to understand.

• Crumpling does not happen in zero time.
• When velocity reaches zero (but not before), crumpling stops.
• A slow moving car has less crumpling and less force, maybe just a dent.

 

[berkeman]

visualizing a car crumpling

Sorry about your car.

 

[hutchphd]

Although (to contradict myself in #29) I think some of the force limiters in F1 cars don't become effective until higher speeds (tires shear off and some crumple zones only at high force). And when I used to downhill ski the bone-breaking accidents were often the slow speed ones because the bindings didn't release. So just don't run into into stuff.

 

[dibbsy]

The car experiences a change of velocity Δv in a time Δt. The force is then $F=m {\frac {\Delta v}{\Delta t} }$. Δv is obviously larger. Δt is smaller because things are happening faster. So the force is larger.

Sorry, I don't understand that at all.

 

[dibbsy]

You've got some funny ideas, if you don't mind my saying.

There's an old joke that if you jump off a tall building it's not the fall that kills you ... but when you hit the ground.

But, actually, it's not hitting ground that kills you ... it's what happens after you hit the ground.

Sorry, after you hit the ground the damage has already begun! Nothing funny about that idea, if you don't mind my saying ;-)

 

[Frabjous]

Sorry, I don't understand that at all.

Think of it as an average force. The car comes to a stop (the velocity goes to zero) in a short time, Δt.

 

[dibbsy]

OK, but how does that explain why the faster car does the greater damage? No closer to a solution as far as I can see, unless we leave change of velocity right out of it and focus on force varying with kinetic energy. Surely this has been understood since Newton? What am I missing here?

 

[Frabjous]

OK, but how does that explain why the faster car does the greater damage? No closer to a solution as far as I can see, unless we leave change of velocity right out of it and focus on force varying with kinetic energy. Surely this has been understood since Newton? What am I missing here?

The magnitude of the force is roughly proportional to Δv. Δv is equal to the initial velocity of the car. So a faster car generates a larger force. A larger force generates a larger stress (defined in an earlier post) which causes more damage.

 

[dibbsy]

Sure, but this has nothing to do with acceleration! The Delta V here is the difference between the V of the slower car and the V of the faster car, both of which are travelling inertially. Given that, then what you say is fine and completely consistent with what I'm saying: in the inertial case it's all about Delta mv^2, not Delta V/Delta T since there is no Delta T

 

[Frabjous]

Acceleration is fundamentally about changes in velocity.

The Delta V here is the difference between the V of the slower car and the V of the faster car, both of which are travelling inertially. Given that, then what you say is fine and completely consistent with what I'm saying: in the inertial case it's all about Delta mv^2, not Delta V/Delta T since there is no Delta T

No it is not. There are two Δv's, one for the fast car and one for the slow car. I gave you an equation with a Δt and it confused you, so I simplified further.

 

[256bits]

not Delta V/Delta T since there is no Delta T

You ever run, play softball and slide into second base, ride a bicycle.
How do you ever slow down to a stop?
Certainly not instantaneously.
The slowing down is called deceleration ( which can be called negative acceleration ).
There is a change in speed, in a certain time period.

A car crash is the same thing, except the change in speed happens much quicker.
the deceleration ( negative acceleration ) can be in milliseconds.

 

[dibbsy]

Acceleration is fundamentally about changes in velocity.No it is not. There are two Δv's, one for the fast car and one for the slow car. I gave you an equation with a Δt and it confused you, so I simplified further.

The equation you gave does not explain what I'm wanting explained. As I said, there is no acceleration, so the greater force of the faster car cannot be explained by acceleration. Simple as that.

 

[dibbsy]

You ever run, play softball and slide into second base, ride a bicycle.
How do you ever slow down to a stop?
Certainly not instantaneously.
The slowing down is called deceleration ( which can be called negative acceleration ).
There is a change in speed, in a certain time period.

A car crash is the same thing, except the change in speed happens much quicker.
the deceleration ( negative acceleration ) can be in milliseconds.

OK, but nothing you have said explains why the faster car causes the greater damage! How could deceleration possibly explain this?

 

[Ibix]

The point you seem to be missing (or, more accurately, vigorously denying) is that there is acceleration: the car comes to a stop (or at least slows down if the wall is flimsy). And if the car is initially going faster the acceleration will be larger.

 

[dibbsy]

Sorry, but don't you see how absurd that sounds? If the car is going faster, the deceleration is larger. Call it 'negative acceleration' if you like, but it no more explains the larger force than my going backwards faster than you explains any damage to what is in front of me. Absurd!

 

[weirdoguy]

Sorry, but don't you see how absurd that sounds?

You clearly do not understand what acceleration is. Change in velocity means there is acceleration, per definition.

 

[Frabjous]

Sorry, but don't you see how absurd that sounds? If the car is going faster, the deceleration is larger. Call it 'negative acceleration' if you like, but it no more explains the larger force than my going backwards faster than you explains any damage to what is in front of me. Absurd!

Then you have no understanding of what F=ma means.

We are not sitting across a table from one another. Forum communications are much more limited. Given this, you will be better served by trying to figure why we are right, not by trying to have us refute your position.

I am tapping out.

 

[Ibix]

Sorry, but don't you see how absurd that sounds? If the car is going faster, the deceleration is larger. Call it 'negative acceleration' if you like, but it no more explains the larger force than my going backwards faster than you explains any damage to what is in front of me. Absurd!

Do recall when describing things as "absurd" that you are talking about the principles of physics used to design and build the car in the first place. There are literally thousands of pieces of hard evidence around you that those principles are not absurd, however you might feel.

You yourself quoted that force is mass times acceleration, so you presumably accept that there's a force now you accept that there's an acceleration. The reason that the force has a negative sign is that it's pointing in the opposite direction to the car's motion - from the wall into the car. That's why the car crumples and slows.

Now remember Newton's third law. We've been talking about the force on the car, but the car exerts an equal and opposite force on the wall. That's the one that does the damage to the wall.

 

[weirdoguy]

Assuming that time of collision is roughly the same you have
$$F_1=m\frac{\Delta v_1}{\Delta t},
F_2=m\frac{\Delta v_2}{\Delta t}$$
If $\Delta v_1>\Delta v_2$ then $F_1>F_2$.

Why LateX is not working?
EDIT: it works now, yay.

 

[Ibix]

Why LateX is not working?

You're the first to use it on this page. Refresh and it'll be fine.

 

[dibbsy]

You clearly do not understand what acceleration is. Change in velocity means there is acceleration, per definition.

No, change in velocity can be acceleration or deceleration. Direction actually matters. I still do not see how, if A decelerates into W with greater magnitude than B decelerates into W, that explains why A does greater damage to W than B does. It has to be about kinetic energy, not deceleration.