Should I Consider Momentum or Work Done in Calculating Car Collision Damage?

[venkatesh]

I am trying to find the kinematics of a car collision with a wall. To calculate the damages, should I take into account the momentum or work done?
In other words, is this proportional to velocity or square of the velocity?
Thanks,
Venkatesh

 

[Gara]

i think its something like, velocity squared by mass.

as in:

10 kg weight going 10 m/s
5 kg weight going 40 m/s

have the same kinetic energy.

>i think< that's what the squared thingie means.

also, surface area of inpact makes a big differnce too. the smaller the more damage.

also note that cars are designed to crumble. the more parts of the cars crumble, the more kinetic energy is "lost" in the inpact, as well as directed around the driver.

a 1000 kg brick into a car would just smoosh the car. but a 1000 kg car into another car, would do a lot less damage, cos it has crumble zones.

 

[enigma]

Welcome to the forums, Venkatesh!

I think you need to define what you mean by "damage"

Like Gara pointed out, a car's frame is designed to dissipate the energy around the passenger compartment, resulting in a slower deceleration (which is what hurts the soft squishie parts)

 

[ShawnD]

How do you know when to use momentum and when to use kinetic energy? I've had problems in the past trying to figure out which one to use.

 

[MathematicalPhysicist]

How do you know when to use momentum and when to use kinetic energy? I've had problems in the past trying to figure out which one to use.

from what i learned usually you are using kinetic energy when the collision is plastic or non-elastic.

 

[ShawnD]

What does elastic mean though?

I can't find a definition on the web other than "Capable of resuming original shape after stretching or compression; springy".

 

[Chen]

from what i learned usually you are using kinetic energy when the collision is plastic or non-elastic.

There is not much point in using the energy equations in non-elastic collisions because energy is not conserved. It's only useful in fully elastic collisions, in which all energy is conserved.

ShawnD:
http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html

 

[expscv]

umm momentum = Impulse mathmatically

momentum = mass*velocity and equal to impulse =force * time
both unit in Ns

That The lower the force , the low the impact on car

and if reduce the velocity would help. mv=Ft F= (mv)/t

and if increase the time of crash it will experience less force (compare to same amount of momentum) so the use of croumple zone is built to do such thing
which allow the head of car be squashed so increase the time to reach to the person sits in. P=Ft F=P/t


P= momentum f=Foce t=time v= velocity

Mementum = Impluse not work done i apologize and momentum = Mv not Mv^2

 

[ShawnD]

momentum = mass*(velocity square) and equal to work done =force * time
both unit in Ns

Uh... no it isn't.

$$mv^2 = Ft$$

units:

$$(kg)(\frac{m^2}{s^2}) = Ns$$

$$(kg)(\frac{m^2}{s^2}) = (kg)(\frac{m}{s^2})(s)$$

$$(kg)(\frac{m^2}{s^2}) = (kg)(\frac{m}{s})$$

That doesn't work out.

 

[Chen]

You got it all wrong, expscv. "Work" is defined as the http://omega.albany.edu:8008/calc3/dot-product-dir/cornell-lecture.html of the force and the path along which it works. "Kinetic energy" is defined as the product of the velocity of an object squared and its mass, divided by two. "Momentum" is defined as the product of the velocity of an object and its mass. "Impulse" is defined as the product of the force and the time period.

 

[expscv]

all yeah ... change work to Impulse so sorry I apologise ~~~

work= Fd impulse = Ft F=force d=distace t=time

would that work now?

opps that's a big mistake

 

[expscv]

and momentum = mass*velocity

relationship between Momentum and impulse is the mass*velocity=force*time
force = (mass*velocity)/time

and the formula Kinetic energy = mass * velocity ^2

is too help understand the damage to the car after the crash which the kinetic energy become 0 that means the KE is converted into other form of energy like sound heat or damage to the car.

 

[ShawnD]

It doesn't really matter if you use momentum or kinetic energy

kinetic energy:

$$Fd = \frac{mv^2}{2}$$

$$F = \frac{mv^2}{2d}$$


momentum:
find acceleration

$$Vf^2 = Vi^2 + 2ad$$

$$a = \frac{Vf^2 - Vi^2}{2d}$$

force:

$$F = ma$$

$$F = \frac{m(Vf^2 - Vi^2)}{2d}$$


How interesting.

 

[Chen]

Did you really expect to get different results, ShawnD? ;)

 

[MathematicalPhysicist]

There is not much point in using the energy equations in non-elastic collisions because energy is not conserved. It's only useful in fully elastic collisions, in which all energy is conserved.

ShawnD:
http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html

as i said this is what i learned.
from this link you will see what i mean (it's in hebrew so i think only you can read it (-: ):
http://tch.netanya.k12.il/SubParagraph/HighSchool/StudyingOrganization/Nature/phis/11_M_5.doc

 

[Stingray]

I am trying to find the kinematics of a car collision with a wall. To calculate the damages, should I take into account the momentum or work done?
In other words, is this proportional to velocity or square of the velocity?
Thanks,
Venkatesh

Neither momentum nor energy alone give a good estimate of what's going on here. It matters how quickly the car is decelerated, which depends on the specifics of the car and the rigidity of the wall. You also need to define "damages." Its maybe best to ask what happens to the occupants. In that case, you can make estimates in the following way:

Assume the car will crush uniformly, and is of length L. There's more than one way to define "uniformly," but I'll take the simplest definition - that the deceleration is constant. Also assume the wall is perfectly solid. The occupant has to completely decelerate within a distance L/4 or so (otherwise he'll be crushed and certainly die). So use work:

W=1/2m*v^2<F*(L/4)=(ma)*(L/4)

a>2*v^2/L

This estimate is very weak, and does not replicate reality well at all. The average acceleration is not necessarily the most useful measure (i.e. the seatbelt's concentrated pressure could crack ribs), acceleration is not necessarily constant, the drivers legs are usually crushed with a far weaker collision than would be required to otherwise cause serious injury, and most importantly the crush distance would in reality be a very complicated function of v, among other things. There's a reason that companies invest large amounts of money physically crashing their cars to improve safety.

 

[Chen]

as i said this is what i learned.
from this link you will see what i mean (it's in hebrew so i think only you can read it (-: ):
http://tch.netanya.k12.il/SubParagraph/HighSchool/StudyingOrganization/Nature/phis/11_M_5.doc

You should read it again. It clearly says $$\Delta E_k <> 0$$ for non-elastic and plastic collisions.

 

[ShawnD]

Did you really expect to get different results, ShawnD? ;)

Of course I did
https://www.physicsforums.com/showthread.php?t=16992

On that impact question (the first one he asks), momentum and energy give completely different answers. Momentum gives 2600N, energy gives something like 3200N if I remember correctly.

 

[Chen]

Of course I did
https://www.physicsforums.com/showthread.php?t=16992

On that impact question (the first one he asks), momentum and energy give completely different answers. Momentum gives 2600N, energy gives something like 3200N if I remember correctly.

Only because the energy method was applied incorrectly.

 

[ShawnD]

Wait a second... not it wasnt. That momentum method on the other question did not take gravity into account. Gravity applies force at all time so the force should have been from the change in velocity (from 4 to 0) but also include the force of gravity. The mass of the guy was 65kg. 65 x 9.8 = 637. 2600 + 637 = 3237
Dang, I think i got to go post in that other one.

 

[MathematicalPhysicist]

You should read it again. It clearly says $$\Delta E_k <> 0$$ for non-elastic and plastic collisions.

so there still can be conversation of energy, it can be $$\Delta E_k=(1/2)kx^2$$ the change in energy can be equal to the elastic energy or the work done by the body.