f=ma ?

[YT2095]

probably a dumb question but I read that as Force = Mass x Acceleration.

 

shouldn`t it be force = mass x Velocity (at point of impact)?

 

or is it for something else entirely?

[BenSon]

We were taught force = mass X acelleration

The thing that gets me about that equation is what if the acceleration is so great that it accelerates up to 99.999c dosn't the mass change? I thought that this equation assumes that mass not should be variable on the value of a. Thats what bugs me about it...

 

~Scott

[Janus]

No, Mass x velocity is momentum.

 

The force imparted in an impact is related to the "stopping" distance of that impact, which in turn is related to the stopping time, and that in turn is a measure of acceleration.

 

I.E. It takes ten times more force to stop a bullet over a distance of 1 mm then it does to stop it over 1 cm. Of course it takes the same energy to stop the same bullet in both instances, as energy is force times distance.

[Spyman]

Newtons second law is "Force = Mass x Acceleration".

 

Acceleration is at which rate a body changes Velocity caused by the Force.

 

For example in an impact the bodys don't change velocity at instant, they accelerate.

(but sometimes very fast)

 

The new velocity is thus depending how long time the force acts on the body.

[YT2095]

erm... lets no go the speed of light and stuff eh, I really did want to keep this straightforward and in Laymans terms as much as possible!

 

so Janus, if a projectile is say 1 gram and moving at a rate of 500 metres per second and then hits a brick wall, what force does it hit with ?(please ignore drag coefficients of air resistance and other trivia I can`t even begin to think of).

 

as well as the answer can you tell me what formula you used and show your working out please.

 

Thanks

[Janus]

erm... lets no go the speed of light and stuff eh' date=' I really did want to keep this straightforward and in Laymans terms as much as possible!

 

so Janus, if a projectile is say 1 gram and moving at a rate of 500 metres per second and then hits a brick wall, what force does it hit with ?(please ignore drag coefficients of air resistance and other trivia I can`t even begin to think of).

 

as well as the answer can you tell me what formula you used and show your working out please.

 

Thanks [/quote']

 

Again, the force would depend on how far the projectile penetrates into the wall before in comes to a stop.

[BenSon]

Sorry YT i'll start another thread

 

~Scott

[YT2095]

Benson, No worries mate, it`s just that I wanted to keep this one as clear and as conscise as possible, largely for my own edification/sanity

 

 

 

Again, the force would depend on how far the projectile penetrates into the wall before in comes to a stop.

 

 

the wall is not capable of penetration, how much power is dissipated.

the distance prior to the hit is irellevent, it hits with 1gm at 500m/sec

[swansont]

Force is actually dp/dt, where p is momentum.

 

So F = d(mv)/dt = m dv/dt + v dm/dt

 

The second term is important in a situation like a rocket, where you are ejecting mass. But if that term is zero, then you have the familiar F = m dv/dt = ma

 

So you need to know over what time interval the speed was brought to zero, to find the average force.

 

There are other ways of attacking it, too.

 

using the chain rule, a = dv/dt = dv/dx dx/dt = v dv/dx

 

if you know F (and thus a) is constant, then ad = v²/2 (assuming it comes to rest) so you can use the distance travelled to find the acceleration. Not surprisingly, you can see that this formula is the basis of the work-energy equation for a constant force: W= Fd = 1/2 mv²

[YT2095]

Swansont,

 

Mega appreciated seriously! but I`m a practical sort, and non of this really makes much sense

 

use my example 1 gm 500m/sec thinggy, employ that with your formula, that way I`ll see the patern and be able to replicate it when I need to, at the moment it`s all code to me without meaning or rellevance.

 

Cheerz

[swansont]

Sorry - left out one step.

 

you get a final equation of Ft = mv, of F = mv/t

 

so if your mass took a millisecond to stop, it felt a force of 500N (500 m/s * 10^-3 kg/ 10^-3sec)

 

it also took .25 m to come to a stop

 

(500 m/s)²/2*500N

[YT2095]

so from that I gather that 1m/sec = 1 newton?

 

actualy what units are we working in here? is 10^-3 = 1g?

 

and .25m - 25cms?

I said it was a solid wall, incappable of being penetrated.

 

ok maybe it`s better if I ask what power in watts does a projectile of 1gm traveling at 500metres per second have?

 

I think it`s my wording

[J.C.MacSwell]

so from that I gather that 1m/sec = 1 newton?

 

 

1 kilogram(?) metre / second squared

[YT2095]

1 m per second squared sounds like it`s going faster over time though?

 

if time didn`t factor into it 1 m/ sec would be a constant wouldn`t it? (I`m only guessing here).

edit: Aha! I gather that Newtons is a form of Thrust then, It`s just occured to me!

[J.C.MacSwell]

1 m per second squared sounds like it`s going faster over time though?

 

if time didn`t factor into it 1 m/ sec would be a constant wouldn`t it? (I`m only guessing here).

 

sorry I edited, I missed the mass part:

 

1 kg metre/ second squared

[YT2095]

ok, fair enough, but even so, per second ^2 still sounds like speed increase?

 

like a falling body at 1g at standard RTP 32 foot per second per second (something like that anyway), untill it reaches "terminal velocity".

 

I`m enquiring about that speed, a projectile, no longer powered hitting a solid inpenetrable taget at 500m/sec the projectile at 1 gram

 

how many watts does this projectile have in kinetic energy, whenit slams into the target?

 

 

edit: oh Sh!t sorry all

I wrote 1g in some post meaning 1 gram and in this post wrote the same meaning 1 Gravity

sinscere applologies! I feel such an idiot now

[ed84c]

KE=1/2MV^2

 

(one day ill learn latex, really, i will)

[YT2095]

KE=1/2MV^2

 

explain it in real terms using the example provided.

[J.C.MacSwell]

".

 

I`m enquiring about that speed' date=' a projectile, no longer powered hitting a solid inpenetrable taget at 500m/sec the projectile at 1 gram

 

how many watts does this projectile have in kinetic energy, whenit slams into the target?

 

 

[/quote']

 

If it was 1 kg at 500 m/s and it stopped in 1 second that would take 500 newtons. (so 1 gram would take half a newton)

 

The average velocity would be 250 m/s so it would travel 250 metres

 

I think a watt is a newton metre so your 1 gram would have 125 watts of K.E.

[YT2095]

 

1) If it was 1 kg at 500 m/s and it stopped in 1 second that would take 500 newtons. (so 1 gram would take half a newton)

 

2) The average velocity would be 250 m/s so it would travel 250 metres

 

3) I think a watt is a newton metre so your 1 gram would have 125 watts of K.E.

 

this is aproaching the sort of answer I was after, I`ve added the 1 2 and 3 to your post for clarity.

 

1. that makes sense totaly.

 

2. I`m uncertain about this as I stated 500 m/sec, how is an Avg worked out and why?

 

3. 125w sounds about right, I can`t be certain tho?

 

eitherway, you`ve cleared SOME stuff up for me, Thnx

[J.C.MacSwell]

this is aproaching the sort of answer I was after' date=' I`ve added the 1 2 and 3 to your post for clarity.

 

1. that makes sense totaly.

 

2. I`m uncertain about this as I stated 500 m/sec, how is an Avg worked out and why?

 

3. 125w sounds about right, I can`t be certain tho?

 

eitherway, you`ve cleared SOME stuff up for me, Thnx [/quote']

 

It's right I think but I think I made 2 errors. Watts is a rate of work

[YT2095]

I was thinking in terms of power dissipation in watts upon the projectile hitting the target, some will be sound, some heat, some in projectile fragmentation etc...

 

so watts is a good unit to use (I think?).

[J.C.MacSwell]

I was thinking in terms of power dissipation in watts upon the projectile hitting the target' date=' some will be sound, some heat, some in projectile fragmentation etc...

 

so watts is a good unit to use (I think?).[/quote']

 

So you need twice the watts to stop it in half the time.

[swansont]

ok' date=' fair enough, but even so, per second ^2 still sounds like speed increase?

[/quote']

 

A decrease, in this case.

 

As for the impenetrable barrier, you'd instead have a compression of the projectile. Of course, there's always the possibility that it would bounce instead, or in addition, which complicates matters.

 

Compression of the projectile also indicates that the force would not be constant in time, and there's really no such thing as an impenetrable barrier. But fear not - this is physics. We deal with "frictionless surfaces" and "elephants whose mass may be ignored" all the time.

[Callipygous]

ok, fair enough, but even so, per second ^2 still sounds like speed increase?

 

not so much that it is a speed increase as it will cause a speed increase. newton is a measure of force. just like gravity (a force) is written m/s(squared), so are newtons. except gravity is immune to the mass of the object, with newtons the acceleration depends on the mass, so you have Kg m/s^2 . remember, surplus force causes acceleration.

 

 

as for your whole impenetrable barrier thing, physically impossible. one of them has to give. the force required to stop something instantly is infinite, which means the closer you get to both of them being impenetrable the greater force exerted on both surfaces (it hurts more to get hit with a bowling ball than a basket ball, the bowling ball exerts a larger force on your head, but over a shorter time, the basketball gives, so its a lesser force over more time)

 

actually, both of them have to give, the ball bending doesnt help slow down the atoms at the front edge of the ball, so the wall has to give at least enough to slow down the front edge.