f=ma , p=mv , EK=1/2 x m x v^2


by tommyleehutch
Tags: ek1 or 2
tommyleehutch
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#1
Apr23-09, 12:17 PM
P: 8
f=ma
p=mv
ek=1/2 x m x v^2


Not exactly homework, but some guys on the net are wondering about what formula to use for a boxer throwing a punch.
I know its the EK one.
But some dude said F=MA and I don't have a clue how to explain he is wrong.
Also I've just been doing this stuff (linear motion) in physics at school and I need to know things like this for my end of year exams.
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Gorz
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#2
Apr23-09, 12:19 PM
P: 29
Depends on what you are trying to find. You will more than likely have to use a combination of these.
tommyleehutch
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#3
Apr23-09, 12:26 PM
P: 8
Yeah I figured as much.
But what are the things trying to be found?
Force? that has the units KGms^-2. So it tells the acceleration of the mass away from its initial location.
P=ms^-1 that is just the speed it moves away from its initial location.
EK.... energy, joules. hm....???

Gorz
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#4
Apr23-09, 12:57 PM
P: 29

f=ma , p=mv , EK=1/2 x m x v^2


Quote Quote by tommyleehutch View Post
Yeah I figured as much.
But what are the things trying to be found?
Force? that has the units KGms^-2. So it tells the acceleration of the mass away from its initial location.
P=ms^-1 that is just the speed it moves away from its initial location.
EK.... energy, joules. hm....???
Force is measured in Newtons and is a vector quantity(has magnitude & direction)
P is impulse and is measured in kgm/s, so is a measure of mass*velocity per second
Kinetic Energy is measure in Joules yes.

Im still lost as to what youre trying to find out.
tommyleehutch
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#5
Apr23-09, 07:14 PM
P: 8
I think P is momentum. ^_^

Just ah... hmm ... hard to explain if you don't get it... :(
xxChrisxx
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#6
Apr23-09, 07:21 PM
P: 2,032
well for practicalities sake the most useful would be p=mv and f=ma

The faster he throws a punch the more momentum he has. So when it connects, and decelerates (which brings in f=ma) the opponents face feels a force.

Surely the most useful thing you want to know is the force that a boxer hits with.

A real world thing such as a punch cant really be dealt with with such trivial equations, the physics of punching something is ever so complicated.
tommyleehutch
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#7
Apr24-09, 05:29 PM
P: 8
In physics class our teacher taught us that EK is to do with inflicting damage on stuff.
xxChrisxx
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#8
Apr24-09, 05:33 PM
P: 2,032
Well from a physics point of view he's wrong.

All he meant was, the faster you throw a punch the more energy it has. Thats not the thing doing tha damage, the damage is to do with how much force you impart.

The reason why a punch with more KE does more damage, is that the force is higher.

So although from a physics point of view its not strictly correct, from a practical point of view, saying more KE causes more damage is fine.

I've just thought of a decent example to explain this:

You throw a punch with a certain speed, so it has a set KE.

You punch a large peice of foam - it doesnt hurt.
You punch a wall, and it hurts.

Thats becuase the wall decelerates your fist down much quicker, imparting a higher force.
tommyleehutch
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#9
Apr24-09, 05:48 PM
P: 8
But doesn't the 1/2 x m x v^2 equation help understand it better?
Because it tells that V is much more important than M?

I don't think F=MA really shows that, especially to people who I wouldn't expect to be too good at physics.
Even I don't see that so obviously when looking at the F=MA equation. ^^
xxChrisxx
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#10
Apr24-09, 06:00 PM
P: 2,032
Well this is a tricky one. Is one of those yes and no answers.

As all the energy shows you is how much you had to do to get it moving at that speed.
When you double the speed you have 4x the KE. But that doesnt mean you with 4x the force.

The thing that determines force is rate of momentum change which is p=mv. the rate of change is f=ma by the way.

So you throw a punch twice as fast, but it stops in the same time, you will imparting twice the force. But it takes 4 times the energy to do.
tommyleehutch
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#11
Apr28-09, 01:34 AM
P: 8
But in car crashes aren't we taught that if you double the speed the damage will be 4x greater???
xxChrisxx
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#12
Apr28-09, 04:27 AM
P: 2,032
No, thats a misunderstanding.

Think about what is always conserved in a collision. Momentum. And what deforms cars - Force. How you get between the two is time.

F = m ((vst - vend)/time)

Vend = 0, V st is velocity at start of crash. You have a set m.
You double the speed you are travelling, you double Vst. Which as everything else is the same doubles the force acting on the car.

Acceleration is the dominant factor in crashes, this is the reason why they are commonly quoted in g's.

Like the old saying goes. Speed doesnt kill, its the sudden stop that does.
TVP45
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#13
Apr28-09, 06:12 AM
P: 1,130
Going back to the boxer, where are you hitting him? If, for example, it's a blow to the head, you want as much acceleration of his head as possible so that you damage his brain. If it's a solar plexus shot, you want to move him as far as possible. So, which quantity do you think describes each type of punch?
Dweirdo
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#14
Apr28-09, 01:15 PM
P: 174
Well ,
I was told that the best punch is when You exert the highest force on a body(seems logical) , so fast punch=best , and by fast i mean not just the swing, you need to get your hand back fast as well, like spanking :P
cause F*dt=Impulse=change in momentum
so as the the dt goes down, F goes up=>hurts more.
If that's what You were looking for.
minger
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#15
Apr28-09, 03:15 PM
Sci Advisor
P: 1,498
From a purely physics standpoint, most of what has been said is correct. However, I would disagree with you Chris on the energy point.

When talking about damage, breaking, containment of flying object, energy is the property that used to determine is something may be safe or not. In the gas turbine world, there are often times in which proper measures need to be taken to contain a turbine blade which breaks off the of disk.

When we make this computation, it is often helpful to use strategies such as "Just-Constrained Energy" and the like. These formulas describe the amount of energy that a shell can withstand before complete failure.

You've also said that in a crash momentum is conserved. However, energy is conserved as well. In short, every problem in the world (well anything that can be solved with Newtonian physics) can be solved using only three equations

1) mass in = mass out + time rate of change of mass (continuity)
2) Conservation of Forces/Momentum (Newton's second law) note here: that (f=ma) = (p=mv) they are the same equation
3) Conservation of Energy

However, the difference I was pointing out is purely semantical, and in the "book"-world, often times things can be simplified much more.
xxChrisxx
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#16
Apr28-09, 04:39 PM
P: 2,032
Hehe, I kind of guessed this could be brought up about bits flying off after failure in a crash, which is where what i've said falls down i'll admit. In that case its fair enough to use energy.

So the idea of using KE to describe the damage caused by an impact from say a railgun slug is fine, as this is more easily comparable to the energy released from a chemical explosive. As keeping track of the momentums of all the little bits is impossible.

If comparing this to how much a punch would hurt using energy is slightly useless. As the thing that makes it hurt is the force, which is determined by acceleration and is largely independent of KE.

The problem I find is that using energy to describe impacts is it leads to the false idea of double the speed = 4x times the force of impact. Its a common thing i've seen on the forums in homework questions, you get an impact loading question and people immediately try to apply the energy equation and get confused when they cant find the force.
Dweirdo
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#17
Apr29-09, 04:07 AM
P: 174
anyway most of the energy in your punch converted to heat when you punch some one
in anyway you punch, making the impact time lower will increase the force and thus it will hurt more.
minger
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#18
Apr29-09, 07:45 AM
Sci Advisor
P: 1,498
True, the real-world problem of falling impact, or for that matter any type of impact is EXTREMELY difficult to do when you're not given a pretty impulse/time curve like you do in a dynamics textbook.


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