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chipmeisterc

F=MA

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So something troubled me when i was driving earlier, and pondering various thoughts, if F = MA...and i have 0 acceleration I would, by result of that equation have 0 force? ..So if i am a bullet at terminal velocity and my acceleration is 0 i have no force. This is clearly wrong. I remembered also that F = MV2 which would certainly make the bullet hurt a bit more!!...But if acceleration is M/S2 then how does F = MA = MV2 ? Am i being stupid ? Im sure someone can clear this up pretty quickly :)

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F = MA...and i have 0 acceleration I would, by result of that equation have 0 force?
Your thinking is muddled. An object does not "have" force. A force is applied to an object, accelerating that object. The bullet has acceleration zero because there is no force being applied to it (which is, of course, wrong; air resistance is forcing it to decelerate).

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I remembered also that F = MV2

You remember incorrectly. Perhaps you're thinking of kinetic energy, KE=1/2*M*V2? The bullet sure does have a lot of kinetic energy.

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Yes, F = MA, you are right, but objects don't have 'force'. Force is zero because it isn't accelerating, but it still has momentum:

p = mv

(I think that's it...) p is momentum, and therefore you would have much greater momentum depending on your velocity, which has drastic impacts on collisions (pardon the pun). Computer programs, and physicists, often use impulse based systems to determine collision details from speeds and angles and whatnot. Impulse, from memory, depends on momentum, and so the faster you're going, the harder you are to stop.

f = m * v^2

I don't think I've ever heard that before... f = m dv/dt, is that what you're thinking of? Derivatives? I have no idea where you got that equation from, although from memory in Mathematics C we were resolving banking (like roller coasters, not finance institutions) and pendulum-like systems which used to include v^2 in there somewhere...

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..So if i am a bullet at terminal velocity and my acceleration is 0 i have no force. This is clearly wrong.


You are almost right. You have an acceleration of 0, so your NET force is zero. That means when you add up all of the forces acting on the object, the total, or net, is equal to zero. So you have your force of gravity going one way, and the force from air resistance is going the other. At terminal velocity, the two forces (gravity and air resistance) equal each other, and hence there is no more acceleration.

Quote:
I remembered also that F = MV2 which would certainly make the bullet hurt a bit more!!...But if acceleration is M/S2 then how does F = MA = MV2 ?


Perhaps you are (mis)remembering the equation for force centripetal, which is MV^2/R. This is used in many situations where you have an object moving in a circular path (such as a roller coaster going around a track, or a planet orbiting the sun).

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Thanks guys, now youve explained it, I realise that I was just getting my terms confused, force is just "applied" to an object to give a resulting momentum!

As for the second equation, it would appear I was getting confused with KE! Which is probably what sparked my confusion in the first place! Thanks for clearing that up!

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So if im driving my car along, at a fixed velocity ( no acceleration ) and i collide with another car, resulting in a perfectly elastic collision, how would you caluculate the force of the impact?
Presumably this is based velocity and deceleration? Then the force of the impact is divided into both bodys?

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Original post by chipmeisterc
So if im driving my car along, at a fixed velocity ( no acceleration ) and i collide with another car, resulting in a perfectly elastic collision, how would you caluculate the force of the impact?

A perfectly elastic collision? Well, assuming the cars bounce off each other "immediately", very very high. If they sort of deform like rubber balls and take a while to bounce, somewhat lower. One doesn't speak of the "force of a collision" as a whole. That's not a meaningful phrase. "Impulse" is a more common term in that scenario.

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Even if you know that the collision is completely elastic (highly unlikely, except in Chinese SUVs), there are a bunch of factors that go into determining the result of a collision: the cars' locations, velocities, rotational velocities, masses, rotational inertias, and probably a few more that I've forgotten.

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body A(mass ma, velocity va)
body B(mass mb, velocity vb)
cor = coefficient of restitution.
n = collision normal.
i = collision impulse (what you want to find).

conservation of momentum equation gives : (va' - vb').n = -e * (va - vb).n
change of momentum of body A after collision : va' = va + (j / ma) * n
change of momentum of body B after collision : vb' = vb - (j / mb) * n

(va - vb).n + (n.n * (j / ma + j/mb)) = -e * (va - vb).n
-(1 + e) * (va - vb).n = n.n * (j / ma + j / mb)
-(1 + e) * (va - vb).n = n.n * j * (1 / ma + 1/mb)

so you get

j = -(1 + e) * (va - vb).n / (n.n * (1 / ma + 1/mb))

That's an approximation of how momentum is transfered during a colision.

If the two bodies can rotate, then you have to consider angular momentum, and it's a little bit more complicated.

check out chris hecker docs for more info.

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