Someone just wrote: "Net force on an object is equal to its rate change of momentum." Is this the same as f=ma
In keeping with community guidelines, I can't use your name and you don't take e-mail. It's been awhile since I had these memorized. But your way of expressing Newton's 2nd law of motion didn't ring a bell. Are these two statements equivalent?
Well, thank you very kindly, folks. Lindajune didn't answer but y'all stood in nicely.
- Dr BobLv 61 decade agoBest Answer
Yes. F=ma is a famous equation, but physicists often put it in terms of the rate of change of momentum:
F = dp/dt
(F, p, v, and a are all vectors.)
Since momentum p is mv (mass times velocity), the force F is
F = d (mv) /dt
= m (dv/dt) [because m is constant]
Therefore, the two are the same.
These equations hold for Newtonian mechanics, but we need to be careful when we talk about special relativity. In special relativity, the equation
F = dp/dt
is still correct, but F=ma is not.
In short, F=ma is useful when working with speeds small compared to that of light, but F=dp/dt is the correct expression of Newton's second law for all speeds.
-- comment on sparc77's answer
He's correct that if the mass of the object is changing (such as a rocket ejecting fuel), then F=ma and F=dp/dt are not equivalent. F=dp/dt is still correct, but we must use the chain rule, as NBL mentioned:
F = dp/dt
= m*dv/dt + v*dm/dt
(I'm ignoring special relativity here.)
I'm not sure I see the point of the last sentence of his answer, however. It's true that there are lots of formulas that define force under various circumstances (such as F=-kx for an object attached to a spring), but that just tells you what to substitute on the left-hand-side of the equation F=dp/dt. It doesn't affect the validity of the equation.
- TroasaLv 71 decade ago
I hope the person responds. In the meantime, your equation is correct. This applies when mass is constant and velocity does not approach the speed of light. A different equation is used in special relativity.
Observed from an inertial reference frame, the net force on a particle is proportional to the time rate of change of its linear momentum: F = d[mv] / dt. Momentum is the product of mass and velocity. This law is often stated as F = ma (the net force on an object is equal to the mass of the object multiplied by its acceleration).
This can also be stated as net force on an object is equal to its rate change of momentum
Notice that the second law of motion only holds when the observation is made from an inertial reference frame.
- FaessonLv 71 decade ago
Uncle Ike's Law Part Deux
"Observed from an inertial reference frame, the net force on a particle is proportional to the time rate of change of its linear momentum: F = d[mv] / dt. Momentum is the product of mass and velocity. This law is often stated as F = ma (the net force on an object is equal to the mass of the object multiplied by its acceleration)."
- sparc77Lv 71 decade ago
Dr. Bob Has the best answer so far. F=ma only under certain conditions. F=dP/dt is correct. When mass is a constant then this reduces to F=ma.
But mass is not always a constant. Every hear people talk about rocket science? As a rocket burns fuel, it loses mass. So F=ma does not work for a rocket, but F=dP/dt does. Also keep in mind that F=-bv (motion through a fluid, such as a boat through water) and also F=kx (for springs).
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- NBLLv 61 decade ago
Yes it is the same thing, and it's actually the best way to state it.
So the change of rate of momentum is dp/dt, that is the derivative of momentum with respect to time.
momentum, p=mv, where we v=v(t), since it could change with time, and mass is just a constant m, since the mass of an object never changes.
dp/dt= m*(dv/dt) + v*(dm/dt), which is just,
dp/dt=m*(dv/dt) + v*(0), so dp/dt=m*(dv/dt), but we know that the rate of change of velocity with respect to time, is just acceleration.
- Steve HLv 51 decade ago
Yep. And I think this way is actually Newton's original way of stating the 2nd law, too.
- MarkLv 61 decade ago
Newton's Lex II phrase was in Latin, and translated it referred to the change in "the quantity of motion."
- 1 decade ago
yes, because momentum is mass * velocity,
The rate of change of momentum would therefore be mass*v^2, or m*acceleration#.