Understanding the Proof for F=dp/dt: Common Mistakes and Troubleshooting

[Enjoicube]

Ok, so I have a provblem with this.
s=distance
t=time

P=MV
P=MS/T
dP/dT=

(MS)(dP/dT)=F'(1/T)
1/t=t^-1
(MS)(dP/dT)=-T^-2
dp/dt=-MS/t^2
dp/dt=-MA
Where did i go wrong?

 

[Cyrus]

Well, for one thing:

\frac {dP}{dt} = M\frac{d(V)}{dt}+ V \frac{d(M)}{dt}

If the mass of the body does not change with time that term drops to zero.

Then you are left with:

\frac {dP}{dt} = M\frac{d(V)}{dt}

But:

\frac{d(V)}{dt} = A

So,

F=MA

Your flaw: DO NOT take derivatives of units.

 

[Enjoicube]

ok, so product rule is used here.

 

[Cyrus]

Yes, it is always used.

 

[Jimmy Snyder]

P=MS/T
dP/dT=(MS)(dP/dT)

You have gone wrong here too.

 

[StatMechGuy]

I didn't realize you could prove Newton's second law? Why would you go about proving something that's simply stated as fact?

 

[masudr]

He's not proving anything.

What he's doing is deriving the simpler

F=m\frac{d^2x}{dt^2}

from the more general F=dp/dt, assuming constant mass.

 

[Einstein Mcfly]

Well, for one thing:

\frac {dP}{dt} = M\frac{d(V)}{dt}+ V \frac{d(M)}{dt}

If the mass of the body does not change with time that term drops to zero.

If we allow the mass to change in some way depending on the velocity, does the "force" correspond to anything that looks like a relativistic "force"? What I mean is, can Newton's law in the form \frac {dP}{dt} be shown to give the correct relativistic increase from the rest mass if \frac{d(M)}{dt} is done correctly (ie, with some special version of M)?

 

[Cyrus]

To be honest, I don't know a damn thing about relativity. I'm not studying physics, sorry. Doc Al or Van Esh can give you an answer to that question.

 

[masudr]

@Einstein McFly: the answer to your question is no.

Accounting for variable mass is correct still only non-relativistically.

To look at the force law for relativistic mechanics, you have to look at four-force and it's relation to four-momentum.

 

[brp1387]

s=vt is used for a body moving with constant speed.by taking derivative your are implying that acceleration is already present i.e. speed is not constant.

 

[Jheriko]

@Einstein McFly: the answer to your question is no.

Accounting for variable mass is correct still only non-relativistically.

To look at the force law for relativistic mechanics, you have to look at four-force and it's relation to four-momentum.

If you use relativistic mass then Newton's F=\frac{dp}{dt} still holds, just with the modified form for momentum: p=m_{r}v, where m_{r} is the relativistic mass \gamma m_{0}, where m_0 is the rest mass and \gamma is the usual Lorentz factor.

Doubt this will help. Relativistic mass equations just confuse things in my experience... it is much better to use four-vectors for everything as suggested above, it then becomes inherently more difficult to break things with bad Newtonian logic.

 

[Enjoicube]

s=vt is used for a body moving with constant speed.by taking derivative your are implying that acceleration is already present i.e. speed is not constant.

Taking the derivative w/respect to time, not velocity.

 

[brp1387]

I am saying that s and t will change in such a manner that s/t is constant.so taking its derivative with respect to anything won't affect it.
Therefore p=ms/t.dp=m(d(s/t))for a non changing mass.
dp/dt=m(d(s/t))/dt.since d(s/t)=0,f=dp/dt=0.which is very true since the formula s=vt is used(valid only for constant speed motion).

 

[nivlem9]

My question is similar. I hope I'm posting this in the right place.


If F = ma, we can integrate to get P=mv, and integrate again
to get Ek = 1/2 mv^2 . Since a=dv/dt, and v=ds/dt, where s
is distance, can we replace P=mv with P=m(ds/dt) and integrate
to get Ek = ms? This would make kinetic energy a function of distance,
not time. Mathematically, I can't see any errors in those
assumptions, but my instinct tells me that we can't just get rid
of the time variable altogether. It would make sense if it were potential energy as a function of distance, though. Is there some connection?
Does the law of conservation of energy come in somewhere
here?

 

[masudr]

No. In the first instance you have integrated with respect to velocity and in the second instance you have integrated with respect to time. There is no reason why these should be equal.Explicitly:

<br /> \begin{array}{rcl}<br /> mv &amp;=&amp; m\frac{ds}{dt} \\<br /> \int mv\,dv &amp;\ne&amp; \int m\frac{ds}{dt}\,dt<br /> \end{array}<br />