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I was studying Newton's laws(law2) and encountered a problem,

In my textbook(Selina publications),

it is written:

Force applied is directly proportional do the rate of change of momentum, and the units are chosen as such that the constant of proportionality is 1

so with the correct choice units and according to my text, $$F = \frac{\Delta p}{\Delta t}$$

Now as $p(t) = m(t) \cdot v(t)$ where $m$ and $v$ and mass and velocity functions of time[dont interpret dot as dot product, it is a scalar product],

and also replacing concrete $\Delta$ changes to infinitesimal $d$ changes, and using all of these facts ,

$$F = \frac{d}{dt} (m(t)v(t))$$ and using the chain rule, $$F = m \frac{dv}{dt} + v \frac{dm}{dt}$$

Now here, the problem arises, I also have a Physics handbook by Yavorsky and Detlaf, in there, [2.8.1]

$$\frac{d}{dt}(mv) = F + v_1 \frac{dm}{dt}$$

where $F$ is the resultant of all the forces acting on the body $v_1$ is the velocity of the added masses before joined to the body(if $\frac{dm}{dt} > 0$), or that of the detracted mass after being separated from the body (if $\frac{dm}{dt} < 0$)

here, in one side(The first cited equation )$F$ is directly equal to $\Delta p$ while on the other(the handbook cited equation), there is an extra term with $F$ to equate it with $\Delta p$. Can anyone tell me which one is right??

I am a ninth grader, and my knowledge may be incomplete in some areas of physics, so please answer right from the basics...

Qmechanic
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Rutajit45adude
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1 Answers1

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They are equivalent, so no need to worry. The two $F$'s just represent two different things.

  • In the first-mentioned definition that leads to $$F_1 = \frac{d}{dt} (mv),$$ the $F_1$ is the total force taking into account both any mass that is being added/removed and the speed changing.

  • In the second-mentioned definition with $$\frac{d}{dt}(mv) = F_2 + v \frac{dm}{dt},$$ the $F_2$ is the force only on the current mass, whose speed might be changing. That definition of force does not take into account any changes of the mass. So, if any mass is being added/removed to/from the system, then we here must add a correction term.

What should this correction term be? You found from the first-mentioned definition that $F = m \frac{dv}{dt} + v \frac{dm}{dt}$, and here the first term is the force contribution in case of changes in speed (acceleration), while the second term is the contribution in case of changes in the mass. So, it is precisely this latter term that is missing from the second-mentioned definition. In other words, $F_2 = m \frac{dv}{dt}$.

So, with my labels as shown, the two definitions are actually related like this:

$$F_1=F_2 + v \frac{dm}{dt}.$$

You can say that the first-mentioned definition is more general.

In fact, very often you will see Newton's 2nd law described under the assumption of constant mass, so with the second-mentioned definition, in which case it simplifies to the well-known form:

$$F=\frac{\mathrm dp}{\mathrm dt}= m \frac{dv}{dt} + v \underbrace{\frac{dm}{dt}}_0=m \frac{dv}{dt}=ma\quad\text{, so }\quad F=ma,$$ where $a$ is the acceleration. When the mass isn't constant, then you will rarely hear people refer to it as Newton's law.

  • This version is very useful because the mass very often is constant. Push on a block, and you must apply force to accelerate it. But the mass stays constant. So, the force is described by $F=ma$. We use the term isolated about a system/object that does not gain nor lose mass, and isolated systems are very common.
  • On the other hand, if you push a cart with constant speed and somebody drops a bag of flour into it, then you suddenly must apply more force to maintain the constant speed - this force would now be described by $F = v \frac{dm}{dt}$ which is valid under constant speed but changing mass. This cart is a non-isolated system.
  • Now imagine that somebody pushes the cart and then lets go, and now it rolls freely forwards. Suddenly the flour bag is dropped into it. With noone to maintain its speed, the speed will now drop. It has now gained mass and the speed changed. Both are happening simultaneously, so now you must use the full version to find the force: $F = m \frac{dv}{dt} + v \frac{dm}{dt}.$ In this case, the force calculation often becomes far too complicated to do, and you would instead rely on other laws of nature, such as the momentum conservation law.

All in all, the lesson learned here is that sometimes the same symbol is used for different things by different authors. So, we shouldn't lock our understanding in on the choice of symbols.

Steeven
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