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Context:

In my textbook it is given: 'momentum' short for 'linear momentum':

Mass = $m$, momentum is $p=mv$. In time $\Delta t$, momentum changes by $\Delta p$, the rate of change of momentum is:

$$\frac{\Delta p}{\Delta t} = \frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$

My Doubts:

  1. Isn't a $\Delta$ sign missing beside the $t$ in the second fraction, and thus it should be $\frac{\Delta(mv)}{\Delta t}$
  2. How did they derive the third fraction from the second. I tried a lot but can't seem to get that.

My Work:

I have looked at this question - How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$?, but it's a totally different equation.

My Final Question:

Can someone please clear my doubts about this equation and help me understand how does:

$$\frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$

Thanks a lot !

Community
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Gaurang Tandon
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5 Answers5

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1) Yes indeed, the absence of a $\Delta$ in the second expression is just a typo.

2) The last expression is derived assuming that mass is a constant. If it helps, just set the mass equal to 4, or something. If we want to know how the quantity $ 4v $ changes, we really only need to know how the quantity $v$ changes. Suppose $v$ changes from $v_1$ to $v_2$. Then

$$ \Delta v = v_2 - v_1 \,,$$

and what will $\Delta(4v)$ be? Why it will be

$$ \Delta(4v) = 4v_2 - 4v_1 = 4(v_2 - v_1) = 4\Delta v \,.$$

So the constant comes out the front, because it doesn't change between the start and end points that you're considering. Hence it can be factored out.

3) It's important that I point out that

$$ \frac{\Delta p}{\Delta t} = m \frac{\Delta v}{\Delta t} $$

is not Newton's second law. It's just a relationship between the rate of change of momentum and acceleration, and can be derived straight from the definitions of these things. Newton's second law cannot be derived, and is a statement of real physical content --- hence it is called a law. Newton's law can be written as either

$$ F = m \frac{\Delta v}{\Delta t} \,, $$

or

$$ F = \frac{\Delta p}{\Delta t} \,,$$

whichever you prefer. You can show these two are equivalent using the argument in your textbook (although in fact they're not quite equivalent, as I established above --- the first equation only holds when mass is constant; the second is more general, and more universally true).

The point is that Newton's second law tells us how the acceleration or rate of change of momentum of an object is related to the force acting on it.

Hope this helps!

gj255
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    Newton's Laws hold only when the mass is constant. The second is not more general in the sense of this question. But it is more general in the sense that it works in both Newtonian physics and special relativity. – garyp Dec 31 '20 at 01:18
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  1. Yes. It should be: $$\frac{dp}{dt}=\frac{d(mv)}{dt}$$ I'm using $d$ instead of $\Delta$ because I am thinking about the limit where the changes in $p$ and $t$ are very small. Then these are called infinitessimal changes, and denoted by a $d$.

  2. Usually, when one considers simple problems in Newtonian mechanics, what one does is study a given object with a fixed and constant mass, $m$. This means that $dm=0$ at all times by definition. We have, using the product rule (which only holds for infinitessimal changes): $$\frac{d(mv)}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$ Can you guess what happens to the first term on the right hand side of the equal sign?

Danu
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You are right, there is a $\Delta$ missing in front of the $t$.

$\Delta v = v_2 - v_1$. If the mass is not changing, then $\Delta (mv) = mv_2 - mv_1 = m(v_2 - v_1) = m\Delta v$. Hope that helps.

The equation that includes $\frac{\Delta m}{\Delta t}$ is not Newton's second law. The second law is valid only for systems of constant mass. An equation like that one does appear in the complete analysis of systems of variable mass (like a rocket with its propellent being exhausted from the rear). It is also sometimes called Newton' second law in incorrect analyses of the rocket problem.

garyp
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    This is not correct: Newton actually wrote $\vec F=d\vec p/dt$ (although without the benefit of modern notation he had to expand to "The alteration of motion [defined earlier as mass times velocity] is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed") – rob Jun 09 '14 at 13:27
  • You wrote "the second law is valid only for systems of constant mass," which is false. You seem to imply that the second law is $\vec F = m\vec a$, which is a usually-good-enough approximation. – rob Jun 09 '14 at 17:33
  • @rob Sorry I erased the comment that led to yours. However, Newton's second law in either form, $F=ma$ or $F=\mathrm{d}p/\mathrm{d}t$ is valid only for systems of constant mass. This is true, although frequently not appreciated. Read the second sentence in this Wikipedia page – garyp Jun 09 '14 at 17:42
  • Hmm. The reference is interesting. The damning statement is that the term $\vec v, dm/dt$ "violates the relativity principle under Galilean transformations … [implying] a "force" $-\vec v, dm/dt$ in a system where the particle moves with velocity $\vec v$." I would interpret this as a failure to account for the momentum carried away by the missing mass, but maybe you consider that a "constant-mass" approach. – rob Jun 10 '14 at 00:09
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    @rob Indeed, if you account for the "missing mass" then the system is closed and the mass is constant. See this answer and for a quick explicit statement, this Wikipedia entry – garyp Jun 10 '14 at 01:35
  • Déja vu, I commented there too :-) – rob Jun 10 '14 at 03:34
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Well that's a result using differentiation and derivation.
Have you studied calculus? If not, there is a simple way to look at it.

$$\frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t}$$ (Yes there should be a $\Delta t$ in the denominator, too.)

Now, what does $\Delta(mv)$ mean? It represents the change in the quantity $mv$. For current situations, your text book assumes that the mass of the body will be constant with respect to time, so it will not change! Therefore $mv$ is like saying $\text{(some constant)}v$. Since this constant won't change with time, we can take it out of the $\Delta$ and write $\Delta(mv)$ as $m\Delta v$.

mikhailcazi
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  • Thanks for help! And no, I haven't studied calculus yet :) – Gaurang Tandon Mar 28 '14 at 11:51
  • Well using calculus, it can be shown that $d(xy) = x\cdot dy + y\cdot dx$. ('d' represents a tiny change in the quantity). Therefore we'd get $d(mv) = mdv + vdm$. Since we're considering $m$ to be constant, $dm$(which is the change in m) will be $0$! Therefore $d(mv) = mdv + 0 = mdv$. :) But you don't need to worry about that. Calculus is something which is taught in the 11th (in India at least), so you got plenty of time. – mikhailcazi Mar 28 '14 at 11:55
  • Oh, I see. Thanks ! And yes, I have time, but I think I will cover it in 10th. – Gaurang Tandon Mar 28 '14 at 11:58
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You can't derive Newton's Second law from within Newtonian physics, even its original, general form of F = d(p)/dt. F = m*a is just a special case, not a "derivation". You can derive it from the action principle, but then you would and should ask whether you can derive the action principle. The answer is you can't. From the action principle you can derive the laws of physics (including general relativity, quantum mechanics, classical and quantum field theories ...) at least separately ;-) ), and these laws are found to hold true.

ClimberT8
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