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Newton formulated his Second Law as such:

$$\sum{\vec{F}} = \frac{\delta \vec{p}}{\delta t}$$

and of course, $\vec{p} = m \vec{v}$.

Why is it that if the net force $\sum \vec{F}$ is constant (which implies that the rate of change of momentum is constant), then

$$\frac{\delta \vec{p}}{\delta t} = \Delta \vec{p}$$

In other words, why does the rate of change of momentum equal to the total change in momentum?

If there were previous answers on this, please inform me! Thanks :)

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The rate of change of momentum is not equal to the total change in momentum. They can't be equal, they have different units. What made you think they were equal? (Did you see this stated in a reference somewhere?) –  David Z May 19 '13 at 4:34

2 Answers 2

A constant net force means:

$$\Sigma\vec{F}=\frac{d\vec{p}}{dt}=C$$

where $C$ is some constant. This means that

$$\int \ dp=p=C\int\ dt=Ct+p_0$$

where $p_0$ is the initial momentum. Now, you can easily verify that

$$p_2-p_1=\Delta p=Ct_2+p_0-Ct_1-p_0=C(t_2-t_1)=C\Delta t$$

In particular, you see that $\Delta p \neq \frac{dp}{dt}$, unless $\Sigma \vec{F}=0$, since then both equal zero.

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Oh yeah! Thanks. –  Yuruk May 19 '13 at 10:50

I think I have figured out the answer, hopefully.

Firstly, let us begin by stating the First Law:

$$\sum \vec{F} = \frac{\delta \vec{p}}{\delta t}$$

When the net force is constant, we it means that there is no change of momentum, in other words,

$$ \frac{\delta \vec{p}}{\delta t} = 0$$

In this case, we know that the function of momentum is a constant, i.e, a straight line function.

This would derive the impulse-momentum theorem:

$$\Delta \vec{p} = \vec{J}$$

Would this be valid?

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