# Newton's second law dp/dt version?

Hello,

I am confused by the momentum version of newtons second law...

So we know
$$\bar{F}=m\bar{a}=m\left(\frac{d\hat{v}}{dt}\right)$$
and that
$$\bar{\rho}=m\bar{v}=m\left(\frac{d\bar{x}}{dt}\right)$$

so is

$$\frac{d\bar{p}}{dt}=m\frac{d\left(\frac{d\bar{x}}{dt}\right)}{dt}$$

What I mean is this bit $$\frac{d\left(\frac{d\bar{x}}{dt}\right)}{dt}$$ somehow equal to $$\bar{a}$$

Thanks

Filip Larsen
Gold Member
Yes, since acceleration is the (first order) time derivative of velocity and velocity is the (first order) derivative of position, the acceleration is said to be the second order derivative of position, which can be written as ##a = \frac{d^2x}{dt^2}##

You can read more about other notations for higher derivatives on [1]

[1] https://en.wikipedia.org/wiki/Derivative

Force F = dp/dt, this result summarizes newton's first/second law, to prove this, we know that F = ma = m*dv/dt, mass is invariant so we can treat it as a constant, which yields to m*dv/dt = d(m*v)/dt = dp/dt, If no force is acting on an object F = 0 = dp/dt, p is constant from which it follows newton's first law and momentum conservation, Good luck :p

Andrew Mason
Homework Helper
Hello,

I am confused by the momentum version of newtons second law...

So we know
$$\bar{F}=m\bar{a}=m\left(\frac{d\hat{v}}{dt}\right)$$
and that
$$\bar{\rho}=m\bar{v}=m\left(\frac{d\bar{x}}{dt}\right)$$
Using the product rule:

$$\vec{F} = \frac{d\vec{p}}{dt}=\frac{d}{dt}(m\vec{v}) = m\frac{d\vec{v}}{dt} + \frac{dm}{dt}v$$

If m is constant, dm/dt = 0 so:

$$\vec{F} = m\frac{d\vec{v}}{dt} = ma = m\frac{d}{dt}v = m\frac{d}{dt}(\frac{d\vec{x}}{dt})$$

AM

Last edited:
Thanks guys, forgot about the chain rule for differentiation.

So in general, whenever there is a $$\frac{d^{2}}{dx}$$ then it can be thought of two separate derivatives, each giving their own result. But the $$^2$$ means it skips the first result and we go right to the second?

Andrew Mason
So in general, whenever there is a $$\frac{d^{2}}{dx}$$ then it can be thought of two separate derivatives, each giving their own result. But the $$^2$$ means it skips the first result and we go right to the second?