Expressing Newton's 2nd law in terms of momentum

[Calpalned]

Homework Statement

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

Homework Equations

$\Sigma \vec F = m \vec a$
$\vec a = \frac {d \vec v}{dt}$
$\vec p = m \vec v$

The Attempt at a Solution

We need to prove that $\frac {d \vec p}{dt} = m \vec a$. When I physicists correctly take the derivative of $\vec p$, they get $m \vec a$. How come taking the derivative doesn't affect $m$ ? If $m$ is constant, shouldn't it go to zero?
I know that I am wrong somewhere, but I want to fully understand how this formula was derived.

 

[Hithesh]

The method to find derivative of such a function is different.

When derivative of 'only' the constant is taken, it is zero.

When a constant is multiplied with some function, the derivative of the resulting function is the constant multiplied with the derivative of the function.
F=dp/dt
F=d(mv)/dt
F=m dv/dt
F=ma.

 

[ToBePhysicist]

Edit: the V is not to be here.

 

[Calpalned]

The method to find derivative of such a function is different.

When derivative of 'only' the constant is taken, it is zero.

When a constant is multiplied with some function, the derivative of the resulting function is the constant multiplied with the derivative of the function.
F=dp/dt
F=d(mv)/dt
F=m dv/dt
F=ma.

 

[Hithesh]

Hope it helped

 

[Calpalned]

Yes, that makes sense.
Normally, if $\vec v$ is a variable and $m$ is a constant, the derivative of $m \vec v$ will become $m$. Is $\vec v$ not a variable here?

 

[Anirudh Sharma]

Second law of motion states that, "The rate of change of momentum is directly proportional to the force applied. "

Now p = mv

We take m as constant, because rate of change of mass in situations involving second law of motion is negligible (you can consider rocket propulsion as an example in which mass varies significantly with time).

taking derivative both sides with respect to time (t),

ΣF ∝ dp/dt
⇒ ΣF ∝ d(mv)/dt
⇒ ΣF ∝ m.dv/dt
⇒ ΣF ∝ ma (a = acceleration = d(v)/dt )
⇒ ΣF = kma (k = proportionality constant = 1, in SI units)
⇒ ΣF = ma

Since, you are taking derivative of velocity with respect to time equals to 1 instead of acceleration, that is why you are getting on that result.

 

[ToBePhysicist]

Yes, that makes sense.
Normally, if $\vec v$ is a variable and $m$ is a constant, the derivative of $m \vec v$ will become $m$. Is $\vec v$ not a variable here?

V has nothing to do here. kick it out.

 

[Hithesh]

Yes, that makes sense.
Normally, if $\vec v$ is a variable and $m$ is a constant, the derivative of $m \vec v$ will become $m$. Is $\vec v$ not a variable here?

V is the variable.
But derivative of mv isn't m. It is m multiplied by derivative of v with respect to time.
Derivative of velocity with respect to time is nothing but acceleration