# I Momentum-Determining External Forces

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1. Jan 31, 2017

### Arman777

I am studying momentum and I just want to check that I understand the idea correctly.

Think theres a system.In this system theres two masses $m_1$ and $m_2$ moving with some velocity $\vec v_1$ and $\vec v_2$ and they exert a forces each other.Lets call the total force acting on $m_1$ is $\vec F_1$ and for $m_2$ is $\vec F_2$.

So If $\vec F_1=\vec F^{ext}+\vec F_{21}$
$\vec F_2=\vec F^{ext}+\vec F_{12}$

Then the change in the total momentum;
$Δ\vec P=\vec {(F_{tot})^{ext}} Δt$

If theres no exteral force then, $Δ\vec P=0$ .So If theres no external force then the change in momentum should be zero .If theres external force and its constant then the change in momentum will be constant but non-zero.$Δ\vec P≠0$

If the change in momentum is zero(No external force) the center of mass of the system will be not moving.If theres external force it will accelerate by $\frac {\vec {(F_{tot})^{ext}} } {M}=a_{com}$ ($M$ here total mass of the system)

Just I am confused with the idea of external force...Think theres two object making projectile motion.and colliding in the air.Is the external force in here is gravity ? (No air drag)

How can we determine the external forces in a given system ?

Most collions happens in a small amount of time ($Δt$ nearly $0$), in this case can we say $Δ\vec P=0$ ? and If we can why ? (There will be external force)

Last edited: Jan 31, 2017
2. Jan 31, 2017

### BvU

Think again: its momentum doesn't change. But it can well be nonzero.

Correct. The reverse process: fireworks ! A rocket shoots off and explodes into a bunch of fireballs. Without air drag the center of mass simply continues the parabola the original rocket was following.

For instance by using $F = mg$ when gravity is the only external force acting !

3. Jan 31, 2017

### Arman777

So in any system The free body diagram will be showing us the external forces ?

4. Jan 31, 2017

### Arman777

If we collide two objects in space.There would be no external force so the $Δ\vec P=0$ Even if theres external forces in small $Δt$ (the collusion time) from the equation of $Δ\vec P=\vec {(F_{tot})^{ext}} Δt$

$Δ\vec P$ should be zero

5. Jan 31, 2017

### Arman777

$Δ\vec V_{com}=0⇔Δ\vec P=0$ $Δ\vec V_{com}$ can have any value bot the change must be zero
Is this true ?

6. Jan 31, 2017

### BvU

Yes, that's about it. We restrict our considerations to constant mass cases, so ${d\over dt} \left ( m\vec v \right ) = m {dv\over dt}$.