Momentum-Determining External Forces

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Discussion Overview

The discussion revolves around the concept of momentum and the role of external forces in determining changes in momentum within a system. Participants explore the implications of external forces, particularly in scenarios involving collisions and projectile motion, while questioning how to identify these forces in various contexts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant suggests that if there is no external force acting on a system, the change in momentum should be zero, implying that the center of mass remains stationary.
  • Another participant agrees that gravity acts as an external force in projectile motion scenarios, affirming that the center of mass continues its trajectory without air drag.
  • There is a discussion about using the equation \( F = mg \) to determine external forces when gravity is the only force acting on the system.
  • One participant posits that during collisions in space, the change in momentum should be zero due to the absence of external forces, despite the presence of forces during the brief collision time.
  • Another participant questions the relationship between the change in center of mass velocity and change in momentum, stating that while the change in velocity can be non-zero, the change in momentum must be zero under certain conditions.
  • There is a confirmation that the relationship \( \Delta \vec V_{com} = 0 \Leftrightarrow \Delta \vec P = 0 \) holds true, with a focus on constant mass cases.

Areas of Agreement / Disagreement

Participants express differing views on the implications of external forces and the conditions under which momentum changes. While some agree on the role of gravity as an external force, others contest the interpretation of momentum changes during collisions, indicating that the discussion remains unresolved.

Contextual Notes

Participants highlight the complexity of identifying external forces in various scenarios, particularly in brief interactions like collisions, where the time frame may complicate the analysis of momentum changes.

Arman777
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I am studying momentum and I just want to check that I understand the idea correctly.

Think there's a system.In this system there's 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 there's no exteral force then, ##Δ\vec P=0## .So If there's no external force then the change in momentum should be zero .If there's 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 there's 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 there's 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)
 
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Arman777 said:
If the change in momentum is zero(No external force) the center of mass of the system will be not moving
Think again: its momentum doesn't change. But it can well be nonzero.

Arman777 said:
Think there's two objects making projectile motion and colliding in the air. Is the external force in here is gravity ? (No air drag)
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.

Arman777 said:
How can we determine the external forces in a given system ?
For instance by using ##F = mg## when gravity is the only external force acting !
 
BvU said:
For instance by using F=mgF=mgF = mg when gravity is the only external force acting !

So in any system The free body diagram will be showing us the external forces ?
 
If we collide two objects in space.There would be no external force so the ##Δ\vec P=0## Even if there's 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
 
BvU said:
Think again: its momentum doesn't change. But it can well be nonzero.

##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 ?
 
Arman777 said:
##Δ\vec V_{com}=0⇔Δ\vec P=0##
##Δ\vec V_{com}## can have any value but the change must be zero
Is this true ?
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}##.
 
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