Momentum and net external force

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Homework Help Overview

The discussion revolves around a problem involving momentum and net external forces, specifically in the context of a firecracker scenario. Participants are analyzing the effects of gravity on momentum changes and the implications of directional definitions in momentum calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between net external forces and momentum changes, questioning the correctness of reasoning regarding the final momentum of the system. There are discussions about the directionality of momentum and how it affects calculations. Some participants also raise questions about the role of gravity and the implications of using specific values in momentum equations.

Discussion Status

The discussion includes attempts to clarify the reasoning behind momentum calculations and the effects of external forces. Some participants affirm the correctness of certain reasoning, while others seek further clarification on specific points, indicating an ongoing exploration of the concepts involved.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for discussion. There is an acknowledgment of the need to consider vector directions in momentum calculations, as well as the role of external forces like gravity in the system.

Joe Armas
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Homework Statement


In the picture

Homework Equations


In the picture attached.
0219151551.jpg


The Attempt at a Solution


For the firecracker problem, I believe the only net external force is gravity. Therefore using, Fnet * delta t = delta p. Change in momentum is 12 kg * m/s. Thus, the final momentum of the system is zero. Is this reasoning correct? Please let me know of any errors in my reasoning. Thank you.
 

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Joe Armas said:
For the firecracker problem, I believe the only net external force is gravity. Therefore using, Fnet * delta t = delta p. Change in momentum is 12 kg * m/s.
Right.

Joe Armas said:
Thus, the final momentum of the system is zero. Is this reasoning correct?
How did you conclude that?
 
Nathanael said:
Right.How did you conclude that?
The initial momentum of the system is 3kg * -4m/s = -12 kg * m/s. The change in momentum from gravity is 12 as I explained.
Pf = Pi + Delta P
Pf = -12 + 12 = 0
 
Joe Armas said:
The initial momentum of the system is 3kg * -4m/s = -12 kg * m/s. The change in momentum from gravity is 12 as I explained.
Pf = Pi + Delta P
Pf = -12 + 12 = 0
By saying "-4 m/s" you have defined a direction for positive and negative.
The magnitude of the change in momentum is 12 kg.m/s, but Δp is a vector; what is it's direction?
 
Nathanael said:
By saying "-4 m/s" you have defined a direction for positive and negative.
The magnitude of the change in momentum is 12 kg.m/s, but Δp is a vector; what is it's direction?

Gravity then should be negative since it too acts downward. So, delta p should be -30 * 0.4 or -12? So final momentum of system is actually -24 kg * m/s. The bottom piece of mass 2kg must therefore be moving at -14 m/s, since the top piece of mass 1kg is moving at +4m/s, and the sum of these two final momenta must add up to -24 kg * m/s.
 
Is this reasoning correct?
 
Yes, it is correct.
 
Nathanael said:
Yes, it is correct.
0222151508.jpg

I know in part b there is a net external force (gravity) that does positive work, fnet > 0, thus dp/dt is positive. Does the spring in part c increase or decrease the momentum of the system? I am not too sure of my answer.
 

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