Momentum and net external force

AI Thread Summary
The discussion centers on the firecracker problem, where the only net external force considered is gravity. The initial momentum is calculated as -12 kg*m/s, and the change in momentum due to gravity is determined to be 12 kg*m/s, leading to a final momentum of zero. However, the direction of momentum is clarified, indicating that gravity should also be negative, resulting in a final momentum of -24 kg*m/s. The conversation also touches on the net external force in part b, noting that gravity does positive work, and raises a question about the effect of a spring on the system's momentum in part c. The reasoning regarding momentum and external forces is confirmed as correct throughout the discussion.
Joe Armas
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Homework Statement


In the picture

Homework Equations


In the picture attached.
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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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