Momentum and net external force

In summary: In part c, there is a net external force (spring) 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?dp/dt is positive, so the spring in part c increases the momentum of the system.
  • #1
Joe Armas
17
0

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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  • #2
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?
 
  • #3
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
 
  • #4
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?
 
  • #5
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.
 
  • #6
Is this reasoning correct?
 
  • #7
Yes, it is correct.
 
  • #8
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.
 

1. What is momentum?

Momentum is a physical quantity that describes the motion of an object. It is calculated by multiplying an object's mass by its velocity. The unit of momentum is kg*m/s.

2. How is momentum related to net external force?

According to Newton's second law of motion, the net external force acting on an object is equal to the rate of change of its momentum. This means that an object will accelerate in the direction of the net external force applied to it.

3. What happens to an object's momentum if no net external force is applied?

If no net external force is applied to an object, its momentum will remain constant. This is known as the law of conservation of momentum, which states that the total momentum of a system is conserved in the absence of external forces.

4. Can the momentum of an object change direction?

Yes, the momentum of an object can change direction if the net external force acting on it changes direction. This is known as a change in the object's velocity or acceleration.

5. How does the mass of an object affect its momentum?

The momentum of an object is directly proportional to its mass. This means that a heavier object will have a greater momentum than a lighter object moving at the same velocity.

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