Spring momentum conservation problem

  • #1
ChiralSuperfields
1,216
132
Homework Statement
I have successfully solved the problem below by assuming that momentum is conserved and that there is an inelastic collision occurring between the masses. However, I am wondering whether the momentum being conserved is a valid assumption that I have made.
Relevant Equations
##E_i = E_f##
##\vec p_i = \vec p_f##
For this problem,
1692395139491.png

The reason why I am not sure whether it is a valid assumption whether momentum is conserved because during the collision if we consider the two masses to be the system, then there will be a uniform gravitational field acting on both masses, and a spring force that is acting upwards. Therefore, there will be two external forces acting on the system. The only reason I can think of for momentum being conserved in this case is if the forces acting on the both the masses acted over such a short time interval that there was no change in the momentum due to the forces.

However, if we define the system as everything, the two masses, the spring, and the source of the g-field, then I believe everything is internal force pairs so momentum is conserved.

If someone please knows whether momentum is conserved is a valid assumption and why, that would be greatly appreciated!

Many thanks!
 
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  • #2
ChiralSuperfields said:
The only reason I can think of for momentum being conserved in this case is if the forces acting on the both the masses acted over such a short time interval that there was no change in the momentum due to the forces.
This reasoning is correct. Over the time interval that the masses stick together they are not displaced appreciably in the gravitational field.
 
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  • #3
ChiralSuperfields said:
Homework Statement: I have successfully solved the problem below by assuming that momentum is conserved and that there is an inelastic collision occurring between the masses. However, I am wondering whether the momentum being conserved is a valid assumption that I have made.
Relevant Equations: ##E_i = E_f##
##\vec p_i = \vec p_f##

For this problem,
View attachment 330735
The reason why I am not sure whether it is a valid assumption whether momentum is conserved because during the collision if we consider the two masses to be the system, then there will be a uniform gravitational field acting on both masses, and a spring force that is acting upwards. Therefore, there will be two external forces acting on the system. The only reason I can think of for momentum being conserved in this case is if the forces acting on the both the masses acted over such a short time interval that there was no change in the momentum due to the forces.

However, if we define the system as everything, the two masses, the spring, and the source of the g-field, then I believe everything is internal force pairs so momentum is conserved.

If someone please knows whether momentum is conserved is a valid assumption and why, that would be greatly appreciated!

Many thanks!
The momentum of the 1.0 kg mass is clearly not conserved. Neither is the momentum of the 2.0 kg mass. But if you take the 1.0 kg mass, the 2.0 kg mass, and the spring (which is ideal, and thus massless) to be your system, then the momentum is conserved, (If you want to be picky, throw the Earth into this and use the Newtonian gravitational potential energy.)

-Dan
 
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  • #4
topsquark said:
The momentum of the 1.0 kg mass is clearly not conserved. Neither is the momentum of the 2.0 kg mass. But if you take the 1.0 kg mass, the 2.0 kg mass, and the spring (which is ideal, and thus massless) to be your system, then the momentum is conserved, (If you want to be picky, throw the Earth into this and use the Newtonian gravitational potential energy.)

-Dan
That doesn’t really help. The issue is the time interval over which momentum is to be considered.
If you take the time up until the dropped mass reaches its lowest point, clearly momentum of the two mass+spring system is not conserved. Including the Earth is not being picky, it's essential.
To avoid that, we can use a very short time interval, making the assumption that the coalescence is achieved quickly.
In between these extremes, we could model the coalescence as a spring of high constant during compression and zero constant in relaxation.
 
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  • #5
kuruman said:
This reasoning is correct. Over the time interval that the masses stick together they are not displaced appreciably in the gravitational field.
topsquark said:
The momentum of the 1.0 kg mass is clearly not conserved. Neither is the momentum of the 2.0 kg mass. But if you take the 1.0 kg mass, the 2.0 kg mass, and the spring (which is ideal, and thus massless) to be your system, then the momentum is conserved, (If you want to be picky, throw the Earth into this and use the Newtonian gravitational potential energy.)

-Dan
haruspex said:
That doesn’t really help. The issue is the time interval over which momentum is to be considered.
If you take the time up until the dropped mass reaches its lowest point, clearly momentum of the two mass+spring system is not conserved. Including the Earth is not being picky, it's essential.
To avoid that, we can use a very short time interval, making the assumption that the coalescence is achieved quickly.
In between these extremes, we could model the coalescence as a spring of high constant during compression and zero constant in relaxation.
Thank you for your replies @kuruman, @topsquark and @haruspex!
 

1. What is the principle of conservation of momentum?

The principle of conservation of momentum states that in a closed system, the total momentum before an event is equal to the total momentum after the event. This means that the total amount of momentum in a system remains constant, even if individual objects within the system collide or interact.

2. How does momentum conservation apply to a spring?

In a spring momentum conservation problem, the spring is considered to be a closed system. This means that the total momentum of the spring before and after an event, such as a collision, will be equal. This allows us to use the principle of conservation of momentum to solve for unknown variables.

3. What is the equation for calculating momentum?

The equation for momentum is: momentum = mass x velocity. Momentum is a vector quantity, meaning it has both magnitude and direction. This means that the direction of an object's momentum will be the same as its velocity.

4. How do you solve a spring momentum conservation problem?

To solve a spring momentum conservation problem, you will need to identify the initial and final momenta of the system, and then use the principle of conservation of momentum to set up an equation. From there, you can solve for any unknown variables using algebraic manipulation.

5. What are some real-life applications of spring momentum conservation?

Spring momentum conservation can be applied in various real-life situations, such as car collisions, sports, and rocket launches. In car collisions, the momentum of the vehicles before and after the collision must be equal, allowing us to calculate the speed or mass of the vehicles. In sports, momentum conservation can be used to analyze the motion of objects, such as a hockey puck or a basketball, after a collision. In rocket launches, momentum conservation is crucial for ensuring that the rocket reaches its desired trajectory and destination.

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