Questions about kinetic energy and momentum

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Kinetic energy and momentum have different relationships with velocity, leading to confusion in their conservation during collisions. Kinetic energy is proportional to the square of velocity, meaning an object moving at 10 m/s has four times the kinetic energy of one at 5 m/s, while momentum is linear, resulting in only double the momentum. In elastic collisions, both energy and momentum are conserved, but the distribution of energy between colliding objects can complicate this balance. Additionally, when a constant force is applied, the distance over which it acts increases as the object accelerates, resulting in greater energy gain despite the force remaining constant. Understanding these principles clarifies how energy and momentum interact in physical systems.
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Hi, I am currently in High School physics and we are going over Energy and Momentum. I have some questions about how and why they both work.

*all objects are assumed to have the same mass

Since [Kinetic Energy=1/2mv^2], an objects Kinetic energy is proportional to the square of its velocity, and therefor an object moving at 10 m/s has 4 times the energy as an object moving at 5 m/s.

My confusion comes from the equation for momentum [Momentum=MV]. This would suggest that an object moving at 10 m/s would only have twice the momentum as an object moving at 5 m/s, yet 4 times the amount of Kinetic energy.

If my goal was to stop this object from moving, I could place an identical object in its path and observe an elastic collision. This would maintain conservation of energy and momentum. However, if I placed two objects side by side in its path, assuming that each object was impacted equally and received half of the original objects Kinetic energy, I do not see how Energy and Momentum could be maintained. In order to maintain momentum, each object would have to move away at half the speed of the original object. In order to maintain Kinetic Energy, each object would have to move away at about .7 times the speed of the original object. I'm clearly missing something here, and I would appreciate it if someone could explain how Kinetic energy can have a quadratic relationship and momentum can have a linear relationship to velocity.

Another significant result of E being proportional to V^2 is that the change from 5 to 10 m/s requires more energy than the change from 0 to 5 m/s. However, since f=ma, a constant force will result in a constant acceleration. How is it possible that a constant force can add increasingly high amounts of energy to an object? If I am looking at a spaceship undergoing constant acceleration from burning a constant amount of fuel, I will see it to be gaining kinetic energy at ever increasing rates. How can it be gaining all this extra Kinetic energy if it is only burning through its fuel (chemical energy) at a constant rate?

For those that got this far, thank you for reading and I hope you can help me out!
 
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Two thoughts for you:

1) You're assuming that the original object is at rest after the collision. You can find out the answer by solving for the two unknown velocities (after the collision) under the two constraints (momentum and energy).

2) The energy gain under constant force, hence constant acceleration, is F \cdot d. As the object speeds up, applying the constant force for the same amount of time means applying it over a (much) longer distance, hence much more energy gained.
 
1)Ok thanks that makes sense. If the initial object is moving backwards it can help balance energy and momentum.

2)I am still a bit confused on this one (edit: thought about it some more, think I get it). I understand an example used with gravity, where an object starts with potential energy and ends with purely kinetic. The faster it falls, the faster it gets kinetic energy, and the faster it loses potential energy.

edit: as for the spaceship example above, I was thinking of the force incorrectly. I had assumed that a constant energy applied over time would speed up at a constant rate, without considering that the same amount of "push" would be spread out over a much larger distance, making its actual force much less.
 
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