Momentum scales linearly with velocity and kinetic energy at the square, correct?
If a car travelling at 20 mph requires 20 feet to come to a stop, the same car travelling at 40 mph will require 80 feet to come to a stop, correct?
It will also take the car about 4 times as much time to accelerate to 40 mph than it would to accelerate to 20 mph, correct?
So, why does kinetic energy scale at the square of velocity but linearly with mass where momentum scales linearly with both velocity and mass? What physics-related reasons are there for kinetic energy to behave like that when it comes to velocity?
Does the different relationship mean that a high momentum/low kinetic energy object impacting another will tend to merely push it whereas a low momentum/high kinetic energy object impacting another will tend to result in penetration/shattering because it broke intermolecular links? Why is velocity better at breaking intermolecular links than mass?
This has been an ongoing arguement in the archery forums for many years. A heavier slower arrow with more momentum will penetrate an animal better than a faster lighter arrow with equal kinetic energy but less momentum. I believe momentum is more directional where Kinetic energy can disperse and deflect more. Waiting for the science guys for a proper answer.
Correct, assuming a constant force causing the braking (a reasonable approximation).
Incorrect, assuming a constant acceleration, but correct, assuming a constant engine power, and I’m not sure which is the more reasonable approximation here (I suspect constant acceleration).
As to the meat of your question, it’s a little hard to see just what you’re asking. If it’s just why energy and momentum scale differently with velocity, the pat answer is because they’re different quantities. If they scaled the same, we wouldn’t bother to define two different names. But we’ve found that a quantity that scales linearly with the velocity, and a different quantity that scales with the square of the velocity, are both useful in some sorts of calculations.
Alternately, you can think of momentum as being force times time, and energy as being force times distance. But distance is speed times time, so energy ends up having an extra factor of the speed in it.
Incorrect, assuming a constant acceleration, but correct, assuming a constant engine power, and I’m not sure which is the more reasonable approximation here (I suspect constant acceleration).
If your foot is holding the accelerator at a fixed position, then the engine is being held to a relatively constant load (unless you’ve got your foot way into it, at which point the RPM’s will vary quite a bit between shifts, and then the shape of the engine’s torque curve matters); acceleration will be relatively constant within any one gear, but acceleration will drop off as the transmission shifts up through the gears. If the vehicle is equipped with a CVT that holds engine RPM constant, then the engine will absolutely be held to a fixed power setting, and acceleration will smoothly roll off as speed increases (because the CVT is gradually adjusting the drive ratio). This is true even before aerodynamic drag gets factored in.