Same old question, I still don't get it: E=1/2mv^2 - more E with V?

[jbriggs444]

The change in momentum for the wall is twice the initial momentum of the wall ball, so it's "big" in that sense.

 

[PeroK]

This is important. Consider an elastic collision of a ball bouncing on the ground. If we neglect the Earth, them it looks like energy and magnitude of momentum are conserved. It's only when we note that momentum has direction that we see that conservation of momentum is manifestly violated.

Likewise, for an inelastic collision, it looks like neither kinetic energy nor momentum is conserved. Whereas, momentum ought to be conserved in all collisions.

Including the Earth in the calculations explains everything.

 

[A.T.]

That philosophy would limit greatly the questions you could set on momentum.

It's not a "philosophy", just a mathematical consequence for the magnitudes of the error you make vs. the quantity you are interested in.

I'm not sure if that's an answer - maybe you can come up with a punch line for it.

I replied in the other thread:
https://www.physicsforums.com/threa...ast-confusing-explanation.896869/post-7136120

 

[sophiecentaur]

What @A.T. says (quoted above) is rather important,

This has turned into a 'how long is a piece of string' thread. A change of momentum of a massive elastic object could be very high but the express train could have hundreds of impacts but none would necessary be 'important' enough to slow down the locomotive noticeably, although the ricochetting bullets could be killing people.
The same algebraic formula would cover all cases of masses and speeds and each would be judged on the actual quantities.

 

[A.T.]

This has turned into a 'how long is a piece of string' thread.

It should be obvious, that the acceptable error of an approximation depends on how it compares to the magnitude of the result. That has nothing to do with some philosophical nonsense question.

 

[gleem]

TL;DR Summary: Q: "does it take more energy to accelerate at higher speeds"
A: The 'usual answer': "Your kinetic energy grows exponentially"....yeah, but how much FUEL do you burn?

Here's the dilemma. Yes, kinetic energy grows exponentially with speed. So a 1kg object changing from 0 to 1m/s increases by .5J. And from 4 to 5m/s increases from 8J to 12.5J = 4.5J.

But the question is NOT how much kinetic energy does a body have as it accelerates. The question is, how much energy does it REQUIRE to accelerate? This is not the same question.

This question cannot be answered here. Increasing a velocity by a given amount does not say anything about acceleration. We do not know the time or the distance over which the energy was added and the velocity was increased.

 

[Philip Koeck]

TL;DR Summary: Q: "does it take more energy to accelerate at higher speeds"
A: The 'usual answer': "Your kinetic energy grows exponentially"....yeah, but how much FUEL do you burn?

I still don't get this and most answers all over kind of miss the point and just kind of recite without really explaining.
Here's the dilemma. Yes, kinetic energy grows exponentially with speed. So a 1kg object changing from 0 to 1m/s increases by .5J. And from 4 to 5m/s increases from 8J to 12.5J = 4.5J.

But the question is NOT how much kinetic energy does a body have as it accelerates. The question is, how much energy does it REQUIRE to accelerate? This is not the same question.

I would say it is.

I'll give you a very clean thought experiment where the work done becomes very obvious and distractions are avoided. Hope it helps. And also study previous posts!!

You have a shuttle weighing 2 kg gliding with almost no friction on a long track.
You also have an ultralight robot (1 mg) that can push the shuttle and can also move next to the track, again with almost zero friction. The point of this is that we can neglect all energy losses due to friction and also the kinetic energy of the robot when it moves.

Now you let the robot do 3 experiments:

1: The robot just pushes the shuttle, which is initially standing still in front of it, with a force F during a time Δt and accelerates it from 0 to 1 m/s. So the kinetic energy of the shuttle changed by 1 J.

2: Now the shuttle is already moving at 10 m/s past the robot. The robot applies the same force F during the same time Δt. This leads to the same change of momentum and speed as in experiment 1, but the change in kinetic energy is 21 J now. It goes from 100 to 121 J.
The reason for this is that the path over which the force was applied is 21 times as long as in experiment 1 (since the average speed of the shuttle was 0.5 m/s in experiment 1, but 10.5 m/s in experiment 2).

3: You think you can beat the system by letting the robot move alongside the shuttle while it pushes. So both the robot and the shuttle move at 10 m/s initially. The robot now performs exactly the same push as in experiment 1 so you might think it's only done 1 J of work, whereas the shuttle has clearly gained 21 J of kinetic energy according to what you observe.
So, where do the remaining 20 J come from?
You have to take into account that the robot is moving against your rest frame (the earth with your lab on it) while it's pushing so it covers the same total distance during the push as in experiment 2.

Take home message: However you jump around between reference frames you can't get more change of kinetic energy than you put in as work.