not if the mass is turning around !
on a track.
We're treating it as a 1-dimensional turnaround. Like a spring or something. Alternatively it's the same as having a proper turning track but only considering the longitudinal component of the velocity. In both cases the velocity we use does indeed go to zero - the mass stops.
If you absolutely must analysise it in 2D, then do that. Of course then there'll be sideways forces too, which you must include. Just makes it harder. And it won't change the fact that the velocity of the 100kg mass changes by 80m/s during 1 second. So it'll end up exactly the same as my calculations.
you should read up on the issue before replying.
He's hit the nail on the head.
I questioned kallogs use of the 80m/s * 100kg to get
a force of 8000N , that he applied to the pipe to cause
the pipe to stop and then move in the other direction.
I didn't do that. Check my calculation again. I also divided it by the turnaround time. Then I got the 8000N.
total change in velocity is 80 m/s
its still not a force , its a change in direction.
Who said it was a force? Only you. It's not a force, it's a change in velocity. It does cause a force to be applied, but we have to actually calculate that force, it's not 80N.
the total change in velocity durring the 1 hour
is 100 miles per hour , but you never drove your
car over 50 miles an hour.
Sure. And I never said the mass travels over 40m/s at any point. Where did you get that idea from?
the change in velocity of the car required no extra gasoline than it would if you just
kept driving north for an entire hour.
The mass's turnaround requires no burning of gas or use or electricity or anything. It's a passive device too. Again I never said we need to put energy into it to make it turn.
In the car example you could turn off the engine and let it cost around a 180deg curved track. In the absence of friction it'll get a 50mph change in velocity too. And of course it'll apply a force to the track, despite not burning any fuel and not losing any speed.
