The mass spends 1s turning around.
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
It does that change in 1s.
Acceleration = (change in velocity) / (time taken)
Acceleration = 80m/s / 1s = 80m/s^2
The _mass_ must experience a +ve force to turn around. That force is:
F=ma
F=100kg * 80m/s^2
F=8000N
Furthermore, the pipe must experience the same force that the mass does, but in the opposite direction. That force is
F_pipe = -8000N
momentum is not a force !
the masses velocity does not change.
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It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
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its momentum changes , momentum is not a force
momentum is just mass multiplied by speed.
so you cannot say that a 8000N force is placed on the
pipe by the change in momentum of the mass.
if you are so certain that the pipe will experience a
-8000N force then where does the force come from , it must
come from somewhere.
the pipe only has a velocity of 4 m/s its mass is only
1000kg so its total force available is only 4000N
4 m/s/s * 1000 kg = 4000N
note: I added time as you require
where does the extra 4000N come from?
have you ever seen this happen in the real world , could
you give an example of this extra force in a turn?
does a ball dropped from 10 feet bounce to a height of 20 ft because of this principle?
because if that extra force is available in a turn then the ball bouncing off of the ground should also get a extra ummph because the earth is hard to move with a ball.
or does the earth also feel a force that is twice the
mass of the ball * its acceleration?
the reason I say twice the mass is because we know the velocity does not increase , so the mass of the ball
must increase.
am I right in assumming that that is what happens in the turn?
if we go in the middle of the turn and catch the mass we could have twice the mass? we could get rich by using gold masses.
at which point in the turn should we capture the gold before it has a chance to shrink its mass?
That means after 1s in the turnaround, it's reached a velocity of 0m/s (in the longitudinal direction, of course it's still doing 40m/s speed, but now sideways).
0m/s in the longitudinal direction isn't turned around, so that 4000N force is incompatible with a 1s turnaround time.
the mass travels all the way through the 40 meter
turn in 1 second because it has a velocity of 40 m/s
40 meters /40 m/s =1s , so in 0.5 seconds the mass is half way through the turn.
its traveling longitudinal and pressing against the
turn.
the velocity/speed of the mass does not decrease.
again you are using the change in momentum but describing it as a change in velocity.
I thought we were going to leave momentum out of this
discussion , so why are you impersonating momentum with velocity?
momentum is not a force so lets not pretend that it is.
velocity is also not a force , btw.
and a change in velocity is also not a force.
just use the actual forces that would actually apply
a force to the pipe or the 100kg mass.
