Struggling with this problem....

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Fab Foodie

Fab Foodie

hanging-on in quiet desperation ...
Location
Kirton, Devon.
For those interested (and some good thinking has been shown), the answer can be found here:

https://www.bbc.co.uk/programmes/articles/4VvcGv74FRMjxDGdZrMV7Z0/today-puzzle-756
 
My gut reaction is that it makes no difference as all momentum transfer is internal to the egg timer.

A thought experiment to explore:

Imagine a very light, very tall cylinder, say Nelson's column size but near zero mass and containing a perfect vacuum. The cylinder has a cannonball at its base, and is instantaneously upended, and placed on a weigh scale. What would the scales observe?

I think:

1. Initially, just the weight of the cylinder. No force exists between cannonball and cylinder, so it's existence cannot be detected by the weigh scale.
2. A very high spike as the cannonball hits the base, transferring momentum. Much more than the weight of the cannonball itself.
3. A constant weight equal to the cylinder plus cannonball.

So I change my gut reaction and propose that the weight does, indeed, reduce by the weight of the sand in flight.

Now replace the cannonball with a very heavy feather...
 

Rocky

Hello decadence
For those interested (and some good thinking has been shown), the answer can be found here:

https://www.bbc.co.uk/programmes/articles/4VvcGv74FRMjxDGdZrMV7Z0/today-puzzle-756
I hate to disagree, Fabbers but the link I posted shows it weighs more.
 

midlife

Guru
Hmm, If there is perfect vacuum in the hourglass and the sand is falling at a a perfectly steady rate with zero friction, then if there is conservation of momentum the weight on the scales should be the same. As the sand starts to fall at the beginning the weight will decrease but increase again a the falling sand hits the bottom of the hourglass if elastic collisions......now my head has exploded !
 
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