### Lunar Leaper

Gravity on the Moon is about 1/6th that on the Earth. A pole-vaulter 2 metres tall can clear a 5 metres pole on the Earth. How high a pole could he clear on the Moon?

### Which Twin Is Older?

A simplified account of special relativity and the twins paradox.

### Whoosh

A ball whooshes down a slide and hits another ball which flies off the slide horizontally as a projectile. How far does it go?

# Moving Stonehenge

##### Age 16 to 18Challenge Level

If transporting the stones dry, the minimum volume of wood required:

$(V_{stone}\rho_{stone} + V_{wood}\rho_{wood})g = V_{wood}\rho_{water}g$
$\therefore V_{wood}(\rho_{water} - \rho_{wood}) = V_{stone}\rho_{stone}$
$\therefore V_{wood} = \frac{V_{stone}\rho_{stone}}{\rho_{water} - \rho_{wood}} = 13.05m^3$
$V_{wood}/(\pi r_{tree}^2) = length_{tree} = 415.4m$

That's nearly half a kilometer of sizeable trees!

If the stones could be transported wet, which would of course require a river about 2 feet deeper, then less wood would have been required:

$(V_{stone}\rho_{stone} + V_{wood}\rho_{wood})g = (V_{wood} + V_{stone})\rho_{water}g$
$\therefore V_{wood}(\rho_{water} - \rho_{wood}) = V_{stone}\rho_{stone}$
$\therefore V_{wood} = \frac{V_{stone}(\rho_{stone} - \rho_{water})}{\rho_{water} - \rho_{wood}} = 8.55m^3$
$V_{wood}/(\pi r_{tree}^2) = length_{tree} = 272.2m$

That's still a lot of trees, but considerably fewer, maybe 25 large trees. That is the absolute minimum value though, at which the object will have neutral buoyancy, i.e. will have the overall density of water.