# Brain Teasers are Back!

If the up-going elevator is fairly close to the top of the shaft, the brick might not have reached terminal velocity and thus cause little damage. If the down-going elevator is near the bottom of the shaft, the brick at terminal velocity could cause considerable damage.

I presume we are dealing with a brick at terminal velocity colliding with the elevator at full speed, regardless of whereabouts they are in the shaft ? I’m visualizing two such identical bricks, simultaneously hitting a full-speed up-going elevator and a full-speed down-going elevator as the elevators pass each other.

Another easy one to provide “hope” while we contemplate the elevators and bricks. The other easy one is the fulcrum.

One amount added to a quarter of that amount becomes 15. What is the amount ?

IanG

Don

1d

Assuming the tether is fixed and can’t slip around the fence the tether should be a bit longer than the radius - if it is exactly the radius it will inscribe an area less than the semicircle. I’ve tried to work out how much longer but don’t see a way to calculate it without some ugly and therefore tedious maths which is no fun. Hoping for some inspiration on the drive home.

Well, the tether IS FIXED and can’t slip and there is a way to calculate it, to any number of significant figures, however, it does involve ELEGANT maths to start, followed by some very tedious maths to get a solution. Nowadays, Excel can help to reduce the tedious part.

So just to remind others, the grass field is circular, let’s say radius 100m. The goat is tethered to a fixed point on the perimeter. The tether (rope) is just long enough to allow the goat to eat half the grass. How long is the rope. Ignore the distance between end of rope and mouth of goat !

As kenc would say “enjoy”

12 unless I’m missing a subtlety.

Hi Eoink,

There was nothing subtle about it and you’re not missing anything.

12 it is and well done ! (even if it was “too” easy !)

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When I did my degree there were 2 papers in the final year called the Comprehensives, which covered fairly basic Physics, A’Level and 1st year degree, the idea being to set “real world” type problems and see if we could actually apply the stuff we’d learnt, rather than just regurgitate it. Each paper was 10% of your whole degree marks. I still remember doing some past papers and working this through, I suspect I got to a less sophisticated version of the paper Ian linked to.

I don’t know if I ever said this on the old board Don, but thanks a lot. I enjoy the occasional stretching of the little grey cells, and I really appreciate the effort you put in to give us these challenges.

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I should have added you can ignore the effect of air resistance so the brick never reaches terminal velocity but just accelerates due to gravity.

Thank you Eoink.

I enjoy working through them before posting.

Ah !

And the elevators move at a constant speed, or if starting from rest, they accelerate far more slowly than the falling brick - well, that’s the idea behind an elevator, I hope

Thus the down-going elevator provides room for the brick to accelerate even faster, whilst the up-going elevator reduces the room for the brick to accelerate etc.

Yes the elevators move at constant speed. I think you’ve got the picture now. If it helps you can think of the elevator being a height h below the brick when it begins to fall, but the value of h is not important.

Maybe a hint will help with the falling brickwork. With an initial separation h between the top of the shaft where the brick falls from and the elevator position at t=0 when the brick starts falling, calculate the time of impact. Clearly this will be longer when the elevator is moving downwards that if the elevator is moving upward. From there you should hopefully be able to calculate a relative velocity at impact - where a surprise awaits.

As with the clock problem I mentioned on the old forum, if you look at this problem from the correct frame of reference you remove the need for any real calculations at all !

Ah ! Ian.

I was visualising (for the sake of clarity), that there were two lifts (elevators). One going up, one going down. Both travelling at the same speed. Also two bricks, each falling to one of the two elevator pods.

I had visualised the bricks dropped simultaneously such that they each collided with their respective elevator, just as the elevators passed each other, one going up, the other going down. The bricks would each be going at the same speed on impact.

I further assumed that the bricks do not deform, but the elevator structures are perfectly elastic and behave like a large brick with a spring attached, which compresses on impact and absorbs kinetic energy.

Is this at variance with your picture of events ?

Yes this is what I have in mind.

This is not quite what I have in mind. Both elevators are at the same height at the moment the brick starts falling. Impact happens first for the rising elevator and later for the descending one.

Yes as you say the damage is proportional to the absorbed kinetic energy which depends on the relative velocity of the brick and elevator.

Hope that helps.

I think we are now on the same wavelength. Many thanks.

ok, a short grubby interlude with excel leads me to the tether being 1.1587 times the radius of the field, which seems plausible.

Gone very quiet in here, have people found out there are other things in life than internet forums !

Only just back from a holiday !

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Hi Ian,

I’ve given the falling brick a bit of thought, but not as much as I would like …… (the weather was good enough for me to do a couple of training flights yesterday and a test today on a new pilot who is just finishing his training on 737s with Ryan Air – he passed !).

Anyway, back to the falling brick. These are my current thoughts.

There are lots of unknowns; and collisions in the real world are never perfectly elastic nor inelastic. Take the lift material for example. If the “brick” and “Lift” were both made of solid, hard steel, the collision could be close to elastic, with virtually no loss of kinetic energy. The lift is being acted upon by an external force, ie the hoist cable. This, together with the mass of a typical lift, would mean the lift continues at a steady speed regardless of the impact of the brick. There are other considerations but these will serve enough to camouflage my uncertainty about tackling this problem.

I have assumed that the lifts move at constant, steady speed throughout, with no loss of kinetic energy and no change in momentum. I have assumed that the brick comes to rest relative to each lift, then continues to move with the lifts. In doing so, that is where the conservation of momentum occurs and where, to all intent and purpose the entire kinetic energy of the brick is dissipated (as heat and sound when distorting the lift cage).

The loss of kinetic energy is proportional to the (relative) speed at impact, (actually the speed squared). Now this (relative speed) is where I might be wrong ( ok , I might also be wrong elsewhere !!). The relative speeds at impact are the same for the down-going and up-going lifts

Sometimes it’s worth trying a couple of numerical examples before launching into algebraic equations. So I considered a brick 1kg in freefall at 10m/s/s crashing into (a) an up-going lift 1,000kg travelling at a steady 2m/s and a down-going lift travelling at 2m/s.

First I did it assuming the lifts passed each other 45m below the released brick, then at 80m. These coincide with 3 secs and 4 secs of freefall at 10m/s/s, with the unrestricted brick travelling at 30m/s and 40m/s respectively. In each case, the collision with the up-going brick will occur before that height and time, with the brick travelling more slowly and with the down-going lift after that height and time with the brick travelling slightly faster…

In the 45m case the impact speeds of the brick are 28.07m/s and 32.06m/s whilst in the 80m case they are 38.05m/s and 42.05m/s.

Now clearly, the bricks that hit the descending lift are travelling somewhat faster than those that hit the up-going lifts. So the kinetic energy of the bricks at the down-going lift will always be greater than at the up-going lifts. The usual equations and graphs illustrate and support this.

I’ll give it a bit more thought, but relative speeds seem to be a more realistic option rather than brick-only speeds. At the moment.

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Don, well done, you have captured the solution I had in mind in this paragraph. The cute point is that the relative speed of impact and hence the dent is identical for upward and downward moving lifts irrespective of where they start. I found this a bit surprising when I first came across it.

That the speeds are the same is obvious when you look at it from the right frame of reference - is that how you saw it or did you do the maths?