Archive for the Cute Problems Category

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More Order-of-Magnitude Physics

Posted in Cute Problems, Education with tags , , , on June 27, 2012 by telescoper

A very busy day today so I thought I’d just do a quick post to give you a chance to test your brains with some more order-of-magnitude physics problems. I like using these in classes because they get people thinking about the physics behind problems without getting too bogged down in or turned off by complicated mathematics. If there’s any information missing that you need to solve the problem, make an order-of-magnitude estimate!

Give  order of magnitude answers to the following questions:

  1. What is the tension in a violin string?
  2. By how much would the temperature of the Earth increase if all its rotational energy were converted to heat?
  3. What fraction of the Earth’s water is in its atmosphere?
  4. How much brighter is sunlight than moonlight?
  5. What is the mass of water in a soap bubble?

There’s no prize involved, but feel free to post answers through the comments box. It would be helpful if you explained a  bit about how you arrived at your answer!

21 Comments »

Driving Test

Posted in Cute Problems with tags on May 16, 2012 by telescoper

I’m currently stuck in the office while my third year students are tackling an exam I set. I have to wait by the telephone in case there’s a problem with the paper that I have to sort out.

As a quick diversion I thought I’d give my blog readers a little test of their own. Try this little poser:

28 Comments »

A Little Bit of Nuclear..

Posted in Cute Problems with tags , , on April 26, 2012 by telescoper

It’s been a while since I posted any cute physics problems, so here’s a little one to amuse you this rainy Thursday morning.

In the following the notation A(a,b)B means the reaction a+A→b+B. The atomic number of Oxygen is 8 and that of Fluorine is 9.

The Q-value (i.e. energy release) of the reaction 19O(p,n)19F is 4.036 MeV, but the minimum energy of a neutron which, incident on a carbon tetrafluoride target, can induce the reaction 19F(n,p)19O is 4.248 MeV. Account for the difference between these two values.

6 Comments »

Yet another cute physics problem

Posted in Cute Problems, The Universe and Stuff with tags , , on November 22, 2011 by telescoper

I’ve spent all day either teaching or writing draft grant applications and am consequently a bit knackered, so in lieu of one of my usual rambling dissertations here is another example from the file marked Cute Physics Problems, this time from thermodynamics. It’s quite straightforward. Or is it? Most people I’ve asked this question in private have got it wrong, so let’s see if the blogosphere is smarter:

Three identical bodies of constant  heat capacity are at temperatures of 300, 300 and 100 K. If no work is done on the system and no heat transferred to it from outside, what is the highest temperature to which any one of the bodies can be raised by the operation of heat engine(s)?

34 Comments »

Order-of-magnitude Physics

Posted in Cute Problems, Education with tags , , , on November 14, 2011 by telescoper

A very busy day today so I thought I’d wind down by giving you a chance to test your brains with some order-of-magnitude physics problems. I like using these in classes because they get people thinking about the physics behind problems without getting too bogged down in or turned off by complicated mathematics. I’ve also kept some of these in archaic units just to annoy people who can only do things in the SI system. I think it’s good to practice swapping between systems, especially for us astro-types who use all kinds of bizarre units, so if you don’t know the units, look them up! And if there’s any information missing that you need to solve the problem, make an order-of-magnitude estimate!

Give  order of magnitude answers to the following questions:

  1. What is the mass of a body whose weight is equivalent to the total force exerted by a 40 mph gale on the side of a house 40 ft long and 20 ft high? Express your answer in tons.
  2. What is the power required to keep in the air a helicopter of mass 500 kg whose blades are 3m long? Express your answer in kilowatts.
  3. The base of the Great Pyramid  is 750 ft square and its  height is 500ft. How much work was done building it?  Express your answer in Joules.
  4. How high would the jet of a fountain reach if it were aimed vertically up and supplied by a water main in which the pressure is 3 atmospheres? Express your answer in feet.

There’s no prize involved, but feel free to post answers through the comments box. It would be helpful if you explained a  bit about how you arrived at your answer!

 

22 Comments »

Transfer Orbit

Posted in Cute Problems, The Universe and Stuff with tags , on November 2, 2011 by telescoper

From time to time I like to post nice physics problems on here. Here is a quickie that I used to use in my first-year Astrophysical Concepts course which has now been discontinued, so I don’t need to keep it to myself it any longer.

A simple way to travel from one planet in the solar system to another is to inject a spacecraft into an elliptical transfer orbit, like the one shown by the dashed curve, which is described by Kepler’s Laws in the same way that the planetary orbits (solid curves) are.

Kepler’s Third Law states that the  period of an elliptical orbit is given by P^2 \propto a^3 where a is the semi-major axis of the ellipse. Assuming that the orbits of Earth and Mars are both approximately circular and the radius of Mars’ orbit is 50% larger than Earth’s, and without looking up any further data, calculate the time taken to travel in this way from Earth to Mars.

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Serious Brain Teaser

Posted in Cute Problems with tags , , , on October 28, 2011 by telescoper

This one has been doing the rounds this morning, so I couldn’t resist posting it here:

PS. If anyone knows where this originated please let me know and I’ll give proper credit!

PPS. Note that the Mike Disney option (300%) is missing…

23 Comments »

A Diving Bell Problem

Posted in Cute Problems with tags , on July 14, 2011 by telescoper

You will have noticed that, in recent weeks, I’ve been posting divers physics problems on here. They seem to be quite popular so I thought I’d try another, which I found this morning in an old A-level physics textbook:

A diving bell of internal volume 6 m3 is lowered into a freshwater lake until the volume of the contained air is 4 m3. The height of a water barometer at the surface is 10 m. Assume that the temperature of the air in the bell does not change as the bell is lowered.

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A Simple Problem in Statistical Physics

Posted in Cute Problems with tags , , , , on July 11, 2011 by telescoper

In physics we often have to resort to computer simulations in which continuously varying quantities are modelled on a discrete lattice. We also have recourse from time to time to model physical properties of a system as random quantities with some associated probability distribution. The following problem came up in a conversation recently, and I think it’s rather cute so thought I’d post it here.

Consider a regular three-dimensional Cartesian grid, at each vertex of which is defined a continuous variable x which varies from site to site with the same probability distribution function F(x) at each location. The value of x at any vertex can be assumed to be statistically independent of the others.

Now define a local maximum of the fluctuating field defined on the lattice to be a point at which the value of x is higher than the value at all surrounding points, defined so that in D dimensions there are 3^{D}-1 neighbours.

What is the probability that an arbitrarily-chosen point is a local maximum?

Solution

Well, the most popular answer is in fact the correct one but I’m quite surprised that a majority got it wrong! Like many probability-based questions there are quick ways of solving this, but I’m going to give the laborious way because I think it’s quite instructive (and because I’m a bit slow).

Pick a point arbitrarily. The probability that the associated value lies between x and x+dx is f(x)dx, where f(x)=dF(x)/dx is the probability density function. According to the question there are 3^3-1=26 neighbours of this point. The probability that all of these are less than x isF(x)\times F(x)\times \ldots F(x) 26 times, i.e. [F(x)]^{26}  becauses they are independent. Note that this is a continuous variable so the probability of any two values being equal is zero. The probability of the chosen point being a local maximum with a given value of x is therefore f(x)dx\times [F(x)]^{26}. The probability of it being a local maximum with any value of x is obtained by integrating this expression over all allowed values of x, i.e. \int f(x) dx [F(x)]^{26} . But the integrand can be re-written

\int f(x) dx [F(x)]^{26} = \int dF \times F^{26} = \frac{1}{27} \int d\left(F^{27}\right) = \frac{1}{27},

because  F=1 at the upper limit of integration and F=0  at the bottom.

So you don’t need to know the form of F(x) – but the calculation does rely on it being a continuous distribution.

This long-winded method demonstrates the applicability of the product rule and the process of marginalising over variables, but the answer should tell you a much quicker way of getting there.  The central point and the 26 neighbours constitute a set of 27 points. The probability that any particular one is the largest of the set is just 1/27, as each is equally likely to be the largest. This goes for the central value too, hence the answer.

 

9 Comments »

A Paradox of Galileo

Posted in Cute Problems with tags , , , on June 1, 2011 by telescoper

Going by the popularity of the little physics problem I posted last week, I thought some of you might be interested in this little conundrum which dates back to the 17th Century (to Galileo Galilei, in fact). It’s not a problem to which there’s a snappy answer, so no poll this time, but I think it’s quite a good one to think about – and please try to resist the temptation to google it!

The above figure shows a large circular wheel, which rolls from left to right, without slipping, on a flat surface, along a straight line from P to Q, making exactly one revolution as it does so. The distance PQ is thus equal to the circumference of the wheel.

Now, consider the small circle, firmly fixed to the larger circle with the two centres coincident. The small circle also makes one complete revolution as the wheel rolls, and  it travels from R to S. Similarly, therefore, the distance RS must be the circumference of the small circle.

Since RS is clearly equal to PQ it follows that the circumferences of the large wheel and the small circle must be equal!

Since the radii of the large and  small circles are different,  this conclusion is clearly false so what’s wrong with the argument?

24 Comments »

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