Indiana Jones and the Golden Statue

Task

The following clip shows the famous opening scene of the movie Raiders of the Lost Arc. At the beginning of the clip, Indiana Jones is replacing the golden statue with a bag of sand:

The platform on which the statue is placed is designed to detect the mass of the statue so if the bag of sand has a different mass than the statue then a mechanism triggers the spectacular destruction of the cave.

1. The density of gold is about 19.32 ${\rm g}/{\rm cm}^3$ (at room temperature at sea level). The density of sand can vary but a good estimate is 2.5 ${\rm g}/{\rm cm}^3.$ Assuming the statue is solid gold, can the bag of sand and the gold statue have the same mass? Explain.
2. Assuming the statue is about 1000 ${\rm cm}^3$ in volume, what would its mass be if it were solid gold? Is this consistent with the way the statue is handled and tossed around in the video clip?

IM Commentary

Much has been written, from the scientific point of view, about this opening scene from Raiders of the Lost Arc. Here are two examples:

• http://antoine.frostburg.edu/chem/senese/101/measurement/faq/indiana-jones.shtml
• This same problem can be found in college chemistry texts, such as Zumdahl and Zumdahl, Chemistry, Houghton Mifflin Company, 2000, page 35, problem number 74.

The goal of this task is to provide a introduction to the sometimes subtle use of density and units related to density, in a simple and fun context with minimal geometric complexity. It would be ideal if this activity could be linked with work in a science class where the practical aspects of the density of matter are considered. Gold is of course very valuable and so large quantities to experiment with will not be available. Lead, however, has a very similar density to gold so if the teacher has access to a few samples of lead this could help students physically grasp how heavy small amounts of a dense substance can be. It would also be useful to have a one liter bottle of water so that students can get a close-up visual of the volume 1000 ${\rm cm}^3$. Finally, it is worth noting that one explanation for what happens in the video is that the golden statue is hollow so the teacher may wish to add as an assumption that the statue is solid gold.

The first solution presented is hands-on and proceeds by estimating volumes and then using the density provided to convert to masses. In order to do this effectively, students will need measuring tapes and ideally an assortment of objects that they use to approximate the statue (different sized water bottles, a can of tennis balls, etc.). The bag of sand should be easier to replicate using dirt or sand or whatever is readily available. If students use this method, they will either need to measure in centimeters or else know the conversion 1 inch = 2.54 centimeters.

This task provides an opportunity for working with MP4, Modeling with Mathematics. Most people watch this video clip without surprise or questioning. It is only when we begin to study what is happening through the lens of mathematics that something appears to be amiss.

Solutions

Solution: 1 Estimating the volume and mass of objects

1. We can find an approximate estimate for the volume of the sand and statue and then use the given density to obtain the masses. The bag of sand appears to fit nicely in Indiana Jones' hand. When he places the bag down on the podium it is quite flat, maybe 2 inches high. If we imagine the bag of sand in the shape of a rectangular prism, 8 inches by 8 inches would be a reasonable estimate for the the base. With these numbers we find a volume of about $$ 8 \text{ inches} \times 8 \text{ inches} \times 2 \text{ inches} = 128 \text{ inches}^3. $$ These estimates for the volume of sand are not very precise and so it would be more appropriate to use 130 cubic inches (or even 100 cubic inches): we will use 130 cubic inches in the calculations below.

   Next for the statue. We can also try to approximate this with a rectangular prism. It's height also appears to be comparable to the length of Indiana Jones' hands and we used 8 inches for this when we estimated the volume of the bag of sand. It is not as wide or deep as it is high. Assuming that the statue is $\frac{1}{3}$ as wide as it is high and $\frac{1}{2}$ as deep as it is high (which seems reasonable from the pictures) this means that the width is $\frac{8}{3}$ inches and the depth is $4$ inches. This gives a total volume of $$ 8 \text{ inches} \times \frac{8}{3} \text{ inches} \times 4 \text{ inches} \approx 85 \text{ inches}^3. $$

   To determine the mass of the sand and the statue with the given information, we can convert our volume estimates to cubic centimeters. There are 2.54 centimeters in an inch and so 2.54$^3$ cubic centimeters in a cubic inch. For the sand we get an estimate of about $$ 130 \text{ in}^3 \times 2.54^3 \frac{\text{cm}^3}{\text{in}^3} \approx 2000 \text{ cm}^3. $$ For the statue we have an estimate of about $$ 85 \text{ in}^3 \times 2.54^3 \frac{\text{cm}^3}{\text{in}^3} \approx 1400 \text{ cm}^3. $$ Now to find the mass, we can use the given information for the density of sand and for gold. First for the sand, 2000 cubic centimeters will have a mass of about $$ 2000\, {\rm cm}^3 \times 2.5 \frac{{\rm g}}{{\rm cm}^3} = 5000\, g. $$ This is 5 kilograms or a little more than 10 pounds. Next for the statue, $$ 1400\, {\rm cm}^3 \times 19.32 \frac{{\rm g}}{{\rm cm}^3} = 27,000\, g. $$ This is 27 kilograms. These two values are not close to one another and so, based on these estimates, it is extremely unlikely that the sand and the statue have the same mass.

   While not exact, our estimates of the volume are close to one another and this is reasonable looking at the video: it is hard to tell whether or not the sand or the statue occupies more space. In fact, in the video the statue looks like it might take up more space than the sand. Given our calculations above, this makes it essentially impossible that the two have the same mass.

2. To find the mass of the statue, we can multiply its volume by the density of gold: $$ 1000\, {\rm cm}^3 \times 19.32 \frac{{\rm g}}{{\rm cm}^3} = 1932\, g. $$ This is 19.32 kg. Since a kilogram is about 2.2 pounds this is more than 40 pounds! Certainly the way the statue is handled is not consistent with this kind of weight.

   The estimate we made in part (a) gives an even larger mass close to 60 pounds. Even a very strong person would not be able to manage an object this heavy in the way Indiana Jones does!

Solution: 2 Usuing the meaning of density

1. The given density of gold, 19.32 ${\rm g/cm}^3$, is nearly 8 times the given density of sand, 2.5 ${\rm g/cm}^3$. So we would need an amount of sand nearly 8 times the volume of the statue in order to have the same mass as the gold statue. Looking at the video, this is clearly not the case. If anything, it appears that the volume of sand is less than the volume of the statue. So unless the statue is hollow, the sand will weigh far less than the golden statue and Indiana Jones is doomed to failure.

2. To find the mass of the statue, we can multiply its volume by the density of gold: $$ 1000\, {\rm cm}^3 \times 19.32 \frac{{\rm g}}{{\rm cm}^3} = 1932\, g. $$ This is 19.32 kg. Since a kilogram is about 2.2 pounds this is more than 40 pounds! Certainly the way the statue is handled is not consistent with this kind of weight.