TOPIC 1A: FORCES and MOTION

Why is it that a steel block is heavier than a wood block? It obviously depends on how big the blocks are, but what about if the blocks are the same size?

The answer is that in a steel block, the particles are packed tightly together, so there is more matter in a small space. When we say 'matter', we mean solids, liquids or gases. For some substances, like gold, the particles themselves are more massive as well as being tightly packed. We say that gold is a very **dense** substance. There is a large quantity of mass in a small space.

**Figure 1. Wood and stone blocks
**These are approximately the same size -which one is heavier?

The blocks in figure 1 looks about the same size, but how would you check this? The correct term for the total 'size' of the block is the volume. For a regular shaped block like the wood, this is quite straight forward, but for the stone block there are special techniques, covered later in this section.

To compare any two substances, we need to have an identical volume to make it a fair test. Rather than cutting the blocks until they are the same volume, we can do some simple maths to compare them accurately. We can work out how massive they would be in a **single unit of volume**.

For example, if the mass of the wooden block is 400 g and has a volume of 100 cm^{3}, then we can do a simple division and say that it has a mass of 4 g per 1 cm^{3}. This is how density is measured, in units of grams per centimetre cubed (g/cm^{3}). For much larger objects, we can switch to larger units and use kilograms per metre cubed (kg/m^{3}), and this is the official international unit of density. However, a metre cubed is an enormous volume, so for most objects in a school lab, g/cm^{3} are commonly used.

Here is the formula for density:

density (g/cm^{3}) = |
mass (g) |

volume (cm^{3}) |

ρ = |
m |

V |

**Notes:**

- The letter used for density is the Greek letter
**ρ**('rho'). - Volume is given a capital 'V'.
- Take care trying to convert units: To convert cm
^{3}into m^{3}, remember there are 100 cm in a metre, so there are 100 x 100 x 100 = 1,000,000 cm^{3}in a m^{3}.

**Example:**

The wooden block shown in Figure 1 has dimensions of 8cm x 10 cm x 25cm, and has a mass of 1.5 kg.

Calculate the density of the block.

**Answer:**

The volume of the block is 8 x 10 x 25 = 2000 cm^{3}. If we want to use units of g/cm^{3} for density, we need to convert 1.5 kg into grams, which is 1500 g.

ρ = |
m | and therefore:- | ρ = |
1500 |

V | 2000 |

So ** density ρ = 0.75 g/cm^{3}**.

You will need to know how to carry out a practical to find the density of an unknown substance. To do this, you will need to do the following:

- Firstly, find the mass of the substance with a digital balance. Typically, this will be in grams for a lab-based experiment. The mass of a liquid in a container can also be found using a digital balance, but remember to subtract the mass of the container, or use the 'tare' button on the balance (if you know what this does and how to use it!).
- Then find the volume of the substance using one of the methods below.
- Then find the density using the formula for density:

ρ = |
m |

V |

There are several standard ways to find the volume of a solid or liquid:

If the substance is a regular block with a rectangular cross section, then finding the volume is relatively easy:

- Use a ruler to measure the length, width and height of the block.
- Find the volume using the maths formula volume = length x width x height

For **liquids**, the volume can be found using a measuring cylinder. The volume of a liquid is typically measured in millilitres (ml), but a millilitre is **the same volume** as a cm^{3}.

For irregular solids, for example a small stone, you will need a specialist piece of science apparatus called a **eureka can,** as shown in figure2:

**Figure 2. Using a eureka can to find the volume of a stone**

The can is filled with water until it leaks out of the spout at the side. Once the leaking has stopped, put an empty measuring cylinder under the spout and then place the stone in the can. The stone will **displace** water - effectively pushing it out of the way, and this displaced water can be collected in the measuring cylinder. The volume of the water collected is the **same as the volume of the stone**.

**Questions:**

1. A block of green plastic is shown here.

- a) State the formula used to calculate density.
- b) Using information from the image, calculate the density of the block.

a) The formula is:

ρ = |
m |

V |

b) The volume of the block is length x height x width, *
*V = 10 x 8 x 8 = 640 cm

To find the density we use the formula from part (a), so:

ρ = |
768 g |

640 cm^{3} |

2. The stone in figure 2 shown above in the eureka can has a mass of 71 g. Using measurements shown in the image, calculate the density of the stone.

The volume of water shown in the cylinder in figure 2 is 30 ml. (Each small division is 5 ml).

We know that
30 ml = 30 cm^{3} volume. The mass of the stone is given as 71 g. Therefore:

ρ = |
m |

V |

ρ = |
71 |

30 |

3. A customer orders are large block of aluminium for an art project.

- a)The block must have a length of 1.2 m, a width of 2.0 m, and a height of 0.8 m. If the density of aluminium is 2700 kg/m
^{3}, calculate the mass of this block. - b) The customer suggests supplying the block in two halves of length 0.6 m, to make it easier to transport. What is the density of the aluminium in one of these smaller blocks?

a) The volume of this large block will be:

V = 1.2 x 2 x 0.8

V = 1.92 m^{3 }

The density formula is:

ρ = |
m |

V |

So m = *ρ* x V,

m = 2700 x 1.92

**m= 5 184 kg**, or 5 180 kg to 3 sig figs (5.18 tonnes!)

b) This is a trick question! The density of a substance does not depend on the dimensions of the object! The density of aluminium is **still 2700 kg/m ^{3}**. You could go through the maths and halve the volume, but remember to halve the mass as well. You will then get the same answer for the density.

The density of pure water is exactly **1.0 g/cm ^{3}^{}**. (if you are interested, this is not a coincidence - the gram was defined this way). The density does change slightly with temperature, as the water expands. However, for iGCSE we can assume that it is 1.0 g/cm

If you put an object like a piece of wood on the surface of a pure water lake, then:

- It will
**sink**if the density is**above 1.0 g/cm**(the density of water).^{3} - It will
**float**if the density is**below 1.0 g/cm**.^{3}

Most metals and stones sink, so they must have a density above 1.0 g/cm^{3}. What about the huge container ship shown in figure 3? How can it float if it is made of metal?

**Figure 3. A vast container ship - still floating!**

*(NOAA - CC 2.0 licence) *

The answer is that the boat is made of metal on the outside, but has huge empty spaces inside that contain just air with a very low density. This means that the **average** density of the boat is less than 1.0 g/cm^{3} and it will float!

A hot air balloon 'floats' in air, so therefore must have a density less than the surrounding air - caused by the hot air expanding and taking up a larger volume.

A great example of this is when you mix oil and water. Oil is less dense than water, and so floats to the top. As oil does not dissolve in water, they stay separate.

In figure 4, many different liquids of different densities have been combined. From the top of the image we have:

- Olive Oil (least dense)
- Vegetable Oil
- Wine (red)
- Water (blue)
- Washing up Liquid (green)
- Maple Syrup (most dense)

**Figure 4. Liquids of different densities.***Kelvinsong, CC BY 3.0*

**Now test your understanding using these quick, 10 minute questions on this topic from Grade Gorilla:**