Lesson Explainer: Calculating Density Physics • 9th Grade

In this explainer, we will learn how to use the formula 𝜌=𝑀𝑉 to calculate the densities of different materials and objects.

Density is a property of materials and objects that measures how much mass is in a given amount of space.

Imagine two spheres of the same size: one is made of iron and the other is made of polystyrene. Intuitively, we know that the iron sphere will be far heavier than the polystyrene sphere.

The iron sphere will have a greater mass even though it is the same size, so we can say that it has a greater density.

This is why, for example, an iron ball will sink in a pool of water but a polystyrene ball will float.

Even though the two spheres are the same size, the iron ball has a greater density than water, which means it will sink. However, the polystyrene sphere has a much lower density than water, and hence it will float.

This will be the case no matter how big the spheres are. It is the density that determines if they float, not the size. So if we had a very large polystyrene sphere, it would still float in water. If we had a very small iron ball, it would still sink.

The density of an object is usually denoted by the Greek letter 𝜌 (which has the name β€œrho”). This letter looks like the English letter 𝑝, but if you look closely, you can see that it is slightly different.

We define the density, 𝜌, of an object algebraically as the mass, 𝑀, of the object divided by its volume, 𝑉. As an equation, this looks like 𝜌=𝑀𝑉.

Definition: Density

Density is a measure of the mass per unit volume of an object. The mathematical formula for the density of an object, which is denoted 𝜌, is 𝜌=𝑀𝑉, where 𝑀 is the mass of the object and 𝑉 is the volume of the object.

Any object made purely of one material will have the same density as any other object made purely of the same material. For example, a block of iron that has a volume of 1 m3 has the exact same density as a block of iron with a volume of 100 m3. The masses and volumes of the two blocks are very different, but the densities of the blocks are the same.

This is because the density of the block is the ratio of the mass of the block to the volume of the block. As the volume increases, so does the mass, but the ratio of these two quantitiesβ€”the densityβ€”stays the same.

However, some objects are made up of more than one material. If each of these materials has a different density, then the density of the whole object will be different in different parts of the object.

For example, imagine a sphere made partly out of polystyrene and partly out of iron. A cross section of this sphere is shown in the diagram below. The density in the part of the sphere made out of polystyrene is much lower than the density of the part of the sphere made out of iron.

If 𝑀 is the mass of the whole sphere and 𝑉 is the volume of the whole sphere, then 𝜌=𝑀𝑉 will give the average density of the sphere. This density will have a value that is in-between the density of polystyrene and the density of iron.

It is important to remember this difference. For a specific material, the density of this material is a property that can be looked up, and any object made purely of this material will have this density.

However, the density of an object that is made up of several materials is specific to that object. We can use 𝜌=𝑀𝑉, where 𝑀 is the mass of the whole object and 𝑉 is the volume of the whole object, to find the average density of the whole object.

Example 1: Finding the Density of an Object given Its Mass and Volume

A cube has a mass of 30 kg. If the volume of the cube is 0.02 m3, what is its density?

Answer

The density of an object is given by 𝜌=𝑀𝑉, where 𝑀 is the mass of the object and 𝑉 is the volume of the object.

In this question, we are told that the mass of a cube is 𝑀=30kg and the volume of the cube is 𝑉=0.02m.

We need to substitute in the values we know for the mass and volume of the cube into the equation for the density. Doing this gives us 𝜌=𝑀𝑉=300.02.kgm

Let’s split this fraction into the numerical part and the units part. We can calculate the numerical part as 300.02=1500. The units in the fraction can simply be left as kilograms per cubic metre.

As a result, our final answer is 𝜌=1500.kgm

If we need to calculate the density of a cube, we can do so if we know the mass of the cube and the length of one of its sides.

Recall that the volume of a cube with side length 𝑙 is given by 𝑉=𝑙.

Let’s see some examples of using the formula for density.

Example 2: Finding the Density of an Object given Its Mass and Dimensions

A small cube of iron has sides with length 0.15 m. If the mass of the cube is 26.6 kg, what is its density? Give your answer to the nearest kilogram per cubic metre.

Answer

In this question, we are given the mass, 𝑀=26.6kg, of an iron cube as well as its side length, 𝑙=0.15m.

Given this, we are asked to find the density 𝜌 of the cube. Recall that the formula for density is 𝜌=𝑀𝑉.

We know the mass of the cube, but we do not know the volume of the cube. We can work out the volume using the formula for the volume of a cube, which is 𝑉=π‘™οŠ©. So, here we have 𝑉=(15)=0.15=0.003375.mmm

We are now at a point where we can work out the density of the iron cube. We need to substitute the mass 𝑀=26.6kg and the volume 𝑉=0.003375m into our density equation. Doing this gives 𝜌=𝑀𝑉=26.60.003375.kgm

We can now simplify this result by first working out the numerical value to be 26.60.003375=7881.48148. We can leave units as they are, which is kilograms per cubic metre. This means 𝜌=7881.48148.kgm

The question asks us to give our answer to the nearest kilogram per cubic metre, which means our final answer is 𝜌=7881.kgm

Example 3: Finding the Volume of an Object given Its Mass and Density

Find the volume of a 54 kg block of aluminum. Use a value of 2β€Žβ€‰β€Ž700 kg/m3 for the density of aluminum.

Answer

In this question, we are given the density of a block of aluminum and asked to find the volume of the block. We are told that the block has a mass, which we will call 𝑀, of 54 kg. We are also told that the density of aluminum, which is what the block is made of, is 𝜌=2700/kgm.

We can rearrange our formula for density, 𝜌=𝑀𝑉, and use it to calculate the volume of this block. If we multiply both sides of the density equation by volume 𝑉, we get π‘‰πœŒ=𝑀.

We can then divide both sides of the equation by the density 𝜌, which gives us 𝑉=π‘€πœŒ.

We now need to substitute the values we know for 𝑀 and 𝜌 into this equation. Doing this gives us 𝑉=542700.kgkgm

Let’s split this fraction into its numerical part and its units. We can work out the numerical part to be 542700=0.02.

For the units, we can first divide the top and bottom of the fraction by kilograms, and we see that all of the kg cancel out from the fraction. Secondly, if we multiply the top and bottom of the fraction by cubic metres, we see that the m3 cancel on the bottom, but we have a factor of m3 on the top. This mean that overall, we have 𝑉=0.02.m

Thus, our final answer is that the volume of the aluminum block is 0.02 m3.

Example 4: Finding the Mass of an Object given Its Volume and Density

It is worked out that a solid gold crown has a volume of 150 cm3. Find the mass of the gold crown, using a value of 19β€Žβ€‰β€Ž300 kg/m3 for the density of gold. Give your answer to one decimal place.

Answer

In this question, we are given the volume 𝑉 and density 𝜌 of a gold crown and are asked to find the mass 𝑀 of the crown.

Let’s start by taking our equation for density, 𝜌=𝑀𝑉, and multiply both sides by the volume, 𝑉, to get 𝑀=πœŒπ‘‰.

This tells us that the mass of the crown is simply the density of the crown multiplied by the volume of the crown.

However, before we calculate the mass, we should notice that the density of the crown is given to us in units of kilograms per cubic metre, while the volume of the crown is given to us in cubic centimetres.

This means that before we start combining these two quantities we need to convert them into the comparable units. Here this means we should convert the volume into cubic metres.

Recall that 100=1cmm. This means that (100)=1cmm, and thus, 1000000=1cmm.

This mean we need to divide the volume in cubic centimetres by 1β€Žβ€‰β€Ž000β€Žβ€‰β€Ž000. The volume of the crown, in cubic metres, is then 𝑉=0.00015.m

We can now substitute in the values we know for 𝜌 and 𝑉 to find the mass of the crown to be 𝑀=ο€½19300×0.00015=(19300Γ—0.00015)×=2.895.kgmmkgmmkg

The question asks for our answer to be given to one decimal place, so our final answer for the mass of the crown is 2.9 kg.

Note that, in this question, we did not need to know the exact shape of the crown. As long as we are told its volume, we are able to use the equation for density to find its mass.

We can calculate the density of a spherical object if we know the mass of the sphere, 𝑀, and the radius of the sphere, π‘Ÿ. If we know these quantities, then we can use the equation for density, which is 𝜌=𝑀𝑉, along with the equation for the volume of a sphere, which is 𝑉=43πœ‹π‘Ÿ.

If we combine these two equations by substituting the expression for the volume of a sphere into the equation for density, then we see that the density of a sphere is given by 𝜌=π‘€πœ‹π‘Ÿ.οŠͺ

If we multiply the top and bottom of the fraction by 3, we then have 𝜌=3𝑀4πœ‹π‘Ÿ, because the factors of 3 cancel out on the bottom of the fraction.

Example 5: Finding the Density of a Sphere given Its Mass and Radius

A bowling ball has a mass of 5.5 kg. The bowling ball is a sphere with a radius of 7 cm. What is the density of the bowling ball? Give your answer to the nearest kilogram per cubic metre.

Answer

This question asks us to calculate the density 𝜌 of a spherical object given its mass and radius.

Here, the mass of the bowling ball is 𝑀=5.5kg and the radius of the bowling ball is π‘Ÿ=7cm. Note that the question asks us to give the density in kilograms per cubic metre, so it will be useful for us to convert the radius, which is given to us in centimetres, into metres before we begin.

This is done by dividing the radius in centimetres by 100, so π‘Ÿ=0.07m.

Since we now know all the necessary information, we can use the equation for the density of a sphere, which is 𝜌=3𝑀4πœ‹π‘Ÿ.

We just need to substitute in the values for 𝑀 and π‘Ÿ for this question. If we do this, we find 𝜌=3Γ—(5.5)4πœ‹Γ—(0.07).kgm

If we work this out, remembering to cube both the number and the units in the denominator, we find 𝜌=3828.07….kgm

We are asked to give our answer to the nearest kilogram per cubic metre, so our final answer is simply 𝜌=3828.kgm

Example 6: Finding the Volume of an Object given Its Mass and Density

A steel ball bearing has a mass of 0.034 g. Find the diameter of the ball bearing in millimetres, rounded to the nearest millimetre. Use a value of 8β€Žβ€‰β€Ž000 kg/m3 for the density of steel.

Answer

In this example, we are given the mass 𝑀=0.034g of a spherical ball bearing, as well as its density, 𝜌=8000/kgm. Given this, we are asked to find the diameter of the ball bearing.

Let’s call the diameter of the ball bearing 𝑑 and recall that the diameter is two times the radius. So, if the radius of the ball bearing is π‘Ÿ, then 𝑑=2π‘Ÿ. With this in mind, let’s first calculate the radius of the ball bearing, using the density equation for a sphere.

Let’s start by rearranging the density equation for a sphere so that we can use it find the radius π‘Ÿ. The density of a sphere is given by 𝜌=3𝑀4πœ‹π‘Ÿ.

Let’s first multiply both sides of this equation by π‘ŸοŠ©. Doing this gives us πœŒπ‘Ÿ=3𝑀4πœ‹.

We can divide both sides by 𝜌 to get π‘Ÿ=3𝑀4πœ‹πœŒ.

Now, the right-hand side of this equation only involves terms that we know the value of.

The only extra step here is to convert the mass of the ball bearing, which is given to us in grams, into kilograms. We want to do this because the density of steel is given to us in units of kilograms per cubic metre.

We do this by diving the mass in grams by 1β€Žβ€‰β€Ž000, so the mass of the ball bearing is 𝑀=0.000034.kg

It will be much easier to work with this number if we write it in scientific notation. Doing this gives us 𝑀=3.4Γ—10kg.

We can substitute in the values of𝑀 and 𝜌 that we have now to get π‘Ÿ=3Γ—ο€Ή3.4Γ—104πœ‹(8000/).kgkgm

Notice that the units of kg cancel, and overall, the units become m3. We can also calculate the numerical part of this fraction to be 3Γ—3.4Γ—104πœ‹Γ—8000=1.0146Γ—10.

This means we have found the value of the radius cubed to be π‘Ÿ=1.0146Γ—10.m

We now need to take the cube root of this expression to find the radius of the ball bearing, which gives π‘Ÿ=0.001004847…,m where we have taken the cube root of the units as well as the number. The question asks us to give our answer in millimetres, so let’s convert our answer (which is in metres) into millimetres. This is done by multiplying by 1β€Žβ€‰β€Ž000, and so π‘Ÿ=1.004847….mm

We need to double the radius π‘Ÿ to find the diameter 𝑑 of the ball bearing, as the question asks. This, to the nearest millimetre, gives 𝑑=2π‘Ÿ=2.mm

This is our final answer: the diameter of the ball bearing is 2 mm.

We may also want to calculate the density of an object with less regular dimensions, such as a rectangular prism with length 𝑙, width 𝑀, and height β„Ž.

In this case, the volume 𝑉 of the rectangular prism is 𝑉=π‘™π‘€β„Ž.

This means that, by combining this volume formula with our equation for the density of an object, we can say that the density for a rectangular prism of a material is 𝜌=𝑀𝑉=π‘€π‘™π‘€β„Ž.

Example 7: Finding the Density of an Object given Its Mass and Dimensions

A brick has a mass of 3.5 kg. It is a rectangular prism with side lengths of 23 cm, 11 cm, and 7 cm. What is the density of the brick? Give your answer to the nearest kilogram per cubic metre.

Answer

In this question, we are given the dimensions and mass of a rectangular prism and are asked to find its density.

It does not matter which of the dimensions we call the length, height, or width. For this example, let’s say the length is the longest dimension, so 𝑙=23cm. Let’s say that the width is the middle length, so 𝑀=11cm. Finally, let’s say that the height is the shortest length, so β„Ž=7cm.

Notice that these distances are all in centimetres, but we are asked to calculate the density in kilograms per cubic metre. It will be simpler to convert the distances into metres before we start calculating the density. This is done by dividing each distance in centimetres by 100, and so, the dimensions of the brick can be written as 𝑙=0.23,𝑀=0.11,β„Ž=0.07.mmm

This means we are ready to calculate the density of the brick. We use these dimensions along with the mass of the brick that we are told in the question is 3.5 kg. This lets us calculate 𝜌=π‘€π‘™π‘€β„Ž=3.5(0.23)(0.11)(0.07).kgmmm

We can first simplify the units, which become kilograms per cubic metre, and then work out the numerical value to be 3.5(0.23)Γ—(0.11)Γ—(0.07)=1976.28….

This means that, to the nearest kilogram per cubic metre, the density of the brick is 𝜌=1976,kgm and this is our final answer.

Key Points

  • Density is a property of materials that measures the mass per unit volume of the material, which can be written as 𝜌=𝑀𝑉, where 𝜌 is the density of the mass, 𝑀 is the mass of the material, and 𝑉 is the volume of the material.
  • For a given material, its density is always the same regardless of the shape of the object made of that material.
  • We can combine the equation for density with the equation for the volume for specific shapes of objects. In particular, for a cube of side length 𝑙, its volume is 𝑉=π‘™οŠ©.
  • For a rectangular prism of length 𝑙, width 𝑀, and height β„Ž, its volume is 𝑉=π‘™π‘€β„Ž.
  • For a sphere of radius π‘Ÿ, the volume of the sphere is 𝑉=43πœ‹π‘ŸοŠ©.

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