### Video Transcript

In this video, we will be learning
about Newton’s third law of motion and how to apply it to various different
scenarios. So let’s start by looking at a
statement of this law. Newton’s third law of motion
considers the interaction between two objects. Specifically if we’re talking about
an object, let’s say object A, somehow interacting with another object, let’s say
object B, in this case, we’re drawing a collision between the two objects. Then, we can say the following. If object A exerts a force on
object B, so let’s say that the force exerted by A onto B during this collision is a
force towards the right, and we’ll call this force 𝐹. Then, according to Newton’s third
law of motion, object B exerts a force that is equal in magnitude or size, but in
the opposite direction on object A.

So as we can see in the diagram
that we’ve drawn, each object has a force of magnitude 𝐹 acting on it. But the two forces are acting in
opposite directions. Now, an often simplified version of
this law of motion is stated as every action has an equal and opposite reaction. So in this situation, we’ve got the
action force, that’s the force 𝐹, exerted by object A onto object B. And the equal and opposite reaction
is the force 𝐹 exerted to the left by object B on object A.

However, this statement of Newton’s
third law of motion “every action has an equal and opposite reaction” is not
specific enough. For one thing, it makes no mention
whatsoever of what the action force and reaction force are actually acting on. And simply memorising this
statement often leads to a common misconception. It is very easy to think that if
every action has an equal and opposite reaction, then because we’ve got two forces
acting in opposite directions and they’re exactly the same in magnitude, that those
two forces should cancel each other out.

Now, what that would mean is that
either every force ever exerted gets cancelled out by the equal and opposite
reaction force. And therefore, nothing in the
universe happens. Or our understanding of Newton’s
third law of motion is incorrect. And like we said earlier, this
statement of Newton’s third law is indeed not specific enough.

However, if we come back to our
original statement of Newton’s third law of motion as well as the diagram that we’ve
drawn here, we can see that, yes, we are dealing with two equal and opposite
forces. But they don’t cancel each other
out because each object only has one force acting on it. Object A only has the force 𝐹
acting towards the left on it. And object B only has the force 𝐹
acting towards the right on it.

Therefore, if we were to consider
each object by itself, we would see that it only has one force acting on it. And so the forces will not cancel
each other out. So let’s ignore this statement here
because it’s not specific enough and instead realise that, in Newton’s third law of
motion, if an object exerts a force on another object, then the second object exerts
a force equal in magnitude but in the opposite direction on the first object.

Now, it’s worth noting that
Newton’s third law of motion holds true every time a force is applied. In the diagram that we’ve drawn,
there’s a collision between object A and object B. However, it’s not necessary for the
force to be a contact force in order for Newton’s third law of motion to apply.

For example, in a situation where
an object is falling towards Earth, we often say that the Earth exerts a
gravitational force on the object. Specifically, that gravitational
force is the weight of the object. But that’s not all that’s going on
here. In reality, the Earth exerts a
force of 𝑤 on the object. And the object exerts an equal and
opposite force 𝑤 onto the Earth. So it’s not only that the Earth is
pulling the object towards itself. It’s more that the object and Earth
are mutually attracting each other. Now, the reason that we can often
ignore the force exerted by the object onto the Earth is seeing if we consider
Newton’s second law of motion.

Newton’s second law of motion tells
us that the net force on an object is equal to the mass of the object multiplied by
the acceleration experienced by that object. And so if we think about the weight
of the object, so more specifically, the gravitational force exerted by the Earth
onto the object, we see that the weight of the object is equal to the mass of the
object multiplied by the acceleration it experiences. But then, more specifically, we can
remember that objects falling in Earth’s gravitational field will experience the
same acceleration. And that acceleration is what we
call 𝑔, the acceleration due to Earth’s gravity or the gravitational field strength
of the Earth.

Now, this acceleration is a pretty
reasonable acceleration. It’s 9.8 meters per second
squared. However, if we then consider the
equal and opposite force 𝑤 acting on the Earth, then we can see that, this time,
although 𝑤 and 𝑤 are equal in magnitude, we are now considering 𝑤 as being equal
to the mass of the Earth multiplied by the acceleration experienced by the
Earth. That’s once again using Newton’s
second law of motion. But then, at this point, we can see
that 𝑤 and 𝑤 are both equal. So we can equate 𝑚𝑔, that’s the
mass of the object multiplied by the gravitational field strength of the Earth, to
the right-hand side of this equation, which is the mass of the Earth multiplied by
the acceleration experienced by the Earth.

Then, if we rearrange to solve for
the acceleration of the Earth, we see that it’s equal to the mass of the object
multiplied by the gravitational field strength divided by the mass of the Earth. And then, we can realise that the
mass of the Earth is absolutely huge. Specifically, the mass of the Earth
is about 5.972 times 10 to the power of 24 kilograms. And so because we have a very large
number in the denominator of this fraction, the overall value of the fraction
becomes very very small. That’s of course assuming that the
mass of the object that we’re dealing with isn’t comparable to the mass of the
Earth.

In other words, the object could be
anything like a tennis ball or a human being or even an aeroplane because the mass
of the object is much much smaller than the mass of the Earth. And so once again, this fraction
can become very very small for objects that are much smaller than the Earth. And in that situation, we therefore
see that the acceleration experienced by the Earth is extremely tiny.

In other words, we can actually
ignore the acceleration of the Earth. And that’s why often when we’re
dealing with an object falling towards Earth in its gravitational field, we
completely ignore the force exerted by the object on the Earth because it has very
little impact on the Earth. However, as soon as the object that
we’re dealing with becomes remotely comparable in mass to the Earth such as, for
example, when we’re thinking about the moon, then we do have to consider the force
exerted on the Earth. And then, we also need to think
about gravity as the interaction between the object and the Earth, rather than just
the Earth pulling the object towards itself.

Now, it’s also worth noting that
when we’re with dealing with relatively small objects, that’s small objects compared
to the mass of the Earth, then ignoring the value of the force exerted by the object
on the Earth is not entirely accurate. But it does simplify our
calculations massively because it gives us one listing to think about. And ignoring this force doesn’t
result in any major consequences to our calculations either. So now that we’ve had a look at a
couple of scenarios where we’d use Newton’s third law of motion, let’s have a go at
an example question.

A rock of mass five kilograms is
dropped onto Earth and decelerates from a speed of one meter per second to rest in a
time of 0.1 seconds. The first part of the question asks
us what magnitude force does the deceleration of the rock exert on Earth?

Okay so first of all, we see that,
in this question, we’ve got the Earth. And we’ve got a rock of mass five
kilograms dropped onto the Earth. We’ve been told that when the rock
hits the Earth, it decelerates from a speed of one meter per second to rest. In other words, previously before
the rocket hit the Earth, it was travelling downward, with a velocity of one meter
per second. Then, as soon as it hit the Earth,
it decelerated to a speed of zero metres per second. In other words, it was at rest. And the time taken for the rock to
change its velocity from one meter per second to rest was 0.1 seconds. So we can say that the time
interval, which we’ll call Δ𝑡, over which the rock lost its velocity was 0.1
seconds.

Then, at this point, we can see
that the rock has a changing velocity. And we know the amount of time over
which this velocity changed. So we can work out the acceleration
of the rock, which we will call 𝑎. And we can recall that the
acceleration of any object is given by finding the change in velocity of the object,
divided by the amount of time taken for that change in velocity to occur. Now, in this case, we’ve already
been given Δ𝑡. That’s 0.1 seconds. And we can work out the change in
velocity. To do this, we can find the final
velocity minus the initial velocity, which will give us the change in velocity. And this ends up being zero metres
per second, the final velocity, minus one meter per second, the initial velocity or,
in other words, negative one meter per second.

Now, the reason that the changing
velocity is negative is because the rock is losing velocity. In other words, we will see that
the value of 𝑎 is going to be negative because Δ𝑣 is negative. And this means that the rock is
decelerating. So we can plug in our values for
Δ𝑣 and Δ𝑡. And when we do so, we find that the
acceleration of the rock is negative 10 metres per second squared. In other words, the deceleration of
the rock is 10 metres per second squared.

Now, another way to think about
this is that the acceleration of the rock is in the opposite direction to its
initial velocity because we’ve said that the initial velocity was in this direction
downward towards the Earth. And so for the block to decelerate
when it hit the Earth, it would have to accelerate in the opposite direction. So what we figured out is the
rock’s acceleration and in which direction the acceleration is occurring.

Now at this point, we can move
forward and recall Newton’s second law of motion. This law tells us that the net
force exerted on an object, 𝐹, is given by multiplying the mass of the object by
the acceleration it experiences. So using this equation, we can work
out the net force on the rock when it decelerates after colliding with the
Earth. We can say that the force on the
rock, 𝐹, is equal to the mass of the rock, which we’ve been told is five kilograms,
multiplied by the acceleration experienced by the rock, which is negative 10 meters
per second squared.

Now, before we evaluate the
right-hand side of this equation, we can see that we’re working in base units,
kilograms and metres per second squared. Those are the base units of mass
and acceleration, respectively. And so whatever answer we find will
be in the base unit of force, which is the newton. Therefore, evaluating the
right-hand side gives us a force of negative 50 newtons.

In other words, because we said the
rock falling downwards was falling with a positive velocity, positive one metre per
second, then this means that the force exerted on the rock must be in the opposite
direction because we see that the force is negative. And hence, we can draw an upward
pointing arrow, that’s the force 𝐹 exerted on the rock causing it to
decelerate. And this force is the force exerted
by the Earth on the rock during the collision.

Now, it’s worth noting that what
we’ve been asked to find here is the magnitude of the force exerted on the Earth due
to the deceleration of the rock. However, what we’ve actually found,
the force 𝐹, is the force exerted by the Earth on the rock. So to work out the force that we’re
trying to find, we can recall Newton’s third law of motion. This law tells us that if an object
A exerts a force on another object B, then B exerts an equal and opposite force on
object A. Now, in this situation, the Earth
is object A. And the rock is object B because we
see that the Earth exerts a force 𝐹 on the rock. And that force is in the upward
direction.

So by Newton’s third law of motion,
we see that the rock exerts a force 𝐹 in the opposite direction on the Earth. And hence if we want to find the
magnitude of the force exerted on the Earth due to the deceleration of the rock,
then this force is going to have exactly the same magnitude as the force we’ve
already calculated. We don’t even need to worry about
directions because all what we’ve been asked to find is the magnitude or size of the
force. And hence, we can say that the
magnitude of force exerted on the Earth due to the deceleration of the rock is 50
newtons. Okay, so let’s look at the next
part of the question.

What magnitude force does the
collision with Earth exert on the rock?

So in this part of the question,
we’re trying to find the force exerted on the rock that caused the rock to
decelerate. However, in trying to find the
answer to the first part of the question, we’ve actually already found the answer to
this part of the question. Specifically, we found the value of
𝐹, which is the force exerted on the rock in order to decelerate it from one meter
per second to zero metres per second, in a time interval of 0.1 seconds. And because we once again only need
to give the magnitude of the force, so that’s the size of the force, we don’t need
to worry about direction or sign. And hence, we can say that the
magnitude of the force the collision with the Earth exerts on the rock is 50
newtons. Moving on to the final part of the
question then.

Which of the following statements
most correctly describes the motion of Earth due to the collision with the rock? A) The magnitude of the
acceleration of Earth due to the collision is equal to the acceleration of the rock
but in the opposite direction. B) The magnitude of the
acceleration of Earth due to the collision is equal to the mass of the rock divided
by Earth’s mass. C) Earth is not accelerated at all
by the collision with the rock. D) The magnitude of the
acceleration of Earth due to the collision is equal to the force applied in the
collision divided by Earth’s mass.

Okay, so there’re four different
options here. And each one of them is relatively
complicated in terms of what it’s trying to say. But what we’re trying to find is
the statement that most correctly describes the motion of Earth due to the collision
with the rock. And specifically, each one of these
statements refers to the magnitude of the acceleration of Earth. So which one of the statements A,
B, C, or D gives us a correct description of how to find the acceleration of Earth
due to this collision?

Well, first of all, we’ve seen
already that there is a force 𝐹 exerted on the Earth. And this force was found due to
Newton’s third law of motion. Because the Earth exerted a force
𝐹 upward onto the rock, the rock therefore exerted a force 𝐹 downward onto the
Earth. And then, we can see, using
Newton’s second law of motion, that if an object experiences a force and it has a
certain mass, then it must experience a certain acceleration. And because the force exerted on
the Earth is nonzero, the acceleration of the Earth can also not be zero. Therefore, we can immediately rule
out option C, which says that the Earth is not accelerated at all by the collision
with the rock.

Then, we can see that the force 𝐹
exerted on Earth, using Newton’s second law of motion, must be equal to the mass of
the Earth, which we’ll call 𝑚 subscript 𝐸, multiplied by the acceleration
experienced by the earth, which we’ll call 𝑎 subscript 𝐸. And therefore, if we want to find
the acceleration experienced by Earth, we divide both sides of the equation by the
mass of Earth. This way, 𝑚 subscript 𝐸 cancels
on the right-hand side. And so what we’re left with is 𝐹
divided by 𝑚 subscript 𝐸 is equal to 𝑎 subscript 𝐸.

In other words, the acceleration
experienced by Earth is equal to the force applied on the Earth divided by the mass
of the Earth. And of course, the force applied on
the Earth during this collision is the same in magnitude as the force applied on the
rock during this collision, but in the opposite direction. However, because each one of the
statements only refers to the magnitude of the acceleration, we therefore don’t need
to worry about the directions in which these forces or accelerations are acting. All we need to do is to find the
statement that tells us that the acceleration experienced by Earth is equal to the
force applied in the collision divided by the mass of Earth.

So looking at statement A, this one
says that the magnitude of the acceleration of Earth due to the collision is equal
to the acceleration of the rock, but in the opposite direction. So what this one is saying is that
the acceleration of Earth is equal to the acceleration experienced by the rock, but
in the opposite direction. Now, of course, the rock is falling
in the Earth’s gravitational field. So the acceleration that the rock
experiences is 𝑔. That’s the acceleration due to
gravity. And what this statement is saying
is that the acceleration of Earth is negative 𝑔 because it’s in the opposite
direction.

Well, this doesn’t match what we’ve
worked out here. And also as we said earlier,
because we’re only thinking about magnitudes, we don’t need to worry about
directions. Therefore, statement A is not the
answer to our question. Statement B then says that the
magnitude of the acceleration of Earth due to the collision is equal to the mass of
the rock divided by Earth’s mass. In other words, this one is saying
that the acceleration of Earth is equal to the mass of the rock 𝑚 divided by the
mass of the Earth 𝑚𝐸. But once again, this is not what we
found here. So option B is not the right
answer.

Therefore, it looks like option D
is the answer to our question. This one says that the magnitude of
the acceleration of Earth due to the collision is equal to the force applied in the
collision divided by the Earth’s mass. And that’s exactly what we found
over here. It’s the force applied in the
collision divided by the Earth’s mass. Therefore, option D is the answer
to our question.

Okay, so now that we’ve had a look
at an example, let’s quickly summarise what we’ve talked about in this lesson.

Firstly, in this video, we saw a
statement of Newton’s third law of motion, which says that if an object A exerts a
force on object B, then B exerts an equal and opposite force on object A. Secondly, we saw that the two
forces mentioned in the above statement do not cancel each other out as they act on
different objects. Therefore, each object only has one
force acting on it. And finally, we saw that Newton’s
Third Law also applies when considering gravity. But we often ignore the force on
the much much larger body as its effects are negligible. Such as, for example, when we’re
considering a ball falling towards Earth, we often consider the weight of the
ball. But we tend to ignore the force
exerted by the ball on the Earth because the acceleration that this force will cause
on the Earth is minuscule. So this is an overview of Newton’s
third law of motion.