### Video Transcript

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.