Video: Newton's Third Law of Motion

A rock of mass 5 kg is dropped onto Earth and decelerates from a speed of 1 m/s to rest in a time of 0.1 s. What magnitude force does the deceleration of the rock exert on Earth? What magnitude force does the collision with Earth exert on the rock? 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.

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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.

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