Question Video: Using Newton’s Third Law of Motion | Nagwa Question Video: Using Newton’s Third Law of Motion | Nagwa

Question Video: Using Newton’s Third Law of Motion Physics • First Year of Secondary School

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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] Earth is not accelerated at all by the collision with the rock. [B] 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. [C] The magnitude of the acceleration of Earth due to the collision is equal to the acceleration of the rock but in the opposite direction. [D] The magnitude of the acceleration of Earth due to the collision is equal to the mass of the rock 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. What magnitude force does the deceleration of the rock exert on Earth?

Okay, so in this question, we’ve got a rock that’s decelerating. What’s happening here is that the rock collides with the surface of Earth while moving with an initial speed of one meter per second before the impact. This impact then causes the rock to decelerate such that it comes to rest over a time of 0.1 seconds. Let’s begin by making a sketch of the situation.

This here is Earth, and we’ll suppose that this pink circle here represents the rock. Since the rock is dropped onto Earth, we know that it’s initially moving vertically downward toward Earth’s surface. If we define this initial downward direction of motion of the rock as the positive direction, then since the rock’s initial speed is one meter per second, we can say that it’s got an initial velocity of one meter per second in the downward direction, or an initial velocity of positive one meter per second. Let’s label this initial velocity as 𝑢.

Since the impact with Earth brings the rock to a rest, we know that its final velocity is equal to zero meters per second, and we’ll label this final velocity as 𝑣. Let’s also label the time interval over which the rock is brought to rest as 𝛥𝑡, and we know that this is equal to 0.1 seconds. The only other information that we’re given is the mass of the rock, which we’re told is equal to five kilograms. Let’s label this mass as 𝑚. Now, in this first part of the question, we’re being asked to work out what magnitude force does the deceleration of the rock exert on Earth.

In order to answer this, we’re going to need to understand why it is that the rock actually exerts a force on Earth in the first place. It turns out that this can all be explained in terms of Newton’s laws of motion. We know that the rock experiences a deceleration, which is an acceleration in the opposite direction to its velocity. Since the initial downward velocity of the rock is positive, then the deceleration of the rock here is a negative acceleration, which acts to slow it down. And since this negative acceleration is acting in the opposite direction to the rock’s velocity, then this means that the acceleration must be acting vertically upward.

Now, Newton’s second law of motion tells us that whenever an object is experiencing an acceleration, then that object must be being acted on by a resultant force. Specifically, this law says that the force 𝐹 acting on the object is equal to the object’s mass 𝑚 multiplied by its acceleration 𝑎. So, Newton’s second law says that because the rock in this question experiences a negative or upward acceleration, then there must be a negative or upward force acting on the rock. Let’s label this force as 𝐹 subscript r. This force 𝐹 subscript r is the force that’s exerted by Earth on the rock in order to decelerate it.

If we look again at this first part of the question, though, we see that we’re not being asked about the force exerted on the rock by Earth, but rather the force exerted by the rock on Earth. To understand this force, we need to recall Newton’s third law of motion, which is often summarized as saying that every action has an equal and opposite reaction. What this means is that if we’ve got two objects labeled as A and B and object A exerts a force on object B that acts to the right and has a magnitude of 𝐹 subscript A, then this force is an action, which Newton’s third law tells us has a reaction that is equal and opposite. This reaction means object B exerting a force on object A. And since this reaction is opposite to the action, this force acts in the opposite direction, which is to the left. We’ll label this force exerted by object B on object A as 𝐹 subscript B.

Now, the action and reaction are not just opposite, but they are also equal. This means that the magnitudes of 𝐹 subscript A and 𝐹 subscript B are the same. So what Newton’s third law is saying is that if object A exerts a force on object B, then object B also exerts a force on object A that’s equal in magnitude but opposite in direction. In this question then, since we know that Earth exerts an upward force on the rock, then Newton’s third law tells us that the rock must also exert a downward force on Earth with the same magnitude. Let’s label this force exerted on Earth as 𝐹 subscript e. And we know that the magnitudes of 𝐹 subscript e and 𝐹 subscript r are equal.

Okay, so at this point, we’ve confirmed that there is a force 𝐹 subscript e exerted by the rock’s deceleration on Earth. To work out the magnitude of this force, our process will be as follows. Recalling that the acceleration of an object is equal to the change in the object’s velocity divided by the change in time over which this velocity change occurs, we can use the two velocity values of the rock along with this value for 𝛥𝑡 in order to work out the rock’s acceleration. If we then use this value of the rock’s acceleration along with its mass 𝑚 in Newton’s second law, we can calculate the force that Earth exerts on the rock in order to decelerate it. So that’s the force 𝐹 subscript r.

Then, because of Newton’s third law, we know that the rock must also exert an oppositely directed force on Earth, which we’ve labeled as 𝐹 subscript e. Since the two forces 𝐹 subscript r and 𝐹 subscript e must be equal in magnitude, we can then use our calculated value of 𝐹 subscript r in order to tell us the magnitude of 𝐹 subscript e. Now that we’ve outlined the steps that we need to take, let’s clear some space on the board so that we can make a start on them.

We’ll begin by substituting our values for the rock’s final velocity 𝑣, its initial velocity 𝑢, and the time interval 𝛥𝑡 over which it decelerates into this equation. When we do this, we get an expression for the rock’s acceleration, which we’ve labeled as 𝑎 subscript r. Specifically, we have that 𝑎 subscript r is equal to zero meters per second minus one meter per second all divided by 0.1 seconds. In the numerator, zero meters per second minus one meter per second is just negative one meter per second. Then, dividing negative one meter per second by 0.1 seconds, we calculate an acceleration 𝑎 subscript r of negative 10 meters per second squared. Now, let’s recall that this negative sign just means that the acceleration is acting upward in the negative direction or in the opposite direction to the rock’s initial velocity.

We are now ready to take this calculated acceleration of the rock along with the mass of the rock that we were given in the question and substitute them into this equation from Newton’s second law. When we do this, we find that the force exerted by Earth on the rock in order to decelerate it, which we’ve labeled as 𝐹 subscript r, is equal to five kilograms multiplied by negative 10 meters per second squared. We can recall that if we’ve got a mass in units of kilograms and an acceleration in meters per second squared, then we’ll calculate a force in units of newtons. In this case, when we do the multiplication, the force 𝐹 subscript r that we calculate is equal to negative 50 newtons. Again, this negative sign just means that the direction of this force is in the negative upward direction.

So the negative sign tells us the direction of the force. And then this number 50 along with the unit of newtons tells us the force’s magnitude. We have then that the magnitude of the force 𝐹 subscript r is equal to 50 newtons. Now, let’s recall that from Newton’s third law, we worked out that the magnitude of 𝐹 subscript r is equal to the magnitude of 𝐹 subscript e, where 𝐹 subscript e is the force that’s exerted by the rock on Earth as it decelerates. This means that the magnitude of 𝐹 subscript e is also equal to 50 newtons. This is the magnitude of the force that the deceleration of the rock exerts on Earth, which is our answer to this first part of the question.

Now, let’s clear some space so that we can move on to the question’s second part.

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

Now, in the first part of the question, we were asked for the magnitude of the force exerted on Earth by the rock as it decelerated. Now, in this second part, we’re asked for the force exerted on the rock by Earth. This is simply the force 𝐹 subscript r that we calculated during the first part of the question using Newton’s second law of motion. We found then that this force has a magnitude of 50 newtons. So then, our answer to part two of this question is that the magnitude of the force that the collision with Earth exerts on the rock is equal to 50 newtons.

At this point, it’s worth briefly reiterating that the fact that these two magnitudes that we calculated during parts one and two of this question are equal is a consequence of Newton’s third law of motion, which says that every action has an equal and opposite reaction. Okay, now let’s clear some space once more to look at the third and final part of this question.

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

Okay, so in this final part of the question, we’re being asked about the motion of Earth as a result of the collision with the rock. Let’s begin by recalling that we already worked out earlier in the question that, in the collision, the rock exerts a force on Earth. We labeled this force as 𝐹 subscript e, and we found that it had a magnitude equal to 50 newtons. In order to find this magnitude, we made use of Newton’s second and third laws of motion. Now, in this third part of the question, we can make use of Newton’s second law once more.

Let’s recall that Newton’s second law says that the force 𝐹 acting on an object is equal to the object’s mass 𝑚 multiplied by the acceleration resulting from that force 𝑎. Now, each of these four possible answers refer to the acceleration of Earth. So then, since Newton’s second law relates acceleration to force and we’ve already got some information about the force exerted on Earth in this collision, we can see that this law will be helpful to us here. If we divide both sides of the Newton’s second law equation by the mass 𝑚, then we can see that on the right-hand side, the 𝑚 in the numerator cancels with the 𝑚 in the denominator. This means we can rewrite the equation as acceleration 𝑎 is equal to force 𝐹 divided by mass 𝑚.

We already labeled the force exerted on Earth as 𝐹 subscript e. So let’s also label Earth’s mass as 𝑚 subscript e and the resulting acceleration of Earth as 𝑎 subscript e. Applying Newton’s second law to Earth in this collision then, we get this equation here. With this equation in mind, let’s have a look at the statement given in answer option (A). This statement claims that Earth is not accelerated at all by the collision with the rock. However, we already found that there is a force acting on Earth as a result of this collision.

Then, looking at this Newton’s second law equation, we see that a nonzero value of the force means that there must be a nonzero value of acceleration. This means that we know that Earth does experience an acceleration as a result of the collision with the rock. So the statement given in option (A) cannot be correct. Each of the three remaining answer options claims that Earth does experience an acceleration. However, they each make different claims about this acceleration’s magnitude.

Now, it turns out that Newton’s second law can help us do more than just eliminate answer option (A). It can also help us work out which of these three remaining options is the correct one. This equation we’ve got from Newton’s second law says that 𝑎 subscript e, the acceleration of Earth as a result of the collision with the rock, is equal to 𝐹 subscript e, the force exerted on Earth by the rock, divided by 𝑚 subscript e, which is Earth’s mass.

Now, technically, this equation is relating the vector quantities of acceleration and force. But since this mass term is a directional scalar quantity, then in order for this equation to be true, the acceleration 𝑎 subscript e and the force 𝐹 subscript e must be in the same direction as each other. And the magnitude of 𝑎 subscript e must be equal to the magnitude of 𝐹 subscript e divided by 𝑚 subscript e.

We can express this equation in words by saying that 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. This matches the statement that’s given here in option (B). And so our answer to this third and final part of the question is that the statement which most correctly describes the motion of Earth due to the collision with the rock is the one given in option (B). 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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