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Lesson Video: Newton’s Second Law: Variable Mass Mathematics

In this video, we will learn how to use differentiation for Newton’s second law of motion of a particle with variable mass.

14:07

Video Transcript

Newton’s Second Law: Variable Mass

In this video, we’ll learn how to use differentiation for Newton’s second law of motion of a particle with variable mass. Let’s begin by recalling Newton’s second law of motion by defining it for a body of constant mass.

When a net force acts on a body, that body accelerates in the direction of the force. The equation that describes the link between the acceleration and the force is 𝐹 equals 𝑚𝑎, where 𝐹 is the force, 𝑚 is the constant mass of the body, and 𝑎 is its acceleration. Now it’s worth noting that we can also express Newton’s second law in terms of the body’s momentum. Now, the momentum of a body 𝑃 is equal to its mass 𝑚 multiplied by its velocity 𝑣. Newton’s second law tells us that the force applied to a body is equal to the rate of change of momentum of the body. So, 𝐹 is equal to d𝑃 by d𝑡.

We can substitute 𝑚𝑣 into this equation. Since we’re looking at Newton’s second law for constant mass, this 𝑚 in our equation will be a constant. And so, we can bring it out to the front of our equation. We now have that 𝐹 is equal to 𝑚d𝑣 by d𝑡. Now, we know that d𝑣 by d𝑡 is simply the rate of change of velocity, which is equal to the acceleration or 𝑎. So, we arrive at 𝐹 equals 𝑚𝑎, which is the equation we’re familiar with seeing.

Now whilst we have used vector quantities here, it is worth noting that if the motion is linear, we’re able to use scalar magnitudes instead of vectors in our calculations. Let’s now look at what would happen if we want to use Newton’s second law with variable mass. And we know that the force 𝐹 is equal to the rate of change of momentum d𝑃 by d𝑡, where the momentum is equal to the mass multiplied by the velocity. This time around, both 𝑚 and 𝑣 are variables. So, in order to differentiate 𝑚𝑣, we’ll need to use the product rule. In doing this, we obtain that 𝐹 is equal to 𝑚d𝑣 by d𝑡 plus 𝑣d𝑚 by d𝑡. We have arrived at the formula which we’ll be using to solve problems with variable masses. We can also use the formula in the stage before we perform the differentiation. So, 𝐹 is equal to d by d𝑡 of 𝑚𝑣.

Another thing we may notice is that in the formula 𝐹 is equal to 𝑚d𝑣 by d𝑡 plus 𝑣d𝑚 by d𝑡. The term 𝑚d𝑣 by d𝑡 is equivalent to the right-hand side for the formula of Newton’s second law with constant mass; that’s 𝐹 equals 𝑚𝑎. When we’re dealing with variable mass, we have this extra term of 𝑣d𝑚 by d𝑡. And this makes sense, since if 𝑚 is constant, then d𝑚 by d𝑡 is equal to zero and this second term vanishes, leaving us with 𝐹 is equal to 𝑚𝑎. Let’s now look at an example of how we can solve a problem involving variable mass.

Fill in the blank. The force acting on a mass varying according to the function 𝑚 of 𝑡 is equal to five plus two 𝑡 kilograms and moving with a constant velocity of four meters per second is blank.

Now, remember that the force acting on a body with varying mass is equal to the rate of change of momentum, so that’s 𝑚d𝑣 by d𝑡 plus 𝑣d𝑚 by d𝑡. Since the velocity is a scalar, we’ll be using the scalar equation of Newton’s second law for variable mass instead of the vector form. The question has given us a function for the mass and a value for the velocity. We have that 𝑚 is equal to five plus two 𝑡 kilograms and 𝑣 is equal to four meters per second. We need to differentiate both of these with respect to 𝑡. Starting with 𝑚, when we differentiate the constant five, we’ll get zero. And when we differentiate the two 𝑡, we’ll get two. Hence, d𝑚 by d𝑡 is equal to two. Now, we have that 𝑣 is equal to four, which is a constant. So, when we differentiate it, we’ll get zero.

We’re now able to substitute these values into our equation for 𝐹. We have that 𝐹 is equal to five plus two 𝑡 multiplied by zero plus four multiplied by two. Since the first time is all multiplied by zero, this will disappear. And so, we are left with 𝐹 is equal to eight newtons. And here, we have reached our solution, which is that a body with an initial mass of five kilograms which increases at two kilograms per second which is moving with a constant velocity of four meters per second must have a constant force acting on it of eight newtons. We can also note here that the initial mass of the body does not affect the force required to maintain this constant velocity. The only thing that matters is the rate of change of the mass.

We will now move on to our second example, where we will see a system where both the mass and velocity are functions of time.

A body moves in a straight line. At time 𝑡 seconds, its displacement from a fixed point is given by 𝑠 is equal to six 𝑡 squared plus nine 𝑡 meters. Its mass varies with time such that 𝑚 is equal to eight 𝑡 plus nine kilograms. Write an expression for the force acting on the body at time 𝑡.

Now, in this question, we’re asked to find the force acting on the body; however, we’re given that 𝑚 is equal to eight 𝑡 plus nine kilograms. That means that 𝑚 is a variable. So, in order to find this force, we’ll need to use Newton’s second law for variable mass. Newton’s second law tells us that 𝐹 is equal to the rate of change of momentum or 𝑚d𝑣 by d𝑡 plus 𝑣d𝑚 by d𝑡. We can start by considering the mass. We have that 𝑚 is equal to eight 𝑡 plus nine. We will also need d𝑚 by d𝑡, so let’s differentiate this.

We have that d𝑚 by d𝑡 is equal to eight. Now, in this question, we’ve been given the displacement of the body 𝑠 instead of its velocity. However, we know that the velocity is equal to the rate of change of displacement, so 𝑣 is equal to d𝑠 by d𝑡. So, in order to find 𝑣, we need to differentiate six 𝑡 squared plus nine 𝑡 with respect to 𝑡. We find that 𝑣 is equal to 12𝑡 plus nine. Now, we’ll need to differentiate this again in order to find d𝑣 by d𝑡. By differentiating 12𝑡 plus nine with respect to 𝑡, we find that d𝑣 by d𝑡 is equal to 12.

We now have all the components to substitute into our formula for finding the force 𝐹. We have that 𝐹 is equal to eight 𝑡 plus nine multiplied by 12 plus 12𝑡 plus nine multiplied by eight. Expanding the brackets, we have 96𝑡 plus 108 plus 96𝑡 plus 72. Simplifying this, we reach our solution, which is that the force 𝐹 is equal to 192𝑡 plus 180 newtons. Note that in this case, 𝐹 is a time-dependent force.

In the next example, we’re going to see how we can reverse this process and use the formula for Newton’s second law with variable mass in order to find the rate of change of mass of a body.

A metallic ball moves in a straight line with a constant velocity of magnitude one meter per second. It enters a dusty medium. If the force acting on the ball at any instant is 10 dynes, find the rate of change of mass of the ball due to the dust adherence to its surface.

The first thing to consider here is what we’ve been asked to find, and that is the rate of change of mass of the ball. This implies that the ball does not have a constant mass; hence, we’ll need to use Newton’s second law for variable masses. Newton’s second law for variable masses tells us that 𝐹 is equal to 𝑚d𝑣 by d𝑡 plus 𝑣d𝑚 by d𝑡. What we have been asked to find is d𝑚 by d𝑡. Let’s consider the pieces of information that the question has given us.

We know that the velocity of the ball is one meter per second, and this is a constant velocity. Therefore, d𝑣 by d𝑡 will be equal to zero. We’ve also been given that the force acting on the ball is 10 dynes. In order to make our calculations easier, we need to convert this to newtons. We have that one dyne is equal to 10 to the negative five newtons. So, our force is equal to 10 multiplied by 10 to the negative five newtons or 10 to the negative four newtons. Now, since d𝑣 by d𝑡 is equal to zero, the first term in our formula for finding the force will also be equal to zero. So, in the case of this question with the constant velocity, we have that the force 𝐹 is equal to the velocity 𝑣 multiplied by the rate of change of the mass d𝑚 by d𝑡.

Substituting in our values for 𝐹 and 𝑣, we can see that 10 to the negative four is equal to one multiplied by d𝑚 by d𝑡. Or we could write d𝑚 by d𝑡 is equal to 10 to the negative four. Now, let’s consider the units of this value. Our velocity 𝑣 is in meters per second, and our force 𝐹 is in newtons. This tells us that this rate of change of mass will be in kilograms per second. Now, we know that one kilogram is equal to 10 cubed grams. Therefore, we can write that d𝑚 by d𝑡 is equal to 10 to the negative four multiplied by 10 cubed grams per second. This simplifies to 10 to the negative one grams per second, which can also be written as d𝑚 by d𝑡 is equal to 0.1 grams per second. And this is the solution to this question; that is, the rate of change of mass of the ball due to the dust adherence to its surface is 0.1 gram per second.

This example has given us an idea of what sort of physical process can cause the increase in the mass of the body. Essentially, the body can accumulate mass that it makes contact with if it is part of the medium which it is traveling through. In a similar way, a body can reduce in mass, for example, a rocket expelling fuel. We will see what that might look like in the next example.

A rocket was ascending vertically, projecting its burnt fuel at 3,600 kilometers per hour vertically downwards. Given that, for every eight seconds, it expelled three kilograms of fuel, find the elevator force generated by the rocket’s engine.

Now, we’ve been asked to find the force generated by the rocket’s engine. And we can also see that the rocket’s mass is varying since it’s expelling fuel. Therefore, we will need to use Newton’s second law for variable mass, which tells us that 𝐹 is equal to 𝑚d𝑣 by d𝑡 plus 𝑣d𝑚 by d𝑡.

Now when we look a bit closer at the question, we may notice that the information we’ve been given is in fact about the fuel and not the rocket. We are told that it is projecting it’s burnt fuel at 3,600 kilometers per hour and that every eight seconds three kilograms of fuel is expelled. This means if we’re calculating using these values, we will in fact be finding the force acting upon the fuel which is expelled from the rocket. This force 𝐹 will be acting vertically downwards.

Now, in order to find the elevator force which propels the rocket vertically upwards, we can use Newton’s third law, which tells us that this force will be equal and opposite to the force of the fuel being projected vertically downwards. Hence, it will be a force with the same magnitude 𝐹 just acting vertically upwards.

So, let’s use Newton’s second law for variable mass to find the force acting on the fuel, which is being expelled vertically downwards. We have that the velocity of the fuel will be 3,600 kilometers per hour. However, since this is acting vertically downwards, we can write this as negative 3,600 kilometers per hour. The units of this velocity are kilometers per hour. However, since we’re going to be wanting to work in kilograms, meters, seconds, and newtons, we need to convert this velocity to meters per second. Since there are 1,000 meters in a kilometer and 3,600 seconds in an hour, we need to multiply the negative 3,600 by 1,000 over 3,600. So, our velocity is negative 1,000 meters per second.

Since 𝑣 is a constant, when we differentiate it to find d𝑣 by d𝑡, we will see that it is equal to zero. Hence, when we substitute it into our formula, this will make the first term be equal to zero. Due to this constant velocity, we can rewrite our formula for the force as 𝐹 is equal to 𝑣d𝑚 by d𝑡. We have just found 𝑣; therefore, the only thing we need to find 𝐹 is d𝑚 by d𝑡. This is the rate of change of the mass.

We’re told that for every eight seconds the rocket expelled three kilograms of fuel. Since we are considering the force acting on the fuel which is being expelled from the rocket, this statement tells us that the mass of fuel which has been expelled is increasing by three kilograms every eight seconds. To find the rate of fuel being expelled every second, we simply need to divide the three kilograms by the eight seconds. So, the rate of change of mass is equal to three over eight kilograms per second.

We can now substitute this value along with the velocity into our equation for the force. The force is therefore equal to negative 1,000 multiplied by three over eight, which simplifies to negative 375 newtons. Here, we have nearly reached our solution. However, this is the force acting upon the fuel which has been expelled from the rocket. The elevator force generated by the rocket’s engines will be equal and opposite to this force. Our solution is that the elevator force is equal to 375 newtons.

We have now seen a variety of examples. Let’s recap some key points of the video.

Key Points.

Newton’s second law of motion for variable mass tells us that the force is equal to the rate of change of momentum. So, 𝐹 is equal to d𝑃 by d𝑡, which is also equal to d by d𝑡 of 𝑚𝑣. Using the product rule, we can also find that 𝐹 is equal to 𝑚d𝑣 by d𝑡 plus 𝑣d𝑚 by d𝑡. When motion is linear and our values are given as scalars not vectors, we can use the scalar formula of 𝐹 is equal to 𝑚d𝑣 by d𝑡 plus 𝑣d𝑚 by d𝑡.

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