### 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.