As an example let us calculate the pressure of a gas filled in a container. Let the mass of each molecule be m and let their average speed be v . The number density of the molecules in the gas is taken to be n . Now consider a surface of the container perpendicular to the x-axis. (see figure 8).

Each molecule, when reflected from the wall imparts a momentum equal to 2mv_x to the wall. The average number of molecules hitting are A of the wall per unit time will be half of those contained in a cylinder of base area A and height v_x (the other half will be moving in the other direction). This comes out to be . Thus from the formula derived above the force on the wall applied by these molecules is

which gives the pressure

This is a result you are already familiar with kinetic theory of gases. But now you know how it comes out. Having done this problem we now deal with another very interesting application of the momentum-force relationship, known as the variable mass problem.

So far we have been dealing with particles of fixed masses. Let us now apply the equation   to a problem when the mass of the system under consideration varies with time. The most famous example of this is the rocket propulsion.

Let a rocket with mass M at time t be moving with velocity . A small mass Δm with velocity comes and gets stuck with it so that the rocket now has mass M + Δm and moves with a velocity (see figure 9 below) after a time interval of Δt. We want to find at what rate does the velocity of the rocket increase? We point out that the word rocket has been used here to represent any system with variable mass .

Let us write the momentum change in time interval Δt and equate this to the total external force on the system (that is the sum of external forces acting on M and Δm) times Δt. That gives

is nothing but the relative velocity of the mass Δm with respect to the rocket.

Dividing both sides of the equation above by Δt then leads to

We now let Δt → 0 . In this limit also goes to zero for continuously varying mass. Further, , the rate of change of the mass of the rocket. Thus the equation for the velocity of a rocket is

Note that both the mass and velocity are now functions of time. For a rocket so that . It is this term that provides the thrust to the rocket. As pointed out above, although this equation has been derived keeping rocket in mind, it is true for any system with variable mass .

Example: We now solve a simple problem involving the rocket equation. A rocket is fired vertically up in a gravitational field. What is its final velocity assuming that the rate of exhaust and its relative velocity remain unchanged during the lift off?

The motion of rocket is one-dimensional. We take the vertically up direction to be positive. Then we have where u is a positive number. Therefore the rocket equation takes the form

which gives

Here we have taken the initial time and initial velocity both to be zero. Even after the fuel has all been burnt, we see if we observe the rocket time t after being fired, its velocity will be given by the formula

assuming g to be a constant.

Finally, although the momentum-force equation can provide answers for the velocities, I would like to urge you to always think about how the internal forces that generate momenta in opposite directions are generated. That helps in understanding the underlying physics better. For example in the rocket problems, we say that provides the thrust to make the rocket move forward. But think about what generates this force? The answer is as follows. In a closed container, gas pressure applies force in all directions and these forces cancel each other. But when a hole is made from where the gas can escape, the force in the opposite direction is unbalanced; and that is what makes the rocket move. If you understand this, you should e able to answer the following question. If we take a closed box with vacuum inside and punch a hole in it. Which way will it move?

We conclude this lecture by summarizing what we have learnt. We studied the conservation of momentum and a related concept of the centre of mass. Using momentum, we then calculated the force on a surface being hit by a stream of particles, or jet of water. Finally we learnt about the variable mass problem and applied it to a rocket taking off. In the coming lecture we will use the conservation of momentum principle along with the conservation of energy and see how this combination becomes a powerful tool in solving mechanics problems.