FHSST Physics/Momentum/System

= The Momentum of a System =

In Chapter 4 Forces the concept of a system was introduced. The bodies that make up a system can have different masses and can be moving with different velocities. In other words they can have different momenta.



Since momentum is a vector, the techniques of vector addition discussed in This chapter must be used to calculate the total momentum of a system. Let us consider an example.

Worked Example 34 Calculating the Total Momentum of a System
Question: Two billiard balls roll towards each other. They each have a mass of 0.3kg. Ball 1 is moving at $$v_1=1\ m.s^{-1}$$ to the right, while ball 2 is moving at $$v_2=0.8\ m.s^{-1}$$ to the left.

Calculate the total momentum of the system.

Answer:

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

all in the correct units!
 * the mass of each ball,
 * the velocity of ball 1, $$\overrightarrow{v_1}$$ and
 * the velocity of ball 2, $$\overrightarrow{v_2}$$,

Step 2 :

What is being asked? We are asked to calculate the total momentum of the system. In this example our system consists of two balls. To find the total momentum we must sum the momenta of the balls,

$$\begin{matrix}\overrightarrow{p}_{total} &=& \overrightarrow{p_1}+\overrightarrow{p_2}\end{matrix}$$

Since ball 1 is moving to the right, its momentum is in this direction, while the second ball's momentum is directed towards the left.



Thus, we are required to find the sum of two vectors acting along the same straight line. The algebraic method of vector addition introduced in this chapter can thus be used.

Step 3 :

Firstly we choose a positive direction. Let us choose right as the positive direction, then obviously left is negative.

Step 4 :

The total momentum of the system is then the sum of the two momenta taking the directions of the velocities into account. Ball 1 is travelling at $$1\ m.s^{-1}\ to\ the\ right$$ or $$+1\ m.s^{-1}$$. Ball 2 is travelling at $$0.8\ m.s^{-1}\ to\ the\ left$$ or $$-0.8\ m.s^{-1}$$. Thus,

Right is the positive direction

$$\begin{matrix}\overrightarrow{p}_{total} &=& m_1\overrightarrow{v_1} +m_2\overrightarrow{v_2}\\&=& (0.3kg)(+1\ m.s^{-1}) + (0.3kg)(-0.8\ m.s^{-1})\\&=& (+0.3\ kg.m.s^{-1}) + (-0.24\ kg.m.s^{-1}) \\&=& +0.06\ kg.m.s^{-1}\\&=& 0.06\ kg.m.s^{-1}\textbf{\ to\ the\ right}\end{matrix}$$

In the last step the direction was added in words. Since the result in the second last line is positive, the total momentum of the system is in the positive direction (i.e. to the right).