## 7.1 Linear momentum

When Isaac Newton was formulating his famous laws of motion, he was not thinking in terms of position, velocity, and acceleration. Instead, he worked with what he called the “quantity of motion.” This was a particular quantity that by itself described the motion of an object. Today we call it *linear momentum,* and usually just refer to it as momentum. An object’s momentum is its mass times its velocity:

\[

\vec{p} = m\vec{v} \tag{7.1}

\]

Notice the arrows on top of the symbols for momentum and velocity: these are vector quantities. When working with one-dimensional motion (only moving forward or backwards along a single axis), a positive or negative sign is all we need to tell us direction, and we drop the vector notation. When we work with two-dimensional motion, formal vector notation becomes important (see chapter 2 to review vectors).

The SI units for momentum are kilograms times meters per second (kg·m/s). There is no special name for this unit.

#### Example 7.1

In a world-record-breaking sprint, Jamaican athlete Usain Bolt ran 100 m in 9.58 s. If his mass was 80 kg at the time, what was his momentum? For simplicity, let’s neglect any fluctuations in his speed; in effect we’re finding his average momentum over this time.

\[

p = mv = \left(80\ \textrm{kg}\right)\left(\frac{100\ \textrm{m}}{9.85\ \textrm{s}}\right) = 835\ \textrm{kg}\cdot\textrm{m/s}

\]

#### Example 7.2

A falling ball is traveling at 8.25 m/s when it hits the ground. It bounces, and leaves the ground with the same speed, but in the exactly opposite direction. The ball is in contact with the floor for 7.3 ms. What was the change in the ball’s momentum, if the mass is 0.75 kg?

Say the ball is initially going in the negative direction, and returns in the positive direction.

\[

\begin{align*}

\Delta p &= p_f – p_i \\

&= mv_f – mv_i \\

&= m\underbrace{(v_f – v_i)}_{\Delta v} \\

&= (0.75\ \textrm{kg})\left((8.25\ \textrm{m/s}) – (-8.25\ \textrm{m/s})\right) \\

&= 12.4\ \textrm{kg}\cdot\textrm{m/s}

\end{align*}

\]

I want to point out two details here:

- When the mass does not change, \(\Delta p = m\Delta v\)
- When the ball was traveling in the negative direction, \(v = -8.25\ \textrm{m/s}\). Including the minus sign when you substitute in the value for \(v_i\) to do the final calculation is
*essential*.

#### Practice 7.1

## 7.1 Conservation of Momentum

Momentum is a conserved quantity. For any system of objects, the total momentum of the system does not change as long as there is no “external agent” acting on the system:

\[

\vec{p}_i = \vec{p}_f \tag{7.2}

\]

#### Example 7.3

A rocket is able to move because of conservation of momentum. Even when there is nothing to push off of, firing a blast still propels the ship forward. The momentum of the system must be constant; since the exhaust gases have momentum in one direction, the ship must have the same momentum in the opposite direction. This means the ship must move forward with some velocity. (Due to it’s much larger mass, the rocket will have a lower velocity than the exhaust gases.)

### 7.1.1 Problem Solving Steps

Since we’re still dealing with a conservation law, solving momentum problems is very similar to solving energy problems (see section 6.5). Conservation of momentum is a useful tool to analyze systems where multiple objects are interacting with each other over a short period of time.

You’ll usually find it helpful to draw a series of pictures in step 1 of the problem-solving process, showing what is going on just before and just after the interaction. The fundamental law that we’ll use in step 2 of the process is conservation of momentum:

\[ \vec{p}_i = \vec{p}_f \]

To solve mathematically (step 3), we’ll identify which objects have momentum before and after an interaction, and go from there.

#### Example 7.4

An astronaut is on a spacewalk and is holding onto a satellite with a mass of 300 kg. The astronaut, with spacesuit on, has a mass of about 95 kg. The astronaut and satellite are moving with a velocity of \(v_\textit{sys} = 0.1\ \textrm{m/s}\) when the astronaut pushes the satellite directly away from herself. After the push, the satellite has a velocity of \(v_s = 0.5\ \textrm{m/s}\). What is the velocity of the astronaut?

**Step 1: Draw a picture**

For problems dealing with momentum, it is often beneficial to draw a series of pictures, one before the objects interact, and one after. I’ll represent the astronaut as a circle, and the satellite as a rectangle.

**Step 2: Apply conservation of momentum**

Work from general to specific, starting from a fundamental principle of physics:

\[ p_i = p_f \]

In the “initial” state, state, you have the astronaut and the satellite traveling together as a single system. In the “final” state, the two are separate. So the expression \( p_i = p_f \) is expressed as

\[ p_\textit{sys} = p_s + p_a \]

**Step 3: Solve**

Now we can apply the specific expressions for determining momentum, work through the algebra, and finally substitute in values to calculate a final answer. I’ll say the positive direction is to the right, and show all the steps from the very beginning:

\[

\begin{align*}

p_i &= p_f \\

p_\textit{sys} &= p_s + p_a \\

(m_a + m_s)v_\textit{sys} &= m_sv_s + m_av_a \\

(m_a + m_s)v_\textit{sys} – m_sv_s &= m_sv_s + m_av_a – m_sv_s\\

(m_a + m_s)v_\textit{sys} – m_sv_s &= m_av_a \\

\frac{(m_a + m_s)v_\textit{sys} – m_sv_s}{m_a} &= \frac{m_av_a}{m_a} \\

\hookrightarrow v_a &= \frac{(m_a + m_s)v_\textit{sys} – m_sv_s}{m_a} \\

&= \frac{\left(95\ \textrm{kg} + 300\ \textrm{kg}\right)\left(0.1\ \textrm{m/s}\right) – \left(300\ \textrm{kg}\right)\left(0.5\ \textrm{m/s}\right)}{95\ \textrm{kg}} \\

&= -1.2\ \textrm{m/s}

\end{align*}

\]

Note that this has a negative value; the astronaut is traveling in the opposite direction as they were before.

#### Example 7.5

A spaceship of mass \(M = 2.0 \times 10^6\ \textrm{kg}\) is cruising at a speed of \(v_0 =5.0 \times 10^6\ \textrm{m/s}\) when the antimatter reactor fails, blowing up the ship in two pieces. One section, having a mass of \(m_1 =5.0 \times 10^5\ \textrm{kg}\), is blown straight backward with a speed of \(2.0 \times 10^6\ \textrm{m/s}\). What is the velocity of the other piece after the explosion?

Your drawing for this problem would look very similar to the drawing for the previous example.

\[

\begin{align*}

p_i &= p_f \\

p_\textit{ship} &= p_1 + p_2 \\

Mv_0 &= m_1v_1 + m_2v_2 \\

\hookrightarrow v_2 &= \frac{Mv_0 – m_1v_1}{m_2} \\

&= \frac{Mv_0 – m_1v_1}{M – m_1}

\end{align*}

\]

Note that the first piece is blown *backwards,* so \(v_1 = -2.0 \times 10^6\ \textrm{m/s}\). We find \( v_2 = 7.33 \times 10^6\ \textrm{m/s}\).