## Linear Momentum and Force

This text goes into the concepts of linear momentum, impulse, and how force is used to change momentum over time.

### Linear Momentum

The scientific definition of linear momentum is consistent with most people's intuitive understanding of momentum: a large, fast-moving object has greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system's mass multiplied by its velocity. In symbols, linear momentum is expressed as

$\mathbf{p}=m \mathbf{v}$

Momentum is directly proportional to the object's mass and also its velocity. Thus the greater an object's mass or the greater its velocity, the greater its momentum. Momentum $p$ is a vector having the same direction as the velocity $v$. The SI unit for momentum is $\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$.

#### Linear Momentum

Linear momentum is defined as the product of a system's mass multiplied by its velocity:

$\mathbf{p}=m \mathbf{v}$

#### Example 8.1 Calculating Momentum: A Football Player and a Football

(a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player's momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s.

#### Strategy

No information is given regarding direction, and so we can calculate only the magnitude of the momentum, p. (As usual, a symbol that is in italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum given in the equation, which becomes

$\mathbf{p}=m \mathbf{v}$

when only magnitudes are considered.

#### Solution for (a)

To determine the momentum of the player, substitute the known values for the player's mass and speed into the equation.

$p_{\text {player }}=(110 \mathrm{~kg})(8.00 \mathrm{~m} / \mathrm{s})=880 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$

#### Solution for (b)

To determine the momentum of the ball, substitute the known values for the ball's mass and speed into the equation.

$p_{\text {ball }}=(0.410 \mathrm{~kg})(25.0 \mathrm{~m} / \mathrm{s})=10.3 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$

The ratio of the player's momentum to that of the ball is

$\frac{p_{\text {player }}}{p_{\text {ball }}}=\frac{880}{10.3}=85.9$.

#### Discussion

Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much greater than the momentum of the football, as you might guess. As a result, the player's motion is only slightly affected if he catches the ball. We shall quantify what happens in such collisions in terms of momentum in later sections.

Source: Rice University, https://openstax.org/books/college-physics/pages/8-1-linear-momentum-and-force This work is licensed under a Creative Commons Attribution 4.0 License.