# 8.1 Linear momentum and force  (Page 3/5)

 Page 3 / 5

## Making connections: illustrative example

In [link] , a puck is shown colliding with the edge of an air hockey table at a glancing angle. During the collision, the edge of the table exerts a force F on the puck, and the velocity of the puck changes as a result of the collision. The change in momentum is found by the equation:

$\text{Δp}=m\text{Δv}=m\text{v'}-m\text{v}=m\left(\text{v'}+\left(-\text{v}\right)\right)$

As shown, the direction of the change in velocity is the same as the direction of the change in momentum, which in turn is in the same direction as the force exerted by the edge of the table. Note that there is only a horizontal change in velocity. There is no difference in the vertical components of the initial and final velocity vectors; therefore, there is no vertical component to the change in velocity vector or the change in momentum vector. This is consistent with the fact that the force exerted by the edge of the table is purely in the horizontal direction.

## Test prep for ap courses

A boy standing on a frictionless ice rink is initially at rest. He throws a snowball in the + x -direction, and it travels on a ballistic trajectory, hitting the ground some distance away. Which of the following is true about the boy while he is in the act of throwing the snowball?

1. He feels an upward force to compensate for the downward trajectory of the snowball.
2. He feels a backward force exerted by the snowball he is throwing.
3. He feels no net force.
4. He feels a forward force, the same force that propels the snowball.

(b)

A 150-g baseball is initially moving 80 mi/h in the – x -direction. After colliding with a baseball bat for 20 ms, the baseball moves 80 mi/h in the + x -direction. What is the magnitude and direction of the average force exerted by the bat on the baseball?

## Section summary

• Linear momentum ( momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity.
• In symbols, linear momentum $\mathbf{p}$ is defined to be
$\mathbf{p}=m\mathbf{v},$
where $m$ is the mass of the system and $\mathbf{v}$ is its velocity.
• The SI unit for momentum is $\text{kg}·\text{m/s}$ .
• Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes.
• In symbols, Newton’s second law of motion is defined to be
${\mathbf{F}}_{\text{net}}=\frac{\Delta \mathbf{p}}{\Delta t}\text{,}$
${\mathbf{F}}_{\text{net}}$ is the net external force, $\Delta \mathbf{p}$ is the change in momentum, and $\Delta t$ is the change time.

## Conceptual questions

An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy?

An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum?

Professional Application

Football coaches advise players to block, hit, and tackle with their feet on the ground rather than by leaping through the air. Using the concepts of momentum, work, and energy, explain how a football player can be more effective with his feet on the ground.

How can a small force impart the same momentum to an object as a large force?

## Problems&Exercises

(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of $7\text{.}\text{50 m/s}$ . (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of $\text{600 m/s}$ . (c) What is the momentum of the 90.0-kg hunter running at $7\text{.}\text{40 m/s}$ after missing the elephant?

(a) $\text{1.50}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$

(b) 625 to 1

(c) $6\text{.}\text{66}×{\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$

(a) What is the mass of a large ship that has a momentum of $1\text{.}\text{60}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{kg}·\text{m/s}$ , when the ship is moving at a speed of $\text{48.0 km/h?}$ (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of $\text{1200 m/s}$ .

(a) At what speed would a $2\text{.}\text{00}×{\text{10}}^{4}\text{-kg}$ airplane have to fly to have a momentum of $1\text{.}\text{60}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{kg}·\text{m/s}$ (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of $\text{60.0 m/s}$ ? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship.

(a) $8\text{.}\text{00}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{m/s}$

(b) $1\text{.}\text{20}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{kg}·\text{m/s}$

(c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be $-0\text{.}\text{0100 m/s}$ , which is probably not noticeable.

(a) What is the momentum of a garbage truck that is $1\text{.}\text{20}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{kg}$ and is moving at $10\text{.}\text{0 m/s}$ ? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck?

A runaway train car that has a mass of 15,000 kg travels at a speed of $5\text{.4 m/s}$ down a track. Compute the time required for a force of 1500 N to bring the car to rest.

54 s

The mass of Earth is $5\text{.}\text{972}×{10}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ and its orbital radius is an average of $1\text{.}\text{496}×{10}^{\text{11}}\phantom{\rule{0.25em}{0ex}}\text{m}$ . Calculate its linear momentum.

Why does earth exert only a tiny downward pull?
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