Linear Momentum - Revision history

Edwynbasa at 20:30, 28 April 2026

2026-04-28T20:30:36Z

Joshykim at 00:47, 3 December 2025

2025-12-03T00:47:18Z

← Older revision		Revision as of 20:47, 2 December 2025
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	claimed by joshua kim fall 2025		claimed by joshua kim fall 2025

	~~This page defines~~ the ~~linear~~ momentum ~~of particles~~ and ~~systems~~.		==Linear Momentum==

			It is a vector quantity that describes an object's motion by combining its mass and velocity. Because mass is a scalar while velocity is a vector, momentum always points in the same directions as an object's motion. Momentum is typically represented by the symbol <math>\vec{p}</math>, and its SI unit is the kilogram meter per second (kg*m/s). Momentum is sometimes referred to simply as "momentum", and the plural form may appear as either "momenta" or "momentums".

	==The Main Idea==		==The Main Idea==

	~~Linear momentum~~ is ~~a vector quantity describing an object's~~ motion~~. It is defined as the product~~ of an object~~'s [[Mass]] (<math>m</math>) and [[Velocity]] (<math>\vec{v}</math>)~~. ~~Note that~~ mass ~~is a scalar while~~ velocity ~~is a vector, so~~ an object~~'s linear~~ momentum ~~is always in the same direction as its velocity~~. ~~Linear momentum is~~ a ~~vector quantity. Linear momentum is represented by the letter <math>\vec{p}</math>~~ and is often ~~referred to as simply "momentum." The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta~~ or ~~momentums~~.		Momentum reflects how difficult it is to stop or change the motion of an object. Increasing either the mass or the velocity of an object increases its momentum. Momentum plays a central role in understanding collision, motion in systems of particles, and many other physical processes because it is often conserved even when forces are complicated or short-lived.

	===A Mathematical Model===		===A Mathematical Model===

			For a single particle, linear momentum is defined by <math>\vec{p} = m\vec{v}</math>, where <math>m</math> is the mass and <math>\vec{v}</math> is the velocity. This formula works well for everyday speeds, but for motion approaching the speed of light, the definition must be updated using relativistic momentum.

			For systems of multiple particles, the total momentum is the vector sum of the momenta of all particles: <math>\vec{p}_{\text{total}} = \sum m_i\vec{v}_i</math>. This allows complex systems - such as a rolling disk with many moving pieces - to be treated as if all their mass were concentrated at a single point.

	====Single Particles====		====Single Particles====
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	====Impulse====		====Impulse====
	Impulse is the change in momentum. It is the integral of force over a time ~~interval. Therefore~~, ~~impulse equals~~ to ~~the force,~~ F~~, multiplied by the change in time,~~ t. ~~The unit for impulse~~ is ~~newton*second. Since impulse is the change~~ in ~~momentum~~, ~~an alternative formula of impulse is mass~~, ~~m, multiplied by~~ change in ~~velocity, v. Both the formulae mentioned are important to remember to solve impulse-related questions~~.		Impulse measures the change in momentum during a time interval. It is defined as the integral of the force over time, and reduces to <math>\vec{J} = \vec{F}\Delta t</math>. This relationship is extremely useful in analyzing collisions, sports physics, and any situation where short-duration forces cause a measurable change in motion.

	====In Relation to Other Physics Topics====		====In Relation to Other Physics Topics====

	~~In the absence of a~~ net force, ~~the~~ momentum of a ~~particle stays constant over time~~, ~~as stated by~~ Newton's ~~first law~~.		When no net external force acts on a particle, its momentum remains constant (Newton's First Law). When a force does act, the rate at which the momentum changes equals the net force (Newton's Second Law). In situations where an impulse is applied, the momentum changes by an amount equal to that impulse. For systems in which the total external force is zero, momentum is conserved even if internal forces act between particles.

			===Momentum in Rocket Propulsion==

	~~When~~ a ~~force~~ is ~~applied~~ to ~~a particle~~, ~~its~~ momentum ~~evolves over time according to Newton's second law~~. ~~For more information~~, ~~see [[Newton's Second Law:~~ the ~~Momentum Principle]]~~.		Rockets operate by expelling high-speed exhaust gases backward, which produces a forward momentum change in the rocket. Even in deep space, where there is "nothing to push against", momentum conservation ensures that ejecting mass at high velocity generates thrust. This highlights why rockets require large amounts of fuel: without a constant supply of propellant, the rocket cannot continue changing its momentum.

	~~When an impulse is applied to a particle, its momentum changes in a specific way~~: ~~the change in momentum <math>\Delta\vec{p}</math> is equal to the impulse <math>\vec{J}</math>. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].~~		===A Common Misconception: Newton's Cradle===

	~~When the net external force on a system~~ of ~~particles is 0~~, the ~~system's~~ momentum is conserved ~~(that is~~, ~~constant over time) even if~~ the ~~particles within~~ the ~~system interact with each other (exert internal forces on each other)~~. ~~For more information, see [[Conservation of Momentum]]~~.		Although Newton's cradle is often use to demonstrate conservation of momentum, the device actually relies on both momentum and energy conservation to produce its familiar motion. If momentum alone were conserved, the balls would not swing back with the same amplitude. The cradle highlights how both conservation laws work together in elastic collisions.

	===A Computational Model===		===A Computational Model===

Joshykim at 22:55, 2 December 2025

2025-12-02T22:55:50Z

← Older revision		Revision as of 18:55, 2 December 2025
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	claimed by ~~jonah yonathan~~ fall ~~2024~~		claimed by joshua kim fall 2025

	This page defines the linear momentum of particles and systems.		This page defines the linear momentum of particles and systems.

Jonahy at 17:10, 3 December 2024

2024-12-03T17:10:08Z

@@ Line 137: / Line 137: @@
     v_i = 23.5 m/s   v_f = 3.8 m/s
-    For <math>\Delta p = -17.24 kg m/s </math> (negative because direction is left)
+    p = -17.24 kg m/s (negative because direction is left)
 . Plug in values and solve for m:
     -17.24 = m(3.8-23.5)
-    m =
+    m = -17.24/-19.7
 ===4. (Difficult)===
 A system is comprised of two particles. One particle has a mass of 6kg and is travelling in a direction 30 degrees east of north at a speed of 8m/s. The total momentum of the system is 3kg*m/s west. The second particle has a mass of 3kg. What is the second particle's velocity? Give your answer in terms of a north-south component and an east-west component.
-[[File:Momentumadditionalhardsmaller.jpg]]
+Steps:
 ==Connectedness==

Jonahy at 16:53, 3 December 2024

2024-12-03T16:53:19Z

← Older revision		Revision as of 12:53, 3 December 2024
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	v_i = 23.5 m/s v_f = 3.8 m/s		v_i = 23.5 m/s v_f = 3.8 m/s

	<math>\Delta p ~~</math>~~ = -17.24 kg m/s (negative because direction is left)		For <math>\Delta p = -17.24 kg m/s </math> (negative because direction is left)

Jonahy at 16:52, 3 December 2024

2024-12-03T16:52:16Z

← Older revision		Revision as of 12:52, 3 December 2024
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	v_i = 23.5 m/s v_f = 3.8 m/s		v_i = 23.5 m/s v_f = 3.8 m/s

	<math>\Delta p = -17.24 kg m/s (negative because direction is left)		<math>\Delta p </math> = -17.24 kg m/s (negative because direction is left)
	~~</math>~~

	2. Plug in values and solve for m:		2. Plug in values and solve for m:

Jonahy at 16:51, 3 December 2024

2024-12-03T16:51:23Z

← Older revision		Revision as of 12:51, 3 December 2024
Line 137:		Line 137:
	v_i = 23.5 m/s v_f = 3.8 m/s		v_i = 23.5 m/s v_f = 3.8 m/s

	~~For~~ <math>\Delta p = -17.24 kg m/s (negative because direction is left)		<math>\Delta p = -17.24 kg m/s (negative because direction is left)
			</math>

	2. Plug in values and solve for m:		2. Plug in values and solve for m:

Jonahy at 16:49, 3 December 2024

2024-12-03T16:49:34Z

← Older revision		Revision as of 12:49, 3 December 2024
Line 137:		Line 137:
	v_i = 23.5 m/s v_f = 3.8 m/s		v_i = 23.5 m/s v_f = 3.8 m/s

	<math>\Delta p = -17.24 kg m/s (negative because direction is left)		For <math>\Delta p = -17.24 kg m/s (negative because direction is left)

	2. Plug in values and solve for m:		2. Plug in values and solve for m:

Jonahy at 16:48, 3 December 2024

2024-12-03T16:48:32Z

@@ Line 107: / Line 107: @@
     p_n = 10,000 N
-    In other words, if the velocity was halved, then momentum would correspondingly be halved.
+    In other words, if the velocity was halved, then momentum would
 ===3. (Difficult)===
 You and your friends are watching NBA highlights at home and want to practice your physics. You notice at the beginning of a clip a basketball ball is rolling down the court at 23.5 m/s to the right. At the end, it is rolling at 3.8 m/s in the same direction. The commentator tells you that the change in its momentum is 17.24 kg m/s to the left. Curious at how many basketballs you can carry, you want to find the mass of the ball.
-[[File:momentumhardaf.jpg]]
+Steps:
 ===4. (Difficult)===

Jonahy at 16:34, 3 December 2024

2024-12-03T16:34:17Z

← Older revision		Revision as of 12:34, 3 December 2024
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	1. Initialize Variables		1. Initialize Variables

	m = ~~70kg~~		m = 70 kg

	v = <1,2,3> m/s		v = <1,2,3> m/s
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	2. Momentum Principle		2. Momentum Principle

	p = mv		p = m * v

	= 70 * <1,2,3>		= 70 kg * <1,2,3> m/s

	= <70,140,210>		= <70,140,210> kg m/s

	Therfore, the momentum of the ball is <70,140,210> m/s.		Therfore, the momentum of the ball is <70,140,210> kg m/s.


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	A car has 20,000 N of momentum.		A car has 20,000 N of momentum.
	How would the momentum of the car change if:		How would the momentum of the car change if:

	a) the car slowed to half of its speed?		a) the car slowed to half of its speed?

	b) the car completely stopped?		b) the car completely stopped?

	c) the car gained its original weight in luggage?		c) the car gained its original weight in luggage?

	~~[[File~~:~~momentummiddling~~.~~jpg]]~~
			Steps:

			1. Initialize variables:

			p = 20,000 N
			p_n = ? (new momentum)

			a. Half speed:

			p = m * v = 20,000 N If velocity was halved:

			v = 1/2 v

			p_n = m * 1/2 v

			= 1/2 * m * v

			= since m * v = 20,000 N, then

			p_n = 10,000 N

			In other words, if the velocity was halved, then momentum would correspondingly be halved.

	===3. (Difficult)===		===3. (Difficult)===