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		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46688</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46688"/>
		<updated>2024-12-03T17:10:08Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 70kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize Variables&lt;br /&gt;
&lt;br /&gt;
   m = 70 kg&lt;br /&gt;
&lt;br /&gt;
   v = &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
2. Momentum Principle&lt;br /&gt;
&lt;br /&gt;
   p = m * v&lt;br /&gt;
&lt;br /&gt;
   = 70 kg * &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
   = &amp;lt;70,140,210&amp;gt; kg m/s&lt;br /&gt;
&lt;br /&gt;
Therfore, the momentum of the ball is &amp;lt;70,140,210&amp;gt; kg m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize variables:&lt;br /&gt;
&lt;br /&gt;
   p = 20,000 N&lt;br /&gt;
   p_n = ? (new momentum)&lt;br /&gt;
&lt;br /&gt;
a. Half speed: &lt;br /&gt;
&lt;br /&gt;
   p = m * v = 20,000 N   If velocity was halved:&lt;br /&gt;
&lt;br /&gt;
   v = 1/2 v   &lt;br /&gt;
&lt;br /&gt;
   p_n = m * 1/2 v&lt;br /&gt;
&lt;br /&gt;
   = 1/2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = since m * v = 20,000 N, then&lt;br /&gt;
&lt;br /&gt;
   p_n = 10,000 N&lt;br /&gt;
&lt;br /&gt;
   In other words, if the velocity was halved, then momentum would &lt;br /&gt;
   correspondingly be halved.&lt;br /&gt;
&lt;br /&gt;
b. Car completely stopped:&lt;br /&gt;
&lt;br /&gt;
   If the car was completely stopped, then its velocity would be 0.&lt;br /&gt;
&lt;br /&gt;
   Since momentum is m * v, then the new momentum would be 0 as well.&lt;br /&gt;
&lt;br /&gt;
c. Gained original weight in luggage:&lt;br /&gt;
&lt;br /&gt;
   Double weight = 2m&lt;br /&gt;
&lt;br /&gt;
   p_n = 2m * v&lt;br /&gt;
&lt;br /&gt;
   = 2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = 40,000 N&lt;br /&gt;
&lt;br /&gt;
   The car&#039;s new momentum would be doubled if the mass was doubled.&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
   1. Initialize Variables:&lt;br /&gt;
   &lt;br /&gt;
   v_i = 23.5 m/s   v_f = 3.8 m/s&lt;br /&gt;
&lt;br /&gt;
   p = -17.24 kg m/s (negative because direction is left)&lt;br /&gt;
 &lt;br /&gt;
   2. Plug in values and solve for m:&lt;br /&gt;
&lt;br /&gt;
   -17.24 = m(3.8-23.5)&lt;br /&gt;
&lt;br /&gt;
   m = -17.24/-19.7&lt;br /&gt;
   &lt;br /&gt;
   = 0.8756 kg (approximately)&lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
   1. Initialize Variables:&lt;br /&gt;
&lt;br /&gt;
   m_1 = 6 kg   v_1 = 8 m/s 30 degrees east of north&lt;br /&gt;
&lt;br /&gt;
   m_2 = 3 kg   p_total = 3 kg m/s west&lt;br /&gt;
&lt;br /&gt;
   2. Break Down Velocities into Components for Particle 1:&lt;br /&gt;
&lt;br /&gt;
   North Component: v_1N = v_1 * cos(30) = 6.9282 m/s&lt;br /&gt;
&lt;br /&gt;
   Eastward Component: v_1E = v1 * sin(30) = 4 m/s&lt;br /&gt;
&lt;br /&gt;
   3. Calculate Momentum Components for Particle 1:&lt;br /&gt;
&lt;br /&gt;
   North Momentum: p_1N = m_1 * v_1N = 6 * 6.9282 = 41.5692 kg m/s&lt;br /&gt;
&lt;br /&gt;
   East Momentum: p_1E = m_1 * v_1E = 6 * 4 = 24 kg m/s&lt;br /&gt;
&lt;br /&gt;
   4. Express Total Momentum Components:&lt;br /&gt;
&lt;br /&gt;
   North total momentum:&lt;br /&gt;
   &lt;br /&gt;
   p_total,N = 0 kg m/s (no north-south component in total momentum)&lt;br /&gt;
&lt;br /&gt;
   Eastward total momentum (negative because west is negative east):&lt;br /&gt;
   &lt;br /&gt;
   p_total,E = -3 kg m/s &lt;br /&gt;
&lt;br /&gt;
   5. Conservation of Momentum: &lt;br /&gt;
   &lt;br /&gt;
   p_total,E = p_1E + p_2E&lt;br /&gt;
&lt;br /&gt;
   Particle 2 North Momentum: 0 = 41.5692 + p_2N, p_2N = -41.5692 kg m/s&lt;br /&gt;
&lt;br /&gt;
   Particle 2 East Momentum: -3 = 24 + p_2E, p_2E = -27 kg m/s&lt;br /&gt;
&lt;br /&gt;
   6. Calculate Velocity Components:&lt;br /&gt;
&lt;br /&gt;
   North Velocity: v_2N = p_2N / m_2 = -13.8564 m/s&lt;br /&gt;
&lt;br /&gt;
   East Velocity: v_2E = p_2E / m_2 = -9 m/s&lt;br /&gt;
&lt;br /&gt;
   Therefore, the second particle&#039;s velocity components are:&lt;br /&gt;
&lt;br /&gt;
   13.86 m/s to the South&lt;br /&gt;
&lt;br /&gt;
   9.00 m/s to the West&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46687</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46687"/>
		<updated>2024-12-03T16:53:19Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 70kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize Variables&lt;br /&gt;
&lt;br /&gt;
   m = 70 kg&lt;br /&gt;
&lt;br /&gt;
   v = &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
2. Momentum Principle&lt;br /&gt;
&lt;br /&gt;
   p = m * v&lt;br /&gt;
&lt;br /&gt;
   = 70 kg * &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
   = &amp;lt;70,140,210&amp;gt; kg m/s&lt;br /&gt;
&lt;br /&gt;
Therfore, the momentum of the ball is &amp;lt;70,140,210&amp;gt; kg m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize variables:&lt;br /&gt;
&lt;br /&gt;
   p = 20,000 N&lt;br /&gt;
   p_n = ? (new momentum)&lt;br /&gt;
&lt;br /&gt;
a. Half speed: &lt;br /&gt;
&lt;br /&gt;
   p = m * v = 20,000 N   If velocity was halved:&lt;br /&gt;
&lt;br /&gt;
   v = 1/2 v   &lt;br /&gt;
&lt;br /&gt;
   p_n = m * 1/2 v&lt;br /&gt;
&lt;br /&gt;
   = 1/2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = since m * v = 20,000 N, then&lt;br /&gt;
&lt;br /&gt;
   p_n = 10,000 N&lt;br /&gt;
&lt;br /&gt;
   In other words, if the velocity was halved, then momentum would &lt;br /&gt;
   correspondingly be halved.&lt;br /&gt;
&lt;br /&gt;
b. Car completely stopped:&lt;br /&gt;
&lt;br /&gt;
   If the car was completely stopped, then its velocity would be 0.&lt;br /&gt;
&lt;br /&gt;
   Since momentum is m * v, then the new momentum would be 0 as well.&lt;br /&gt;
&lt;br /&gt;
c. Gained original weight in luggage:&lt;br /&gt;
&lt;br /&gt;
   Double weight = 2m&lt;br /&gt;
&lt;br /&gt;
   p_n = 2m * v&lt;br /&gt;
&lt;br /&gt;
   = 2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = 40,000 N&lt;br /&gt;
&lt;br /&gt;
   The car&#039;s new momentum would be doubled if the mass was doubled.&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
   1. Initialize Variables:&lt;br /&gt;
   &lt;br /&gt;
   v_i = 23.5 m/s   v_f = 3.8 m/s&lt;br /&gt;
&lt;br /&gt;
   For &amp;lt;math&amp;gt;\Delta p = -17.24 kg m/s &amp;lt;/math&amp;gt; (negative because direction is left)&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
   2. Plug in values and solve for m:&lt;br /&gt;
&lt;br /&gt;
   -17.24 = m(3.8-23.5)&lt;br /&gt;
&lt;br /&gt;
   m = &lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46685</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46685"/>
		<updated>2024-12-03T16:52:16Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 70kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize Variables&lt;br /&gt;
&lt;br /&gt;
   m = 70 kg&lt;br /&gt;
&lt;br /&gt;
   v = &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
2. Momentum Principle&lt;br /&gt;
&lt;br /&gt;
   p = m * v&lt;br /&gt;
&lt;br /&gt;
   = 70 kg * &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
   = &amp;lt;70,140,210&amp;gt; kg m/s&lt;br /&gt;
&lt;br /&gt;
Therfore, the momentum of the ball is &amp;lt;70,140,210&amp;gt; kg m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize variables:&lt;br /&gt;
&lt;br /&gt;
   p = 20,000 N&lt;br /&gt;
   p_n = ? (new momentum)&lt;br /&gt;
&lt;br /&gt;
a. Half speed: &lt;br /&gt;
&lt;br /&gt;
   p = m * v = 20,000 N   If velocity was halved:&lt;br /&gt;
&lt;br /&gt;
   v = 1/2 v   &lt;br /&gt;
&lt;br /&gt;
   p_n = m * 1/2 v&lt;br /&gt;
&lt;br /&gt;
   = 1/2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = since m * v = 20,000 N, then&lt;br /&gt;
&lt;br /&gt;
   p_n = 10,000 N&lt;br /&gt;
&lt;br /&gt;
   In other words, if the velocity was halved, then momentum would &lt;br /&gt;
   correspondingly be halved.&lt;br /&gt;
&lt;br /&gt;
b. Car completely stopped:&lt;br /&gt;
&lt;br /&gt;
   If the car was completely stopped, then its velocity would be 0.&lt;br /&gt;
&lt;br /&gt;
   Since momentum is m * v, then the new momentum would be 0 as well.&lt;br /&gt;
&lt;br /&gt;
c. Gained original weight in luggage:&lt;br /&gt;
&lt;br /&gt;
   Double weight = 2m&lt;br /&gt;
&lt;br /&gt;
   p_n = 2m * v&lt;br /&gt;
&lt;br /&gt;
   = 2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = 40,000 N&lt;br /&gt;
&lt;br /&gt;
   The car&#039;s new momentum would be doubled if the mass was doubled.&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
   1. Initialize Variables:&lt;br /&gt;
   &lt;br /&gt;
   v_i = 23.5 m/s   v_f = 3.8 m/s&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\Delta p &amp;lt;/math&amp;gt; = -17.24 kg m/s (negative because direction is left)&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
   2. Plug in values and solve for m:&lt;br /&gt;
&lt;br /&gt;
   -17.24 = m(3.8-23.5)&lt;br /&gt;
&lt;br /&gt;
   m = &lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46684</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46684"/>
		<updated>2024-12-03T16:51:23Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 70kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize Variables&lt;br /&gt;
&lt;br /&gt;
   m = 70 kg&lt;br /&gt;
&lt;br /&gt;
   v = &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
2. Momentum Principle&lt;br /&gt;
&lt;br /&gt;
   p = m * v&lt;br /&gt;
&lt;br /&gt;
   = 70 kg * &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
   = &amp;lt;70,140,210&amp;gt; kg m/s&lt;br /&gt;
&lt;br /&gt;
Therfore, the momentum of the ball is &amp;lt;70,140,210&amp;gt; kg m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize variables:&lt;br /&gt;
&lt;br /&gt;
   p = 20,000 N&lt;br /&gt;
   p_n = ? (new momentum)&lt;br /&gt;
&lt;br /&gt;
a. Half speed: &lt;br /&gt;
&lt;br /&gt;
   p = m * v = 20,000 N   If velocity was halved:&lt;br /&gt;
&lt;br /&gt;
   v = 1/2 v   &lt;br /&gt;
&lt;br /&gt;
   p_n = m * 1/2 v&lt;br /&gt;
&lt;br /&gt;
   = 1/2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = since m * v = 20,000 N, then&lt;br /&gt;
&lt;br /&gt;
   p_n = 10,000 N&lt;br /&gt;
&lt;br /&gt;
   In other words, if the velocity was halved, then momentum would &lt;br /&gt;
   correspondingly be halved.&lt;br /&gt;
&lt;br /&gt;
b. Car completely stopped:&lt;br /&gt;
&lt;br /&gt;
   If the car was completely stopped, then its velocity would be 0.&lt;br /&gt;
&lt;br /&gt;
   Since momentum is m * v, then the new momentum would be 0 as well.&lt;br /&gt;
&lt;br /&gt;
c. Gained original weight in luggage:&lt;br /&gt;
&lt;br /&gt;
   Double weight = 2m&lt;br /&gt;
&lt;br /&gt;
   p_n = 2m * v&lt;br /&gt;
&lt;br /&gt;
   = 2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = 40,000 N&lt;br /&gt;
&lt;br /&gt;
   The car&#039;s new momentum would be doubled if the mass was doubled.&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
   1. Initialize Variables:&lt;br /&gt;
   &lt;br /&gt;
   v_i = 23.5 m/s   v_f = 3.8 m/s&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\Delta p = -17.24 kg m/s (negative because direction is left)&lt;br /&gt;
   &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   2. Plug in values and solve for m:&lt;br /&gt;
&lt;br /&gt;
   -17.24 = m(3.8-23.5)&lt;br /&gt;
&lt;br /&gt;
   m = &lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46683</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46683"/>
		<updated>2024-12-03T16:49:34Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 70kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize Variables&lt;br /&gt;
&lt;br /&gt;
   m = 70 kg&lt;br /&gt;
&lt;br /&gt;
   v = &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
2. Momentum Principle&lt;br /&gt;
&lt;br /&gt;
   p = m * v&lt;br /&gt;
&lt;br /&gt;
   = 70 kg * &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
   = &amp;lt;70,140,210&amp;gt; kg m/s&lt;br /&gt;
&lt;br /&gt;
Therfore, the momentum of the ball is &amp;lt;70,140,210&amp;gt; kg m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize variables:&lt;br /&gt;
&lt;br /&gt;
   p = 20,000 N&lt;br /&gt;
   p_n = ? (new momentum)&lt;br /&gt;
&lt;br /&gt;
a. Half speed: &lt;br /&gt;
&lt;br /&gt;
   p = m * v = 20,000 N   If velocity was halved:&lt;br /&gt;
&lt;br /&gt;
   v = 1/2 v   &lt;br /&gt;
&lt;br /&gt;
   p_n = m * 1/2 v&lt;br /&gt;
&lt;br /&gt;
   = 1/2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = since m * v = 20,000 N, then&lt;br /&gt;
&lt;br /&gt;
   p_n = 10,000 N&lt;br /&gt;
&lt;br /&gt;
   In other words, if the velocity was halved, then momentum would &lt;br /&gt;
   correspondingly be halved.&lt;br /&gt;
&lt;br /&gt;
b. Car completely stopped:&lt;br /&gt;
&lt;br /&gt;
   If the car was completely stopped, then its velocity would be 0.&lt;br /&gt;
&lt;br /&gt;
   Since momentum is m * v, then the new momentum would be 0 as well.&lt;br /&gt;
&lt;br /&gt;
c. Gained original weight in luggage:&lt;br /&gt;
&lt;br /&gt;
   Double weight = 2m&lt;br /&gt;
&lt;br /&gt;
   p_n = 2m * v&lt;br /&gt;
&lt;br /&gt;
   = 2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = 40,000 N&lt;br /&gt;
&lt;br /&gt;
   The car&#039;s new momentum would be doubled if the mass was doubled.&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
   1. Initialize Variables:&lt;br /&gt;
   &lt;br /&gt;
   v_i = 23.5 m/s   v_f = 3.8 m/s&lt;br /&gt;
&lt;br /&gt;
   For &amp;lt;math&amp;gt;\Delta p = -17.24 kg m/s (negative because direction is left)&lt;br /&gt;
&lt;br /&gt;
   2. Plug in values and solve for m:&lt;br /&gt;
&lt;br /&gt;
   -17.24 = m(3.8-23.5)&lt;br /&gt;
&lt;br /&gt;
   m = &lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46682</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46682"/>
		<updated>2024-12-03T16:48:32Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 70kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize Variables&lt;br /&gt;
&lt;br /&gt;
   m = 70 kg&lt;br /&gt;
&lt;br /&gt;
   v = &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
2. Momentum Principle&lt;br /&gt;
&lt;br /&gt;
   p = m * v&lt;br /&gt;
&lt;br /&gt;
   = 70 kg * &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
   = &amp;lt;70,140,210&amp;gt; kg m/s&lt;br /&gt;
&lt;br /&gt;
Therfore, the momentum of the ball is &amp;lt;70,140,210&amp;gt; kg m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize variables:&lt;br /&gt;
&lt;br /&gt;
   p = 20,000 N&lt;br /&gt;
   p_n = ? (new momentum)&lt;br /&gt;
&lt;br /&gt;
a. Half speed: &lt;br /&gt;
&lt;br /&gt;
   p = m * v = 20,000 N   If velocity was halved:&lt;br /&gt;
&lt;br /&gt;
   v = 1/2 v   &lt;br /&gt;
&lt;br /&gt;
   p_n = m * 1/2 v&lt;br /&gt;
&lt;br /&gt;
   = 1/2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = since m * v = 20,000 N, then&lt;br /&gt;
&lt;br /&gt;
   p_n = 10,000 N&lt;br /&gt;
&lt;br /&gt;
   In other words, if the velocity was halved, then momentum would &lt;br /&gt;
   correspondingly be halved.&lt;br /&gt;
&lt;br /&gt;
b. Car completely stopped:&lt;br /&gt;
&lt;br /&gt;
   If the car was completely stopped, then its velocity would be 0.&lt;br /&gt;
&lt;br /&gt;
   Since momentum is m * v, then the new momentum would be 0 as well.&lt;br /&gt;
&lt;br /&gt;
c. Gained original weight in luggage:&lt;br /&gt;
&lt;br /&gt;
   Double weight = 2m&lt;br /&gt;
&lt;br /&gt;
   p_n = 2m * v&lt;br /&gt;
&lt;br /&gt;
   = 2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = 40,000 N&lt;br /&gt;
&lt;br /&gt;
   The car&#039;s new momentum would be doubled if the mass was doubled.&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
   1. Initialize Variables:&lt;br /&gt;
   &lt;br /&gt;
   v_i = 23.5 m/s   v_f = 3.8 m/s&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\Delta p = -17.24 kg m/s (negative because direction is left)&lt;br /&gt;
&lt;br /&gt;
   2. Plug in values and solve for m:&lt;br /&gt;
&lt;br /&gt;
   -17.24 = m(3.8-23.5)&lt;br /&gt;
&lt;br /&gt;
   m = &lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46680</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46680"/>
		<updated>2024-12-03T16:34:17Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 70kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize Variables&lt;br /&gt;
&lt;br /&gt;
   m = 70 kg&lt;br /&gt;
&lt;br /&gt;
   v = &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
2. Momentum Principle&lt;br /&gt;
&lt;br /&gt;
   p = m * v&lt;br /&gt;
&lt;br /&gt;
   = 70 kg * &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
   = &amp;lt;70,140,210&amp;gt; kg m/s&lt;br /&gt;
&lt;br /&gt;
Therfore, the momentum of the ball is &amp;lt;70,140,210&amp;gt; kg m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize variables:&lt;br /&gt;
&lt;br /&gt;
   p = 20,000 N&lt;br /&gt;
   p_n = ? (new momentum)&lt;br /&gt;
&lt;br /&gt;
a. Half speed: &lt;br /&gt;
&lt;br /&gt;
   p = m * v = 20,000 N   If velocity was halved:&lt;br /&gt;
&lt;br /&gt;
   v = 1/2 v   &lt;br /&gt;
&lt;br /&gt;
   p_n = m * 1/2 v&lt;br /&gt;
&lt;br /&gt;
   = 1/2 * m * v&lt;br /&gt;
&lt;br /&gt;
   = since m * v = 20,000 N, then&lt;br /&gt;
&lt;br /&gt;
   p_n = 10,000 N&lt;br /&gt;
&lt;br /&gt;
   In other words, if the velocity was halved, then momentum would correspondingly be halved.&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
[[File:momentumhardaf.jpg]]&lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46672</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46672"/>
		<updated>2024-12-03T05:52:31Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 70kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize Variables&lt;br /&gt;
&lt;br /&gt;
   m = 70kg&lt;br /&gt;
&lt;br /&gt;
   v = &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
2. Momentum Principle&lt;br /&gt;
&lt;br /&gt;
   p = mv&lt;br /&gt;
&lt;br /&gt;
   = 70 * &amp;lt;1,2,3&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   = &amp;lt;70,140,210&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therfore, the momentum of the ball is &amp;lt;70,140,210&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
[[File:momentummiddling.jpg]]&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
[[File:momentumhardaf.jpg]]&lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46671</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46671"/>
		<updated>2024-12-03T05:50:43Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 70kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Initialize Variables&lt;br /&gt;
&lt;br /&gt;
m = 70kg&lt;br /&gt;
&lt;br /&gt;
v = &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
2. Momentum Principle&lt;br /&gt;
&lt;br /&gt;
p = mv&lt;br /&gt;
&lt;br /&gt;
= 70 * &amp;lt;1,2,3&amp;gt;&lt;br /&gt;
&lt;br /&gt;
= &amp;lt;70,140,210&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
[[File:momentummiddling.jpg]]&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
[[File:momentumhardaf.jpg]]&lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46670</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46670"/>
		<updated>2024-12-03T05:49:45Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 70kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
1. Initialize Variables&lt;br /&gt;
m = 70kg&lt;br /&gt;
v = &amp;lt;1,2,3&amp;gt; m/s&lt;br /&gt;
&lt;br /&gt;
2. Momentum Principle&lt;br /&gt;
p = mv&lt;br /&gt;
= 70 * &amp;lt;1,2,3&amp;gt;&lt;br /&gt;
= &amp;lt;70,140,210&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
[[File:momentummiddling.jpg]]&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
[[File:momentumhardaf.jpg]]&lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46669</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46669"/>
		<updated>2024-12-03T05:38:17Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 69kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
[[File:momentumsimple.jpg]]&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
[[File:momentummiddling.jpg]]&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
[[File:momentumhardaf.jpg]]&lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46668</id>
		<title>Linear Momentum</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Linear_Momentum&amp;diff=46668"/>
		<updated>2024-12-03T05:38:03Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by jonah yonathan fall 2024&lt;br /&gt;
This page defines the linear momentum of particles and systems.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Linear momentum is a vector quantity describing an object&#039;s motion. It is defined as the product of an object&#039;s [[Mass]] (&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;) and [[Velocity]] (&amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt;). Note that mass is a scalar while velocity is a vector, so an object&#039;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 &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; and is often referred to as simply &amp;quot;momentum.&amp;quot; The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
====Single Particles====&lt;br /&gt;
The momentum of a particle is defined as follows:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m\vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is the particle&#039;s linear momentum, &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is the particle&#039;s mass, and &amp;lt;math&amp;gt;\vec{v}&amp;lt;/math&amp;gt; is the particle&#039;s velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.&lt;br /&gt;
&lt;br /&gt;
====Multiple Particles====&lt;br /&gt;
&lt;br /&gt;
The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = \sum_i \vec{p}_i&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{system} = M_{tot}\vec{v}_{COM}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small &amp;quot;mass elements,&amp;quot; all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.&lt;br /&gt;
&lt;br /&gt;
====Impulse====&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
====In Relation to Other Physics Topics====&lt;br /&gt;
&lt;br /&gt;
In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton&#039;s first law.&lt;br /&gt;
&lt;br /&gt;
When a force is applied to a particle, its momentum evolves over time according to Newton&#039;s second law. For more information, see [[Newton&#039;s Second Law: the Momentum Principle]].&lt;br /&gt;
&lt;br /&gt;
When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum &amp;lt;math&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt; is equal to the impulse &amp;lt;math&amp;gt;\vec{J}&amp;lt;/math&amp;gt;. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].&lt;br /&gt;
&lt;br /&gt;
When the net external force on a system of particles is 0, the system&#039;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]].&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in &amp;quot;time steps,&amp;quot; or iterations of a loop representing a short time interval. In each time step, the particles&#039; momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle&#039;s momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:&lt;br /&gt;
&lt;br /&gt;
This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/ce43925647&lt;br /&gt;
&lt;br /&gt;
For more information, see [[Iterative Prediction]].&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===1. (Simple)===&lt;br /&gt;
Find the momentum of a ball that has a mass of 69kg and is moving at &amp;lt;1,2,3&amp;gt; m/s.&lt;br /&gt;
&lt;br /&gt;
[[File:momentumsimple.jpg]]&lt;br /&gt;
&lt;br /&gt;
===2. (Middling)===&lt;br /&gt;
A car has 20,000 N of momentum.&lt;br /&gt;
How would the momentum of the car change if:&lt;br /&gt;
a) the car slowed to half of its speed?&lt;br /&gt;
b) the car completely stopped?&lt;br /&gt;
c) the car gained its original weight in luggage?&lt;br /&gt;
&lt;br /&gt;
[[File:momentummiddling.jpg]]&lt;br /&gt;
&lt;br /&gt;
===3. (Difficult)===&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
[[File:momentumhardaf.jpg]]&lt;br /&gt;
&lt;br /&gt;
===4. (Difficult)===&lt;br /&gt;
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&#039;s velocity? Give your answer in terms of a north-south component and an east-west component.&lt;br /&gt;
&lt;br /&gt;
[[File:Momentumadditionalhardsmaller.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: runaway vehicle===&lt;br /&gt;
&lt;br /&gt;
Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The oldest known attempt at quantifying motion using both an object&#039;s speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Mass]]&lt;br /&gt;
*[[Velocity]]&lt;br /&gt;
*[[Vectors]]&lt;br /&gt;
*[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
*[[Impulse and Momentum]]&lt;br /&gt;
*[[Conservation of Momentum]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ol&amp;gt;&lt;br /&gt;
&amp;lt;li&amp;gt;https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&amp;lt;/li&amp;gt;&lt;br /&gt;
&amp;lt;/ol&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Momentum]]&lt;br /&gt;
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html&lt;br /&gt;
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html&lt;br /&gt;
*https://science.jrank.org/pages/4419/Momentum.html&lt;br /&gt;
*https://www.famousscientists.org/john-wallis/&lt;br /&gt;
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Collisions&amp;diff=46667</id>
		<title>Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Collisions&amp;diff=46667"/>
		<updated>2024-12-03T05:35:38Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;A &#039;&#039;&#039;collision&#039;&#039;&#039; is the act in which two or more entities exert forces on each other over a time period, often expressed in a short period. Collisions behave according to the fundamental principles of physics. The nature of collisions allow for assumptions, as discussed later, to solve for unknowns. &lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
If we take the colliding objects as the system, what assumption can we make&lt;br /&gt;
about the total momentum and energy of this system?&lt;br /&gt;
&lt;br /&gt;
In our definition of a collision, we&#039;ve said that interactions preceding&lt;br /&gt;
and succeeding the process are tiny compared to the ones we see during the&lt;br /&gt;
process. The forces that contribute to this sudden spike of interactions is&lt;br /&gt;
internal to the system: they are between the colliding objects. Since we take&lt;br /&gt;
them as the system, we can make the assumption that the outside forces that act on the&lt;br /&gt;
system are negligible.&lt;br /&gt;
&lt;br /&gt;
By this assumption, the change in the total momentum and energy of the system&lt;br /&gt;
(of both the colliding objects) is negligible.&lt;br /&gt;
&lt;br /&gt;
For &amp;lt;math&amp;gt;\Delta p = F * \Delta t&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;F&amp;lt;/math&amp;gt; is negligible, thus&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta p&amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
For &amp;lt;math&amp;gt;\Delta E = W_{surr} + Q&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;W_{surr} = F_{surr} * \Delta x&amp;lt;/math&amp;gt;&lt;br /&gt;
and &amp;lt;math&amp;gt;Q = m * C * \Delta t&amp;lt;/math&amp;gt;. &amp;lt;math&amp;gt;F_{surr}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\Delta t&amp;lt;/math&amp;gt;&lt;br /&gt;
are negligible. Thus &amp;lt;math&amp;gt;\Delta E&amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
Therefore, we can assume that the total momentum and energy of the system (of both the colliding objects) don&#039;t change during the collision. Furthermore, the assumptions of conservation of momentum and conservation of kinetic energy allow us to find the final velocities of each entity in a two-body collision.&lt;br /&gt;
&lt;br /&gt;
Example: Swinging Masses &lt;br /&gt;
&lt;br /&gt;
The conservation of momentum and conservation of energy can be identified in a system which features steel spherical masses that hang side by side in a horizontal line from each other. If a single mass is pulled to the side and released, it follows in the opposing motion and strikes the line. The outer mass swings in the line of movement. Similarly, if two balls are pulled to the side and released, the other two balls swing out in the line of movement. &lt;br /&gt;
&lt;br /&gt;
     p_initial = p_final&lt;br /&gt;
     KE_initial = KE_final&lt;br /&gt;
&lt;br /&gt;
     &#039;&#039;&#039;Case 1:&#039;&#039;&#039;&lt;br /&gt;
     momentum in = momentum out &lt;br /&gt;
     mv = momentum out&lt;br /&gt;
     Kinetic Energy in = Kinetic Energy out&lt;br /&gt;
     (1/2)*mv^2 = kinetic energy out&lt;br /&gt;
    &lt;br /&gt;
     &#039;&#039;&#039;Case 2:&#039;&#039;&#039;&lt;br /&gt;
     momentum in = momentum out &lt;br /&gt;
     2mv = momentum out &lt;br /&gt;
     Kinetic Energy in = Kinetic Energy out&lt;br /&gt;
     (1/2)*2*mv^2 = kinetic energy out &lt;br /&gt;
&lt;br /&gt;
Both the conservation of momentum and conservation of energy are respected in both cases. The conservation of momentum and conservation of energy allow for the understanding of the relationships between internal and external forces. Realistically, there are very few perfectly elastic collisions. Often, kinetic energy is converted into internal energy, or dissipates as sound or heat energy. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Types of Collisions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In [[Elastic Collisions]], the total kinetic energy is conserved. None of the&lt;br /&gt;
kinetic energy of the system is transformed from or into &amp;lt;math&amp;gt;E_{internal}&amp;lt;/math&amp;gt;,&lt;br /&gt;
such as thermal energy and spring potential energy.&lt;br /&gt;
     &lt;br /&gt;
    Elastic collisions are regarded as one-dimensional collisions where the impact is along the line of movement. &lt;br /&gt;
&lt;br /&gt;
An example for such a process is the Newton&#039;s Cradle, where the kinetic energy&lt;br /&gt;
of the colliding ball gets transformed into the kinetic energy of the last&lt;br /&gt;
ball in the cradle. The total kinetic energy is conserved.&lt;br /&gt;
&lt;br /&gt;
In [[Inelastic Collisions]], the total kinetic energy is not conserved. Some/all&lt;br /&gt;
of the kinetic energy of the system can be transformed from or into&lt;br /&gt;
&amp;lt;math&amp;gt;E_{internal}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
    Inelastic collisions are regarded as two-dimensional collisions where the impact is not along the line of movement.&lt;br /&gt;
&lt;br /&gt;
An example for such a process is a car crash, where the kinetic energy of the&lt;br /&gt;
colliding car gets transformed into thermal energy and internal energy contained&lt;br /&gt;
in the crumbled, tense parts of the car chassis.&lt;br /&gt;
&lt;br /&gt;
([[Inelastic Collisions]] where there&#039;s maximum kinetic energy dissipation&lt;br /&gt;
are called [[Maximally Inelastic Collision]]. This doesn&#039;t necessarily mean&lt;br /&gt;
that &#039;&#039;all&#039;&#039; of the kinetic energy gets transformed into &amp;lt;math&amp;gt;E_{internal}&amp;lt;/math&amp;gt;,&lt;br /&gt;
since the total momentum is conserved. Collisions where the objects stick together are&lt;br /&gt;
[[Maximally Inelastic Collision]].)&lt;br /&gt;
&lt;br /&gt;
===Mathematical model===&lt;br /&gt;
By our assumptions, &amp;lt;math&amp;gt;p_f = p_i&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;E_f = E_i&amp;lt;/math&amp;gt;, &lt;br /&gt;
where subscripts &amp;lt;math&amp;gt;f&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;i&amp;lt;/math&amp;gt; mean final and initial&lt;br /&gt;
respectively. These equations hold true for all collisions.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision, &amp;lt;math&amp;gt;K_{total, f} = K_{total, i}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
In an inelastic collision, this equation doesn&#039;t hold, since&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta E = \Delta K + \Delta E_{internal} = 0&amp;lt;/math&amp;gt;, and&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta E_{internal}&amp;lt;/math&amp;gt; might be nonzero.&lt;br /&gt;
&lt;br /&gt;
===Computational model===&lt;br /&gt;
In this model, two objects of different masses are moving towards each other&lt;br /&gt;
with equal magnitude, but opposite direction momenta.&lt;br /&gt;
&lt;br /&gt;
The objects collide, and after the collision, it is observed that the objects&lt;br /&gt;
are embedded within each other with final velocity zero.&lt;br /&gt;
&lt;br /&gt;
This is an example of a [[Maximally Inelastic Collision]]: none of the kinetic&lt;br /&gt;
energy is conserved.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/e600a7595e Computational Model]&lt;br /&gt;
&lt;br /&gt;
Note: The behavior of the colliding objects are hardcoded in the model to&lt;br /&gt;
embed (vs. bouncing, etc.) This is an assumption we make about the&lt;br /&gt;
materials/internal structure of the balls. Correctly modeling deforming&lt;br /&gt;
objects without simplifying assumptions is a very complex task.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A &amp;lt;math&amp;gt;0.5 kg&amp;lt;/math&amp;gt; soccer ball is moving with a speed of &amp;lt;math&amp;gt;5 m/s&amp;lt;/math&amp;gt;&lt;br /&gt;
directly towards a &amp;lt;math&amp;gt;0.7 kg&amp;lt;/math&amp;gt; basketball, which is at rest.&lt;br /&gt;
&lt;br /&gt;
The two balls collide and stick together. What will be their final speed?&lt;br /&gt;
&lt;br /&gt;
[[File:Collision-example-1-mert.png|500px]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;m_1 = 0.5 kg&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;v_1 = 5 m/s&amp;lt;/math&amp;gt;,&lt;br /&gt;
&amp;lt;math&amp;gt;m_2 = 0.7 kg&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;v_2 = 0 m/s&amp;lt;/math&amp;gt;, where&lt;br /&gt;
subscripts &amp;lt;math&amp;gt;1&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;2&amp;lt;/math&amp;gt; signify the soccer ball and basketball respectively.&lt;br /&gt;
&lt;br /&gt;
We&#039;re going to compute &amp;lt;math&amp;gt;v_f&amp;lt;/math&amp;gt; for the collided mass. &lt;br /&gt;
&lt;br /&gt;
We know that momentum is conserved. &amp;lt;math&amp;gt;p_f = p_i \implies p_f = p_1 + p_2&lt;br /&gt;
\implies p_f = (m_1 * v_1) + (m_2 * v_2)&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
We know that the objects stick together, hence their mass &amp;lt;math&amp;gt;m_1 + m_2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Then &amp;lt;math&amp;gt;p_f = (m_1 + m_2) * v_f \implies (m_1 + m_2) * v_f&lt;br /&gt;
= (m_1 * v_1) + (m_2 * v_2)&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If we solve for &amp;lt;math&amp;gt;v_f&amp;lt;/math&amp;gt;, we get&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
    v_f &amp;amp;= \frac{(m_1 * v_1) + (m_2 * v_2)}{m_1 + m_2} \\&lt;br /&gt;
        &amp;amp;= \frac{(0.5 * 5) +(0.7 * 0)}{0.5 + 0.7} \\&lt;br /&gt;
        &amp;amp;= 2.083 m/s&lt;br /&gt;
\end{aligned}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This is a [[Maximally Inelastic Collision]], as the collided objects stuck&lt;br /&gt;
together. &amp;lt;math&amp;gt;K_{tot, f} &amp;lt; K_{tot, i}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A &amp;lt;math&amp;gt;10 kg&amp;lt;/math&amp;gt; asteroid with a velocity of &amp;lt;math&amp;gt;&amp;lt;10, 0, 0&amp;gt; m/s&amp;lt;/math&amp;gt;&lt;br /&gt;
crashes into the Earth in the Sahara Desert and bounces back into the space with&lt;br /&gt;
the same speed and opposite direction.&lt;br /&gt;
&lt;br /&gt;
Nobody is hurt, but some say that they felt the Earth recoil during&lt;br /&gt;
this unusual collision. Find the recoil velocity of the earth to see if it was&lt;br /&gt;
significant enough for people to have felt it. (Keep in mind that the human&lt;br /&gt;
sensory facilities aren&#039;t very precise.)&lt;br /&gt;
&lt;br /&gt;
[[File:Collision-example-2-mert.png|500px]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;m_a = 10 kg&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;m_e = 6 * 10^{24} kg&amp;lt;/math&amp;gt;,&lt;br /&gt;
&amp;lt;math&amp;gt;v_{a, i} = &amp;lt;10, 0, 0&amp;gt; m/s&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;v_{a, f} = &amp;lt;-10, 0, 0&amp;gt; m/s&amp;lt;/math&amp;gt;,&lt;br /&gt;
and &amp;lt;math&amp;gt;v_{e, i} = &amp;lt;0, 0, 0&amp;gt; m/s&amp;lt;/math&amp;gt;, where subscripts &amp;lt;math&amp;gt;a&amp;lt;/math&amp;gt;&lt;br /&gt;
and &amp;lt;math&amp;gt;e&amp;lt;/math&amp;gt; signify the asteroid and the earth respectively.&lt;br /&gt;
&lt;br /&gt;
We&#039;re going to compute &amp;lt;math&amp;gt;v_{e, f}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know that momentum is conserved.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
    &amp;amp;p_f = p_i \\&lt;br /&gt;
    \implies &amp;amp;p_{a, f} + p_{e, f} = p_{a, i} + p_{e, i} \\&lt;br /&gt;
    \implies &amp;amp;(m_a * v_{a, f}) + (m_e * v_{e, f})&lt;br /&gt;
    = (m_a * v_{a, i}) + (m_e * v_{e, i})&lt;br /&gt;
\end{aligned}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Solving for &amp;lt;math&amp;gt;v_{e, f}&amp;lt;/math&amp;gt;, we get&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
    v_{e, f}&lt;br /&gt;
        &amp;amp;= \frac{(m_a * v_{a, i}) + (m_e * v_{e, i}) - (m_a * v_{a, f})}{m_e} \\&lt;br /&gt;
        &amp;amp;= \frac{(10 * &amp;lt;10, 0, 0&amp;gt;) + (6 * 10^{24} * &amp;lt;0, 0, 0&amp;gt;) &lt;br /&gt;
            - (10 * &amp;lt;-10, 0, 0&amp;gt;)}{6 * 10^{24}} \\&lt;br /&gt;
        &amp;amp;= \frac{&amp;lt;100, 0, 0&amp;gt; + &amp;lt;100, 0, 0&amp;gt;}{6 * 10^{24}} \\&lt;br /&gt;
        &amp;amp;= &amp;lt;3.33 * 10^{-23}, 0, 0&amp;gt; m/s&lt;br /&gt;
\end{aligned}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The recoil velocity of the earth is too small for anyone to feel it.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two rocks in space are approaching a third, stationary rock. Rock 1 has a mass&lt;br /&gt;
of &amp;lt;math&amp;gt;30 kg&amp;lt;/math&amp;gt; and moves with a velocity &amp;lt;math&amp;gt;&amp;lt;13, -10, -2&amp;gt; m/s&amp;lt;/math&amp;gt;.&lt;br /&gt;
Rock 2 has a mass of &amp;lt;math&amp;gt;60 kg&amp;lt;/math&amp;gt;, and moves with a velocity&lt;br /&gt;
&amp;lt;math&amp;gt;&amp;lt;-3, 23, 3&amp;gt; m/s&amp;lt;/math&amp;gt;. Rock 3 has a mass of &amp;lt;math&amp;gt;35 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
and is stationary. The rocks collide at Rock 3&#039;s location.&lt;br /&gt;
&lt;br /&gt;
[[File:Collision-example-3a-mert.png|500px]]&lt;br /&gt;
&lt;br /&gt;
After the collision, it is observed that Rock 1 has embedded&lt;br /&gt;
into Rock 3, which is now moving with a new velocity of &amp;lt;math&amp;gt;&amp;lt;3, 5, 2&amp;gt; m/s&amp;lt;/math&amp;gt;,&lt;br /&gt;
and that a chunk of mass &amp;lt;math&amp;gt;15 kg&amp;lt;/math&amp;gt; has broken off of the combined rock.&lt;br /&gt;
&lt;br /&gt;
Rock 2, in the meantime, has a new velocity of &amp;lt;math&amp;gt;&amp;lt;-2, 12, 2&amp;gt; m/s&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Collision-example-3b-mert.png|500px]]&lt;br /&gt;
&lt;br /&gt;
What is the change in the internal energy &amp;lt;math&amp;gt;\Delta E_{internal}&amp;lt;/math&amp;gt; in&lt;br /&gt;
the system as a result of this three-way collision?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Due to the complexity of this question, we are going to use a computer to&lt;br /&gt;
solve this problem. Conceptually, everything remains the same, but instead of&lt;br /&gt;
computing things by hand, we&#039;re getting help from a computer. Feel free to&lt;br /&gt;
solve it by hand if you want to!&lt;br /&gt;
&lt;br /&gt;
    # Define the initial masses/velocities of the rocks.&lt;br /&gt;
    mass_1_i = 30&lt;br /&gt;
    mass_2_i = 60&lt;br /&gt;
    mass_3_i = 35&lt;br /&gt;
    &lt;br /&gt;
    v_1_i = vector(13, -10, -2)&lt;br /&gt;
    v_2_i = vector(-3, 23, 3)&lt;br /&gt;
    v_3_i = vector(0, 0, 0)&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the initial total momentum.&lt;br /&gt;
    total_p_i = (mass_1_i * v_1_i) + (mass_2_i * v_2_i) + (mass_3_i * v_3_i)&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the masses/velocities of the rocks after the collision.&lt;br /&gt;
    mass_chunk = 15&lt;br /&gt;
    &lt;br /&gt;
    mass_2_f = 60&lt;br /&gt;
    mass_3_f = mass_1_i + mass_3_i - mass_chunk&lt;br /&gt;
    &lt;br /&gt;
    v_2_f = vector(-2, 12, 2)&lt;br /&gt;
    v_3_f = vector(3, 5, 2)&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the velocity of the chunk. Remember that the momentum is conserved.&lt;br /&gt;
    v_chunk = (total_p_i - ((v_2_f * mass_2_f) + (v_3_f * mass_3_f))) / mass_chunk&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the initial total kinetic energy.&lt;br /&gt;
    def kinetic_energy(mass, velocity):&lt;br /&gt;
        return 1/2 * mass * mag(velocity) ** 2&lt;br /&gt;
    &lt;br /&gt;
    k_tot_i = kinetic_energy(mass_1_i, v_1_i) \&lt;br /&gt;
            + kinetic_energy(mass_2_i, v_2_i) \&lt;br /&gt;
            + kinetic_energy(mass_3_i, v_3_i)&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the final total kinetic energy.&lt;br /&gt;
    k_tot_f = kinetic_energy(mass_2_f, v_2_f) \&lt;br /&gt;
            + kinetic_energy(mass_3_f, v_3_f) \&lt;br /&gt;
            + kinetic_energy(mass_chunk, v_chunk)&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the change in internal energy.&lt;br /&gt;
    d_e_internal = k_tot_f - k_tot_i&lt;br /&gt;
&lt;br /&gt;
This Python snippet, when evaluated, will show that&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta E_{internal} = 5175 J&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
Collisions are everywhere in our daily lives. Some examples are: a bat colliding with a baseball in your&lt;br /&gt;
neighbor&#039;s backyard, two cars crashing in the highway in front of you, your finger hitting the keyboard, and so&lt;br /&gt;
on.&lt;br /&gt;
&lt;br /&gt;
Depending on the nature of the collision we can usually guess whether it&#039;s&lt;br /&gt;
an [[Elastic Collisions]] or an [[Inelastic Collisions]], and reason about it.&lt;br /&gt;
(If it&#039;s a car crash, things crumble, so there&#039;s increase in thermal energy, etc.)&lt;br /&gt;
&lt;br /&gt;
Implications of collisions may be included in the understanding of the ideal gas law. As temperatures rise within a given sector, the movement of molecules become excited. Molecules of ideal gases will move with perfect elasticity in relation to one another. In other words, the kinetic energy before the initial collision will remain the same as the final kinetic energy after the collision. However, most gases do not behave ideally. More accurately, moments of kinetic energy may dissipate or be absorbed as heat energy and sound energy, respectively. &lt;br /&gt;
&lt;br /&gt;
[[File:Asteroid 2009 VA.png|thumb|right|Asteroid 2009 VA]]Another very interesting application of collisions is in the field of astronomy.&lt;br /&gt;
The universe is ever-changing and we never know what might come our way.&lt;br /&gt;
A large meteor could be flying at the Earth with a very high speed and we could&lt;br /&gt;
use our knowledge of collisions to find out how the Earth will move after the&lt;br /&gt;
collision, what the temperature change of the area would be, and how the rest&lt;br /&gt;
of the Earth would be effected by the collision. By using our collision skills,&lt;br /&gt;
we could evacuate anyone who is prone to harm during the collision.&lt;br /&gt;
&lt;br /&gt;
Since collisions are ubiquitous processes, knowledge about them is instrumental&lt;br /&gt;
for any engineer who deals with the physical world.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
We&#039;ve mentioned that collisions are not special case events in Physics--&lt;br /&gt;
the fundamental principles apply fully for them, and we reason about them&lt;br /&gt;
using these principles. Then it won&#039;t surprise us that the history of collisions&lt;br /&gt;
is tangled with the history of the fundamental principles.&lt;br /&gt;
&lt;br /&gt;
In 1661, Christiaan Huygens, a Dutch physicist, concluded that when a moving&lt;br /&gt;
body struck a stationary object of equal mass, the initial moving body would&lt;br /&gt;
lose all of its momentum while the second object would pick up the same amount&lt;br /&gt;
of velocity that the moving body had before the collision. He also stated that&lt;br /&gt;
the total &amp;quot;quantity of motion&amp;quot; should be the same before and after the&lt;br /&gt;
collision. This was the first thought of the conservation of momentum.&lt;br /&gt;
&lt;br /&gt;
In 1687, Sir Isaac Newton went a little further and stated in his third law of&lt;br /&gt;
motion that the forces of action and reaction between two bodies are opposite&lt;br /&gt;
and equal. Because of this law; we know that the total momentum of a closed system is constant.&lt;br /&gt;
Every force in the system has a reciprocal pair with an opposing direction, which&lt;br /&gt;
implies that their momenta cancel out, keeping the total momentum constant.&lt;br /&gt;
&lt;br /&gt;
In Ernest Rutherford&#039;s famous Gold Foil Experiment of 1899, he found out that the plum&lt;br /&gt;
pudding model of the atom was inaccurate, and that there was a concentrated&lt;br /&gt;
core at the center of the atom (nucleus) and a lot of empty space around.&lt;br /&gt;
He didn&#039;t have access to the sophisticated tools/knowledge that we have now to&lt;br /&gt;
assist in his discovery: he did this through investigating atomic collisions&lt;br /&gt;
between alpha particles and gold foil. He shot alpha particles to a piece of&lt;br /&gt;
thin gold foil, and observed that some of the particles were scattered&lt;br /&gt;
through angles larger than 90 degrees during the collisions. Using collision&lt;br /&gt;
principles, he was then able to reason that the atom had most of its mass&lt;br /&gt;
concentrated in a dense core, rather than spread out evenly throughout the atom&lt;br /&gt;
as the plum pudding model suggested.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
Other pages related to collisions:&lt;br /&gt;
*[[Maximally Inelastic Collision]]&lt;br /&gt;
*[[Inelastic Collisions]]&lt;br /&gt;
*[[Elastic Collisions]]&lt;br /&gt;
*[[Linear Momentum]]&lt;br /&gt;
*[[Energy Principle]],&lt;br /&gt;
*[[Kinetic Energy]]&lt;br /&gt;
*[[Net Force]].&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. 4th ed. Vol. 1.&lt;br /&gt;
Hoboken, NJ: Wiley, 2015. Print. Matter and Interactions.&lt;br /&gt;
&lt;br /&gt;
Ehrhardt, H., and L. A. Morgan. Electron Collisions with Molecules, Clusters,&lt;br /&gt;
and Surfaces. New York: Plenum, 1994. Print.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
Simulation for collisions: [https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
Video on different collision types: [https://www.youtube.com/watch?v=Xe2r6wey26E video]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. 4th ed. Vol. 1.&lt;br /&gt;
Hoboken, NJ: Wiley, 2015. Print. Matter and Interactions.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;The Gold Foil Experiment.&amp;quot; The Gold Foil Experiment. N.p., n.d. Web. 17 Apr.&lt;br /&gt;
2016.&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Collisions&amp;diff=46666</id>
		<title>Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Collisions&amp;diff=46666"/>
		<updated>2024-12-03T05:33:33Z</updated>

		<summary type="html">&lt;p&gt;Jonahy: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;edited by Jonah Yonathan Fall 2024&lt;br /&gt;
&lt;br /&gt;
A &#039;&#039;&#039;collision&#039;&#039;&#039; is the act in which two or more entities exert forces on each other over a time period, often expressed in a short period. Collisions behave according to the fundamental principles of physics. The nature of collisions allow for assumptions, as discussed later, to solve for unknowns. &lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
If we take the colliding objects as the system, what assumption can we make&lt;br /&gt;
about the total momentum and energy of this system?&lt;br /&gt;
&lt;br /&gt;
In our definition of a collision, we&#039;ve said that interactions preceding&lt;br /&gt;
and succeeding the process are tiny compared to the ones we see during the&lt;br /&gt;
process. The forces that contribute to this sudden spike of interactions is&lt;br /&gt;
internal to the system: they are between the colliding objects. Since we take&lt;br /&gt;
them as the system, we can make the assumption that the outside forces that act on the&lt;br /&gt;
system are negligible.&lt;br /&gt;
&lt;br /&gt;
By this assumption, the change in the total momentum and energy of the system&lt;br /&gt;
(of both the colliding objects) is negligible.&lt;br /&gt;
&lt;br /&gt;
For &amp;lt;math&amp;gt;\Delta p = F * \Delta t&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;F&amp;lt;/math&amp;gt; is negligible, thus&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta p&amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
For &amp;lt;math&amp;gt;\Delta E = W_{surr} + Q&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;W_{surr} = F_{surr} * \Delta x&amp;lt;/math&amp;gt;&lt;br /&gt;
and &amp;lt;math&amp;gt;Q = m * C * \Delta t&amp;lt;/math&amp;gt;. &amp;lt;math&amp;gt;F_{surr}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\Delta t&amp;lt;/math&amp;gt;&lt;br /&gt;
are negligible. Thus &amp;lt;math&amp;gt;\Delta E&amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
Therefore, we can assume that the total momentum and energy of the system (of both the colliding objects) don&#039;t change during the collision. Furthermore, the assumptions of conservation of momentum and conservation of kinetic energy allow us to find the final velocities of each entity in a two-body collision.&lt;br /&gt;
&lt;br /&gt;
Example: Swinging Masses &lt;br /&gt;
&lt;br /&gt;
The conservation of momentum and conservation of energy can be identified in a system which features steel spherical masses that hang side by side in a horizontal line from each other. If a single mass is pulled to the side and released, it follows in the opposing motion and strikes the line. The outer mass swings in the line of movement. Similarly, if two balls are pulled to the side and released, the other two balls swing out in the line of movement. &lt;br /&gt;
&lt;br /&gt;
     p_initial = p_final&lt;br /&gt;
     KE_initial = KE_final&lt;br /&gt;
&lt;br /&gt;
     &#039;&#039;&#039;Case 1:&#039;&#039;&#039;&lt;br /&gt;
     momentum in = momentum out &lt;br /&gt;
     mv = momentum out&lt;br /&gt;
     Kinetic Energy in = Kinetic Energy out&lt;br /&gt;
     (1/2)*mv^2 = kinetic energy out&lt;br /&gt;
    &lt;br /&gt;
     &#039;&#039;&#039;Case 2:&#039;&#039;&#039;&lt;br /&gt;
     momentum in = momentum out &lt;br /&gt;
     2mv = momentum out &lt;br /&gt;
     Kinetic Energy in = Kinetic Energy out&lt;br /&gt;
     (1/2)*2*mv^2 = kinetic energy out &lt;br /&gt;
&lt;br /&gt;
Both the conservation of momentum and conservation of energy are respected in both cases. The conservation of momentum and conservation of energy allow for the understanding of the relationships between internal and external forces. Realistically, there are very few perfectly elastic collisions. Often, kinetic energy is converted into internal energy, or dissipates as sound or heat energy. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Types of Collisions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In [[Elastic Collisions]], the total kinetic energy is conserved. None of the&lt;br /&gt;
kinetic energy of the system is transformed from or into &amp;lt;math&amp;gt;E_{internal}&amp;lt;/math&amp;gt;,&lt;br /&gt;
such as thermal energy and spring potential energy.&lt;br /&gt;
     &lt;br /&gt;
    Elastic collisions are regarded as one-dimensional collisions where the impact is along the line of movement. &lt;br /&gt;
&lt;br /&gt;
An example for such a process is the Newton&#039;s Cradle, where the kinetic energy&lt;br /&gt;
of the colliding ball gets transformed into the kinetic energy of the last&lt;br /&gt;
ball in the cradle. The total kinetic energy is conserved.&lt;br /&gt;
&lt;br /&gt;
In [[Inelastic Collisions]], the total kinetic energy is not conserved. Some/all&lt;br /&gt;
of the kinetic energy of the system can be transformed from or into&lt;br /&gt;
&amp;lt;math&amp;gt;E_{internal}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
    Inelastic collisions are regarded as two-dimensional collisions where the impact is not along the line of movement.&lt;br /&gt;
&lt;br /&gt;
An example for such a process is a car crash, where the kinetic energy of the&lt;br /&gt;
colliding car gets transformed into thermal energy and internal energy contained&lt;br /&gt;
in the crumbled, tense parts of the car chassis.&lt;br /&gt;
&lt;br /&gt;
([[Inelastic Collisions]] where there&#039;s maximum kinetic energy dissipation&lt;br /&gt;
are called [[Maximally Inelastic Collision]]. This doesn&#039;t necessarily mean&lt;br /&gt;
that &#039;&#039;all&#039;&#039; of the kinetic energy gets transformed into &amp;lt;math&amp;gt;E_{internal}&amp;lt;/math&amp;gt;,&lt;br /&gt;
since the total momentum is conserved. Collisions where the objects stick together are&lt;br /&gt;
[[Maximally Inelastic Collision]].)&lt;br /&gt;
&lt;br /&gt;
===Mathematical model===&lt;br /&gt;
By our assumptions, &amp;lt;math&amp;gt;p_f = p_i&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;E_f = E_i&amp;lt;/math&amp;gt;, &lt;br /&gt;
where subscripts &amp;lt;math&amp;gt;f&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;i&amp;lt;/math&amp;gt; mean final and initial&lt;br /&gt;
respectively. These equations hold true for all collisions.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision, &amp;lt;math&amp;gt;K_{total, f} = K_{total, i}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
In an inelastic collision, this equation doesn&#039;t hold, since&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta E = \Delta K + \Delta E_{internal} = 0&amp;lt;/math&amp;gt;, and&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta E_{internal}&amp;lt;/math&amp;gt; might be nonzero.&lt;br /&gt;
&lt;br /&gt;
===Computational model===&lt;br /&gt;
In this model, two objects of different masses are moving towards each other&lt;br /&gt;
with equal magnitude, but opposite direction momenta.&lt;br /&gt;
&lt;br /&gt;
The objects collide, and after the collision, it is observed that the objects&lt;br /&gt;
are embedded within each other with final velocity zero.&lt;br /&gt;
&lt;br /&gt;
This is an example of a [[Maximally Inelastic Collision]]: none of the kinetic&lt;br /&gt;
energy is conserved.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/e600a7595e Computational Model]&lt;br /&gt;
&lt;br /&gt;
Note: The behavior of the colliding objects are hardcoded in the model to&lt;br /&gt;
embed (vs. bouncing, etc.) This is an assumption we make about the&lt;br /&gt;
materials/internal structure of the balls. Correctly modeling deforming&lt;br /&gt;
objects without simplifying assumptions is a very complex task.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A &amp;lt;math&amp;gt;0.5 kg&amp;lt;/math&amp;gt; soccer ball is moving with a speed of &amp;lt;math&amp;gt;5 m/s&amp;lt;/math&amp;gt;&lt;br /&gt;
directly towards a &amp;lt;math&amp;gt;0.7 kg&amp;lt;/math&amp;gt; basketball, which is at rest.&lt;br /&gt;
&lt;br /&gt;
The two balls collide and stick together. What will be their final speed?&lt;br /&gt;
&lt;br /&gt;
[[File:Collision-example-1-mert.png|500px]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;m_1 = 0.5 kg&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;v_1 = 5 m/s&amp;lt;/math&amp;gt;,&lt;br /&gt;
&amp;lt;math&amp;gt;m_2 = 0.7 kg&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;v_2 = 0 m/s&amp;lt;/math&amp;gt;, where&lt;br /&gt;
subscripts &amp;lt;math&amp;gt;1&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;2&amp;lt;/math&amp;gt; signify the soccer ball and basketball respectively.&lt;br /&gt;
&lt;br /&gt;
We&#039;re going to compute &amp;lt;math&amp;gt;v_f&amp;lt;/math&amp;gt; for the collided mass. &lt;br /&gt;
&lt;br /&gt;
We know that momentum is conserved. &amp;lt;math&amp;gt;p_f = p_i \implies p_f = p_1 + p_2&lt;br /&gt;
\implies p_f = (m_1 * v_1) + (m_2 * v_2)&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
We know that the objects stick together, hence their mass &amp;lt;math&amp;gt;m_1 + m_2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Then &amp;lt;math&amp;gt;p_f = (m_1 + m_2) * v_f \implies (m_1 + m_2) * v_f&lt;br /&gt;
= (m_1 * v_1) + (m_2 * v_2)&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If we solve for &amp;lt;math&amp;gt;v_f&amp;lt;/math&amp;gt;, we get&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
    v_f &amp;amp;= \frac{(m_1 * v_1) + (m_2 * v_2)}{m_1 + m_2} \\&lt;br /&gt;
        &amp;amp;= \frac{(0.5 * 5) +(0.7 * 0)}{0.5 + 0.7} \\&lt;br /&gt;
        &amp;amp;= 2.083 m/s&lt;br /&gt;
\end{aligned}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This is a [[Maximally Inelastic Collision]], as the collided objects stuck&lt;br /&gt;
together. &amp;lt;math&amp;gt;K_{tot, f} &amp;lt; K_{tot, i}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A &amp;lt;math&amp;gt;10 kg&amp;lt;/math&amp;gt; asteroid with a velocity of &amp;lt;math&amp;gt;&amp;lt;10, 0, 0&amp;gt; m/s&amp;lt;/math&amp;gt;&lt;br /&gt;
crashes into the Earth in the Sahara Desert and bounces back into the space with&lt;br /&gt;
the same speed and opposite direction.&lt;br /&gt;
&lt;br /&gt;
Nobody is hurt, but some say that they felt the Earth recoil during&lt;br /&gt;
this unusual collision. Find the recoil velocity of the earth to see if it was&lt;br /&gt;
significant enough for people to have felt it. (Keep in mind that the human&lt;br /&gt;
sensory facilities aren&#039;t very precise.)&lt;br /&gt;
&lt;br /&gt;
[[File:Collision-example-2-mert.png|500px]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;m_a = 10 kg&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;m_e = 6 * 10^{24} kg&amp;lt;/math&amp;gt;,&lt;br /&gt;
&amp;lt;math&amp;gt;v_{a, i} = &amp;lt;10, 0, 0&amp;gt; m/s&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;v_{a, f} = &amp;lt;-10, 0, 0&amp;gt; m/s&amp;lt;/math&amp;gt;,&lt;br /&gt;
and &amp;lt;math&amp;gt;v_{e, i} = &amp;lt;0, 0, 0&amp;gt; m/s&amp;lt;/math&amp;gt;, where subscripts &amp;lt;math&amp;gt;a&amp;lt;/math&amp;gt;&lt;br /&gt;
and &amp;lt;math&amp;gt;e&amp;lt;/math&amp;gt; signify the asteroid and the earth respectively.&lt;br /&gt;
&lt;br /&gt;
We&#039;re going to compute &amp;lt;math&amp;gt;v_{e, f}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know that momentum is conserved.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
    &amp;amp;p_f = p_i \\&lt;br /&gt;
    \implies &amp;amp;p_{a, f} + p_{e, f} = p_{a, i} + p_{e, i} \\&lt;br /&gt;
    \implies &amp;amp;(m_a * v_{a, f}) + (m_e * v_{e, f})&lt;br /&gt;
    = (m_a * v_{a, i}) + (m_e * v_{e, i})&lt;br /&gt;
\end{aligned}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Solving for &amp;lt;math&amp;gt;v_{e, f}&amp;lt;/math&amp;gt;, we get&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
    v_{e, f}&lt;br /&gt;
        &amp;amp;= \frac{(m_a * v_{a, i}) + (m_e * v_{e, i}) - (m_a * v_{a, f})}{m_e} \\&lt;br /&gt;
        &amp;amp;= \frac{(10 * &amp;lt;10, 0, 0&amp;gt;) + (6 * 10^{24} * &amp;lt;0, 0, 0&amp;gt;) &lt;br /&gt;
            - (10 * &amp;lt;-10, 0, 0&amp;gt;)}{6 * 10^{24}} \\&lt;br /&gt;
        &amp;amp;= \frac{&amp;lt;100, 0, 0&amp;gt; + &amp;lt;100, 0, 0&amp;gt;}{6 * 10^{24}} \\&lt;br /&gt;
        &amp;amp;= &amp;lt;3.33 * 10^{-23}, 0, 0&amp;gt; m/s&lt;br /&gt;
\end{aligned}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The recoil velocity of the earth is too small for anyone to feel it.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two rocks in space are approaching a third, stationary rock. Rock 1 has a mass&lt;br /&gt;
of &amp;lt;math&amp;gt;30 kg&amp;lt;/math&amp;gt; and moves with a velocity &amp;lt;math&amp;gt;&amp;lt;13, -10, -2&amp;gt; m/s&amp;lt;/math&amp;gt;.&lt;br /&gt;
Rock 2 has a mass of &amp;lt;math&amp;gt;60 kg&amp;lt;/math&amp;gt;, and moves with a velocity&lt;br /&gt;
&amp;lt;math&amp;gt;&amp;lt;-3, 23, 3&amp;gt; m/s&amp;lt;/math&amp;gt;. Rock 3 has a mass of &amp;lt;math&amp;gt;35 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
and is stationary. The rocks collide at Rock 3&#039;s location.&lt;br /&gt;
&lt;br /&gt;
[[File:Collision-example-3a-mert.png|500px]]&lt;br /&gt;
&lt;br /&gt;
After the collision, it is observed that Rock 1 has embedded&lt;br /&gt;
into Rock 3, which is now moving with a new velocity of &amp;lt;math&amp;gt;&amp;lt;3, 5, 2&amp;gt; m/s&amp;lt;/math&amp;gt;,&lt;br /&gt;
and that a chunk of mass &amp;lt;math&amp;gt;15 kg&amp;lt;/math&amp;gt; has broken off of the combined rock.&lt;br /&gt;
&lt;br /&gt;
Rock 2, in the meantime, has a new velocity of &amp;lt;math&amp;gt;&amp;lt;-2, 12, 2&amp;gt; m/s&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Collision-example-3b-mert.png|500px]]&lt;br /&gt;
&lt;br /&gt;
What is the change in the internal energy &amp;lt;math&amp;gt;\Delta E_{internal}&amp;lt;/math&amp;gt; in&lt;br /&gt;
the system as a result of this three-way collision?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Due to the complexity of this question, we are going to use a computer to&lt;br /&gt;
solve this problem. Conceptually, everything remains the same, but instead of&lt;br /&gt;
computing things by hand, we&#039;re getting help from a computer. Feel free to&lt;br /&gt;
solve it by hand if you want to!&lt;br /&gt;
&lt;br /&gt;
    # Define the initial masses/velocities of the rocks.&lt;br /&gt;
    mass_1_i = 30&lt;br /&gt;
    mass_2_i = 60&lt;br /&gt;
    mass_3_i = 35&lt;br /&gt;
    &lt;br /&gt;
    v_1_i = vector(13, -10, -2)&lt;br /&gt;
    v_2_i = vector(-3, 23, 3)&lt;br /&gt;
    v_3_i = vector(0, 0, 0)&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the initial total momentum.&lt;br /&gt;
    total_p_i = (mass_1_i * v_1_i) + (mass_2_i * v_2_i) + (mass_3_i * v_3_i)&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the masses/velocities of the rocks after the collision.&lt;br /&gt;
    mass_chunk = 15&lt;br /&gt;
    &lt;br /&gt;
    mass_2_f = 60&lt;br /&gt;
    mass_3_f = mass_1_i + mass_3_i - mass_chunk&lt;br /&gt;
    &lt;br /&gt;
    v_2_f = vector(-2, 12, 2)&lt;br /&gt;
    v_3_f = vector(3, 5, 2)&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the velocity of the chunk. Remember that the momentum is conserved.&lt;br /&gt;
    v_chunk = (total_p_i - ((v_2_f * mass_2_f) + (v_3_f * mass_3_f))) / mass_chunk&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the initial total kinetic energy.&lt;br /&gt;
    def kinetic_energy(mass, velocity):&lt;br /&gt;
        return 1/2 * mass * mag(velocity) ** 2&lt;br /&gt;
    &lt;br /&gt;
    k_tot_i = kinetic_energy(mass_1_i, v_1_i) \&lt;br /&gt;
            + kinetic_energy(mass_2_i, v_2_i) \&lt;br /&gt;
            + kinetic_energy(mass_3_i, v_3_i)&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the final total kinetic energy.&lt;br /&gt;
    k_tot_f = kinetic_energy(mass_2_f, v_2_f) \&lt;br /&gt;
            + kinetic_energy(mass_3_f, v_3_f) \&lt;br /&gt;
            + kinetic_energy(mass_chunk, v_chunk)&lt;br /&gt;
    &lt;br /&gt;
    # Calculate the change in internal energy.&lt;br /&gt;
    d_e_internal = k_tot_f - k_tot_i&lt;br /&gt;
&lt;br /&gt;
This Python snippet, when evaluated, will show that&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta E_{internal} = 5175 J&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
Collisions are everywhere in our daily lives. Some examples are: a bat colliding with a baseball in your&lt;br /&gt;
neighbor&#039;s backyard, two cars crashing in the highway in front of you, your finger hitting the keyboard, and so&lt;br /&gt;
on.&lt;br /&gt;
&lt;br /&gt;
Depending on the nature of the collision we can usually guess whether it&#039;s&lt;br /&gt;
an [[Elastic Collisions]] or an [[Inelastic Collisions]], and reason about it.&lt;br /&gt;
(If it&#039;s a car crash, things crumble, so there&#039;s increase in thermal energy, etc.)&lt;br /&gt;
&lt;br /&gt;
Implications of collisions may be included in the understanding of the ideal gas law. As temperatures rise within a given sector, the movement of molecules become excited. Molecules of ideal gases will move with perfect elasticity in relation to one another. In other words, the kinetic energy before the initial collision will remain the same as the final kinetic energy after the collision. However, most gases do not behave ideally. More accurately, moments of kinetic energy may dissipate or be absorbed as heat energy and sound energy, respectively. &lt;br /&gt;
&lt;br /&gt;
[[File:Asteroid 2009 VA.png|thumb|right|Asteroid 2009 VA]]Another very interesting application of collisions is in the field of astronomy.&lt;br /&gt;
The universe is ever-changing and we never know what might come our way.&lt;br /&gt;
A large meteor could be flying at the Earth with a very high speed and we could&lt;br /&gt;
use our knowledge of collisions to find out how the Earth will move after the&lt;br /&gt;
collision, what the temperature change of the area would be, and how the rest&lt;br /&gt;
of the Earth would be effected by the collision. By using our collision skills,&lt;br /&gt;
we could evacuate anyone who is prone to harm during the collision.&lt;br /&gt;
&lt;br /&gt;
Since collisions are ubiquitous processes, knowledge about them is instrumental&lt;br /&gt;
for any engineer who deals with the physical world.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
We&#039;ve mentioned that collisions are not special case events in Physics--&lt;br /&gt;
the fundamental principles apply fully for them, and we reason about them&lt;br /&gt;
using these principles. Then it won&#039;t surprise us that the history of collisions&lt;br /&gt;
is tangled with the history of the fundamental principles.&lt;br /&gt;
&lt;br /&gt;
In 1661, Christiaan Huygens, a Dutch physicist, concluded that when a moving&lt;br /&gt;
body struck a stationary object of equal mass, the initial moving body would&lt;br /&gt;
lose all of its momentum while the second object would pick up the same amount&lt;br /&gt;
of velocity that the moving body had before the collision. He also stated that&lt;br /&gt;
the total &amp;quot;quantity of motion&amp;quot; should be the same before and after the&lt;br /&gt;
collision. This was the first thought of the conservation of momentum.&lt;br /&gt;
&lt;br /&gt;
In 1687, Sir Isaac Newton went a little further and stated in his third law of&lt;br /&gt;
motion that the forces of action and reaction between two bodies are opposite&lt;br /&gt;
and equal. Because of this law; we know that the total momentum of a closed system is constant.&lt;br /&gt;
Every force in the system has a reciprocal pair with an opposing direction, which&lt;br /&gt;
implies that their momenta cancel out, keeping the total momentum constant.&lt;br /&gt;
&lt;br /&gt;
In Ernest Rutherford&#039;s famous Gold Foil Experiment of 1899, he found out that the plum&lt;br /&gt;
pudding model of the atom was inaccurate, and that there was a concentrated&lt;br /&gt;
core at the center of the atom (nucleus) and a lot of empty space around.&lt;br /&gt;
He didn&#039;t have access to the sophisticated tools/knowledge that we have now to&lt;br /&gt;
assist in his discovery: he did this through investigating atomic collisions&lt;br /&gt;
between alpha particles and gold foil. He shot alpha particles to a piece of&lt;br /&gt;
thin gold foil, and observed that some of the particles were scattered&lt;br /&gt;
through angles larger than 90 degrees during the collisions. Using collision&lt;br /&gt;
principles, he was then able to reason that the atom had most of its mass&lt;br /&gt;
concentrated in a dense core, rather than spread out evenly throughout the atom&lt;br /&gt;
as the plum pudding model suggested.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
Other pages related to collisions:&lt;br /&gt;
*[[Maximally Inelastic Collision]]&lt;br /&gt;
*[[Inelastic Collisions]]&lt;br /&gt;
*[[Elastic Collisions]]&lt;br /&gt;
*[[Linear Momentum]]&lt;br /&gt;
*[[Energy Principle]],&lt;br /&gt;
*[[Kinetic Energy]]&lt;br /&gt;
*[[Net Force]].&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. 4th ed. Vol. 1.&lt;br /&gt;
Hoboken, NJ: Wiley, 2015. Print. Matter and Interactions.&lt;br /&gt;
&lt;br /&gt;
Ehrhardt, H., and L. A. Morgan. Electron Collisions with Molecules, Clusters,&lt;br /&gt;
and Surfaces. New York: Plenum, 1994. Print.&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
Simulation for collisions: [https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
Video on different collision types: [https://www.youtube.com/watch?v=Xe2r6wey26E video]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. 4th ed. Vol. 1.&lt;br /&gt;
Hoboken, NJ: Wiley, 2015. Print. Matter and Interactions.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;The Gold Foil Experiment.&amp;quot; The Gold Foil Experiment. N.p., n.d. Web. 17 Apr.&lt;br /&gt;
2016.&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Jonahy</name></author>
	</entry>
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