Physics Book - User contributions [en]

Ampere-Maxwell Law

2026-04-27T03:36:53Z

Rohitnaras: /* Connectedness */

'''Claimed by Rohit Naras (Spring 2026)'''

The Ampere-Maxwell Law is one of the four [[Maxwell's Equations]] and represents a generalization of Ampere's Law. It states that magnetic fields are produced not only by electric currents but also by time-varying electric fields (the so-called "displacement current"). This addition by James Clerk Maxwell was the missing piece that unified electricity and magnetism, and it directly predicts the existence of electromagnetic waves traveling at the speed of light.

==The Main Idea==

Ampere's original law relates the magnetic field along a closed (Amperian) loop to the conduction current piercing any surface bounded by that loop. This works perfectly for steady currents in continuous wires, but it fails in a famous test case: a charging capacitor. If you draw an Amperian loop around the wire feeding the capacitor and choose a flat surface, you enclose a current ''I''. But if you bulge the surface so it passes between the capacitor plates instead, no conduction current pierces it at all — yet the magnetic field around the loop is the same. Ampere's Law gives two different answers for the same loop, which is a contradiction.

Maxwell resolved this by recognizing that between the plates, the electric field is changing in time, and this changing E-field must itself act as a source of magnetic field. He added a new term — the '''displacement current''' — proportional to the rate of change of electric flux. The corrected law is symmetric with [[Faraday's Law]]: just as a changing magnetic flux produces a curly electric field, a changing electric flux produces a curly magnetic field.

This symmetry is not just aesthetic. When you combine the Ampere-Maxwell Law with Faraday's Law, the equations support self-sustaining oscillations of E and B fields propagating through space at speed <math>c = 1/\sqrt{\mu_0 \epsilon_0} \approx 3 \times 10^8</math> m/s. Light is an electromagnetic wave, and the Ampere-Maxwell Law is half the reason why.

===A Mathematical Model===

The full Ampere-Maxwell Law is:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

where:

* <math>\oint \vec{B} \cdot d\vec{l}</math> is the line integral of the magnetic field around a closed Amperian loop
* <math>I_{enc}</math> is the conduction current passing through any surface bounded by that loop
* <math>\Phi_E = \int \vec{E} \cdot \hat{n}\, dA</math> is the electric flux through that same surface
* <math>\mu_0 = 4\pi \times 10^{-7}</math> T·m/A is the permeability of free space
* <math>\epsilon_0 = 8.854 \times 10^{-12}</math> C²/(N·m²) is the permittivity of free space

The second term, <math>\mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>, is Maxwell's contribution. The quantity <math>I_d = \epsilon_0 \frac{d\Phi_E}{dt}</math> is called the '''displacement current''', though it is not a true current of moving charges, it is a name given to a changing electric flux that has the same magnetic effect as a real current.

==Computational Model==

The simulation below uses VPython/GlowScript to visualize the Ampere-Maxwell Law for a charging parallel-plate capacitor. As the capacitor charges, the electric field between the plates grows, producing a changing electric flux. By the Ampere-Maxwell Law, this changing flux generates a curly magnetic field between the plates — even though no real current flows there.

The simulation shows:
* Two circular capacitor plates being charged by a current ''I''
* A growing electric field <math>\vec{E}</math> (red arrow) between the plates
* The induced curly magnetic field <math>\vec{B}</math> (blue arrows) circulating around the field axis
* Real-time graphs of <math>E(t)</math>, <math>dE/dt</math>, and the resulting <math>B(r)</math> at a fixed radius

You can run the simulation here: [Insert your GlowScript link after publishing — instructions below]

The full GlowScript code is also provided below for transparency and reproducibility.

Link: https://trinket.io/glowscript/faf9ad543e9a

==Examples==

===Simple===

'''Question:''' A 30 A current is charging a parallel-plate capacitor with circular plates of radius R = 2 mm. The plate separation is 5 mm. What is the magnitude of the magnetic field between the plates at a radial distance r = 1 mm from the central axis?

[[File:Example1.png]]

'''Solution:'''

Apply the Ampere-Maxwell Law to a circular Amperian loop of radius ''r'' = 1 mm centered on the axis between the plates. There is no conduction current piercing this loop (we are between the plates, not in the wire), so only the displacement current term contributes:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

The electric field between the plates of a parallel-plate capacitor is uniform across the plate area:

:<math>E = \frac{Q}{\epsilon_0 A_{plate}} = \frac{Q}{\epsilon_0 \pi R^2}</math>

The electric flux through our small Amperian loop of area <math>\pi r^2</math> is:

:<math>\Phi_E = E \cdot \pi r^2 = \frac{Q}{\epsilon_0 \pi R^2} \cdot \pi r^2 = \frac{Q r^2}{\epsilon_0 R^2}</math>

Taking the time derivative and recognizing <math>dQ/dt = I</math>:

:<math>\frac{d\Phi_E}{dt} = \frac{I r^2}{\epsilon_0 R^2}</math>

Substituting back into the Ampere-Maxwell Law:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \cdot \frac{I r^2}{\epsilon_0 R^2} = \frac{\mu_0 I r^2}{R^2}</math>

By the symmetry of the problem, '''B''' is constant in magnitude along the Amperian circle and is everywhere tangent to it, so the line integral simplifies to <math>B(2\pi r)</math>:

:<math>B(2\pi r) = \frac{\mu_0 I r^2}{R^2}</math>

Solving for ''B'':

:<math>B = \frac{\mu_0 I r}{2\pi R^2}</math>

Plugging in ''I'' = 30 A, ''r'' = 0.001 m, ''R'' = 0.002 m:

:<math>B = \frac{(4\pi \times 10^{-7}\ \text{T·m/A})(30\ \text{A})(0.001\ \text{m})}{2\pi (0.002\ \text{m})^2}</math>

:<math>B = \frac{(4\pi \times 10^{-7})(0.03)}{2\pi (4 \times 10^{-6})} = 1.5 \times 10^{-3}\ \text{T}</math>

:<math>\boxed{B = 1.5\ \text{mT}}</math>

===Middling===

Pick a closed, rectangular path in the xz plane with a height h and a width w. Calculate the speed v of the slab.

[[File: AMexample.png]]

At a time change in t, we can calculate the area as the height (h) times the speed over the change in time. ⩟ A = v (⩟ t) h
Because the electric field is constant in this region we can also calculate the change in electric flux over time as Evh(⩟ t) /(⩟ t) which is the same as Evh
Calculating the path integral for the magnetic field we get that ∮B . dl = Bh cos 0 = Bh
An important thing to notice is that there is no current I, so, using the Ampere-Maxwell Law we can see that

Bh = μ. [I+ ε. (vEh)]

but since there is no current,

B = μ. ε. (vE)

From Faraday's law we get that the emf equals the rate of change of the magnetic flux: Eh = Bvh

Substituting E = Bv into our previous equation we get that

B = μ. ε. (v(vB))

Solving for v we get that:

[[File: AMvelocity.png]]

The example shows that the speed of light relates the a time varying electric and magnetic field.

====What this implies====

In the example above we saw that the speed of light relates a time varying electric and magnetic field. This translates to the fact that an electromagnetic wave propagates at the speed of light.

[[File: AMwave.png]]

===Difficult===

[[File:Example3.jpg]]

[[File:Example3b.jpg]]

Question 1: First find the line integral of B around a loop of radius R located just outside the left capacitor plate.

Answer:

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I(t)</math>

Question 2: Now find an expression for the same line integral of B around the same loop located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the Ampère-Maxwell law above.

Answer:

:<math>\Phi_E = \int E \cdot dA = A E(t)</math>

:<math>I = ε_0 d \frac{(AE(t))}{dt} = A ε_0 \frac{dE(t)}{dt} </math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I</math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 A ε_0 \frac{dE(t)}{dt}</math>

Question 3: Express the normal current in terms of the charge on the capacitor plate.

Answer:

:<math>I = ε_0 \frac{d\frac{q}{ε_0}}{dt}</math>

the expression for the normal current is:

:<math>I = \frac{dq(t)}{dt}</math>

the current is directly related to charge on left capacitor plate

===Conceptual Questions===
[[File: AMconceptual.png|500px|Image: 500 pixels]]

The plot above shows a side and a top view of a capacitor with charge Q with electric and magnetic fields E and B at time t. The charge Q is:

1) Increasing in time

2) Decreasing in time

3) Constant in time

4) I don't know

Answer:

The unit vector N points out of the plane. Given that the magnetic field curls clockwise, the electric flux would be positive and decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q.

[[File: AMcapacitor.png|500px|Image: 500 pixels]]

Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is

1) Zero

2) Positive

3) Negative

4) Can’t tell

Answer:

One thing to notice is that there is no enclosed current through the disk. When integrating in the direction shown, the electric flux is positive. Because the plates are charging, the electric flux is increasing. Therefore the line integral is positive.

==Connectedness==
'''How is this topic connected to something that you are interested in?'''

As a computer engineering student, the Ampere-Maxwell Law is foundational to almost every piece of hardware I work with. Every PCB trace carrying a high-frequency signal radiates because of changing electric fields between adjacent conductors, this is exactly the displacement current term in action. Signal integrity engineering, antenna design, and even the GHz-range coupling that causes crosstalk in dense chip layouts are all governed by this equation. Without Maxwell's correction, we would have no theoretical basis for wireless communication, RF circuits, or the high-speed serial links inside modern processors.

'''Interesting industrial application'''

A particularly elegant application is the '''wireless charging''' standard used in modern smartphones (Qi). Two flat coils: one in the charging pad, one in the phone, couple through a time-varying magnetic field. But at the higher-frequency variants (resonant inductive and especially capacitive wireless power transfer), the displacement current term becomes the dominant coupling mechanism. Capacitive wireless power transfer literally relies on the Ampere-Maxwell Law: there is no closed conductor between transmitter and receiver, only a changing electric flux that carries the energy across the gap.

==History==

Andre-Marie Ampere formulated his original law in 1826, relating magnetic fields to steady electric currents. For the next ~35 years it was treated as complete. The problem was that Ampere's Law, in its original form, only worked for closed steady currents — and a charging capacitor is the canonical case where it visibly fails.

In the early 1860s, '''James Clerk Maxwell''' was working to unify the existing laws of electricity and magnetism into a single mathematical framework. While studying Faraday's work on induction, Maxwell noticed the asymmetry: a changing magnetic field produced an electric field (Faraday's Law), but no equivalent law existed for a changing electric field producing a magnetic field. In his 1861 paper ''On Physical Lines of Force'' and more rigorously in his 1865 paper ''A Dynamical Theory of the Electromagnetic Field'', Maxwell introduced the displacement current term to fix Ampere's Law and restore symmetry.

The payoff was immediate and stunning. With the corrected equation, Maxwell could derive a wave equation directly from his four laws, and the predicted wave speed <math>1/\sqrt{\mu_0\epsilon_0}</math> came out to within experimental error of the measured speed of light. Maxwell concluded — correctly — that light itself is an electromagnetic wave. Heinrich Hertz experimentally confirmed the existence of these waves in 1887, and the entire modern world of radio, television, radar, and wireless communication followed.

== See also ==

The Ampere-Maxwell Law relates to the other Maxwell equations:

*[[Gauss's Law]]

*[[Magnetic Flux ]]

*[[Faraday's Law ]]

It might also help to read up on [[Maxwell's Electromagnetic Theory ]]

Also for continuity purposes, it can be helpful to look at the incomplete Ampere's Law:

*[[Ampere's Law]]

===Further Reading===

Matter and Interactions: Volume 2 by Ruth Chabay and Bruce Sherwood (4th Edition)

===External Links===

Maxwell's Equations:
http://study.com/academy/lesson/maxwells-equations-definition-application.html

James Maxwell Biography:
http://www.biography.com/people/james-c-maxwell-9403463

==References==

http://bulldog2.redlands.edu/fac/eric_hill/Phys232/Lectures/Ch%2023%20lect%201.pdf
http://ocw.mit.edu/high-school/physics/exam-prep/electromagnetism/maxwells-equations/
http://ocw.mit.edu/courses/physics/8-02-physics-ii-electricity-and-magnetism-spring-2007/readings/summary_w13d1.pdf
http://www.schoolphysics.co.uk/age16-19/Wave%20properties/Wave%20properties/text/Electromagnetic_radiation/index.html

[[Category:Which Category did you place this in?]]

Ampere-Maxwell Law

2026-04-27T03:36:07Z

Rohitnaras: /* Simple */

'''Claimed by Rohit Naras (Spring 2026)'''

The Ampere-Maxwell Law is one of the four [[Maxwell's Equations]] and represents a generalization of Ampere's Law. It states that magnetic fields are produced not only by electric currents but also by time-varying electric fields (the so-called "displacement current"). This addition by James Clerk Maxwell was the missing piece that unified electricity and magnetism, and it directly predicts the existence of electromagnetic waves traveling at the speed of light.

==The Main Idea==

Ampere's original law relates the magnetic field along a closed (Amperian) loop to the conduction current piercing any surface bounded by that loop. This works perfectly for steady currents in continuous wires, but it fails in a famous test case: a charging capacitor. If you draw an Amperian loop around the wire feeding the capacitor and choose a flat surface, you enclose a current ''I''. But if you bulge the surface so it passes between the capacitor plates instead, no conduction current pierces it at all — yet the magnetic field around the loop is the same. Ampere's Law gives two different answers for the same loop, which is a contradiction.

Maxwell resolved this by recognizing that between the plates, the electric field is changing in time, and this changing E-field must itself act as a source of magnetic field. He added a new term — the '''displacement current''' — proportional to the rate of change of electric flux. The corrected law is symmetric with [[Faraday's Law]]: just as a changing magnetic flux produces a curly electric field, a changing electric flux produces a curly magnetic field.

This symmetry is not just aesthetic. When you combine the Ampere-Maxwell Law with Faraday's Law, the equations support self-sustaining oscillations of E and B fields propagating through space at speed <math>c = 1/\sqrt{\mu_0 \epsilon_0} \approx 3 \times 10^8</math> m/s. Light is an electromagnetic wave, and the Ampere-Maxwell Law is half the reason why.

===A Mathematical Model===

The full Ampere-Maxwell Law is:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

where:

* <math>\oint \vec{B} \cdot d\vec{l}</math> is the line integral of the magnetic field around a closed Amperian loop
* <math>I_{enc}</math> is the conduction current passing through any surface bounded by that loop
* <math>\Phi_E = \int \vec{E} \cdot \hat{n}\, dA</math> is the electric flux through that same surface
* <math>\mu_0 = 4\pi \times 10^{-7}</math> T·m/A is the permeability of free space
* <math>\epsilon_0 = 8.854 \times 10^{-12}</math> C²/(N·m²) is the permittivity of free space

The second term, <math>\mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>, is Maxwell's contribution. The quantity <math>I_d = \epsilon_0 \frac{d\Phi_E}{dt}</math> is called the '''displacement current''', though it is not a true current of moving charges, it is a name given to a changing electric flux that has the same magnetic effect as a real current.

==Computational Model==

The simulation below uses VPython/GlowScript to visualize the Ampere-Maxwell Law for a charging parallel-plate capacitor. As the capacitor charges, the electric field between the plates grows, producing a changing electric flux. By the Ampere-Maxwell Law, this changing flux generates a curly magnetic field between the plates — even though no real current flows there.

The simulation shows:
* Two circular capacitor plates being charged by a current ''I''
* A growing electric field <math>\vec{E}</math> (red arrow) between the plates
* The induced curly magnetic field <math>\vec{B}</math> (blue arrows) circulating around the field axis
* Real-time graphs of <math>E(t)</math>, <math>dE/dt</math>, and the resulting <math>B(r)</math> at a fixed radius

You can run the simulation here: [Insert your GlowScript link after publishing — instructions below]

The full GlowScript code is also provided below for transparency and reproducibility.

Link: https://trinket.io/glowscript/faf9ad543e9a

==Examples==

===Simple===

'''Question:''' A 30 A current is charging a parallel-plate capacitor with circular plates of radius R = 2 mm. The plate separation is 5 mm. What is the magnitude of the magnetic field between the plates at a radial distance r = 1 mm from the central axis?

[[File:Example1.png]]

'''Solution:'''

Apply the Ampere-Maxwell Law to a circular Amperian loop of radius ''r'' = 1 mm centered on the axis between the plates. There is no conduction current piercing this loop (we are between the plates, not in the wire), so only the displacement current term contributes:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

The electric field between the plates of a parallel-plate capacitor is uniform across the plate area:

:<math>E = \frac{Q}{\epsilon_0 A_{plate}} = \frac{Q}{\epsilon_0 \pi R^2}</math>

The electric flux through our small Amperian loop of area <math>\pi r^2</math> is:

:<math>\Phi_E = E \cdot \pi r^2 = \frac{Q}{\epsilon_0 \pi R^2} \cdot \pi r^2 = \frac{Q r^2}{\epsilon_0 R^2}</math>

Taking the time derivative and recognizing <math>dQ/dt = I</math>:

:<math>\frac{d\Phi_E}{dt} = \frac{I r^2}{\epsilon_0 R^2}</math>

Substituting back into the Ampere-Maxwell Law:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \cdot \frac{I r^2}{\epsilon_0 R^2} = \frac{\mu_0 I r^2}{R^2}</math>

By the symmetry of the problem, '''B''' is constant in magnitude along the Amperian circle and is everywhere tangent to it, so the line integral simplifies to <math>B(2\pi r)</math>:

:<math>B(2\pi r) = \frac{\mu_0 I r^2}{R^2}</math>

Solving for ''B'':

:<math>B = \frac{\mu_0 I r}{2\pi R^2}</math>

Plugging in ''I'' = 30 A, ''r'' = 0.001 m, ''R'' = 0.002 m:

:<math>B = \frac{(4\pi \times 10^{-7}\ \text{T·m/A})(30\ \text{A})(0.001\ \text{m})}{2\pi (0.002\ \text{m})^2}</math>

:<math>B = \frac{(4\pi \times 10^{-7})(0.03)}{2\pi (4 \times 10^{-6})} = 1.5 \times 10^{-3}\ \text{T}</math>

:<math>\boxed{B = 1.5\ \text{mT}}</math>

===Middling===

Pick a closed, rectangular path in the xz plane with a height h and a width w. Calculate the speed v of the slab.

[[File: AMexample.png]]

At a time change in t, we can calculate the area as the height (h) times the speed over the change in time. ⩟ A = v (⩟ t) h
Because the electric field is constant in this region we can also calculate the change in electric flux over time as Evh(⩟ t) /(⩟ t) which is the same as Evh
Calculating the path integral for the magnetic field we get that ∮B . dl = Bh cos 0 = Bh
An important thing to notice is that there is no current I, so, using the Ampere-Maxwell Law we can see that

Bh = μ. [I+ ε. (vEh)]

but since there is no current,

B = μ. ε. (vE)

From Faraday's law we get that the emf equals the rate of change of the magnetic flux: Eh = Bvh

Substituting E = Bv into our previous equation we get that

B = μ. ε. (v(vB))

Solving for v we get that:

[[File: AMvelocity.png]]

The example shows that the speed of light relates the a time varying electric and magnetic field.

====What this implies====

In the example above we saw that the speed of light relates a time varying electric and magnetic field. This translates to the fact that an electromagnetic wave propagates at the speed of light.

[[File: AMwave.png]]

===Difficult===

[[File:Example3.jpg]]

[[File:Example3b.jpg]]

Question 1: First find the line integral of B around a loop of radius R located just outside the left capacitor plate.

Answer:

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I(t)</math>

Question 2: Now find an expression for the same line integral of B around the same loop located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the Ampère-Maxwell law above.

Answer:

:<math>\Phi_E = \int E \cdot dA = A E(t)</math>

:<math>I = ε_0 d \frac{(AE(t))}{dt} = A ε_0 \frac{dE(t)}{dt} </math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I</math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 A ε_0 \frac{dE(t)}{dt}</math>

Question 3: Express the normal current in terms of the charge on the capacitor plate.

Answer:

:<math>I = ε_0 \frac{d\frac{q}{ε_0}}{dt}</math>

the expression for the normal current is:

:<math>I = \frac{dq(t)}{dt}</math>

the current is directly related to charge on left capacitor plate

===Conceptual Questions===
[[File: AMconceptual.png|500px|Image: 500 pixels]]

The plot above shows a side and a top view of a capacitor with charge Q with electric and magnetic fields E and B at time t. The charge Q is:

1) Increasing in time

2) Decreasing in time

3) Constant in time

4) I don't know

Answer:

The unit vector N points out of the plane. Given that the magnetic field curls clockwise, the electric flux would be positive and decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q.

[[File: AMcapacitor.png|500px|Image: 500 pixels]]

Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is

1) Zero

2) Positive

3) Negative

4) Can’t tell

Answer:

One thing to notice is that there is no enclosed current through the disk. When integrating in the direction shown, the electric flux is positive. Because the plates are charging, the electric flux is increasing. Therefore the line integral is positive.

==Connectedness==

As a computer engineering student, the Ampere-Maxwell Law is foundational to almost every piece of hardware I work with. Every PCB trace carrying a high-frequency signal radiates because of changing electric fields between adjacent conductors, this is exactly the displacement current term in action. Signal integrity engineering, antenna design, and even the GHz-range coupling that causes crosstalk in dense chip layouts are all governed by this equation. Without Maxwell's correction, we would have no theoretical basis for wireless communication, RF circuits, or the high-speed serial links inside modern processors.

'''Interesting industrial application'''

A particularly elegant application is the '''wireless charging''' standard used in modern smartphones (Qi). Two flat coils: one in the charging pad, one in the phone, couple through a time-varying magnetic field. But at the higher-frequency variants (resonant inductive and especially capacitive wireless power transfer), the displacement current term becomes the dominant coupling mechanism. Capacitive wireless power transfer literally relies on the Ampere-Maxwell Law: there is no closed conductor between transmitter and receiver, only a changing electric flux that carries the energy across the gap.

==History==

Andre-Marie Ampere formulated his original law in 1826, relating magnetic fields to steady electric currents. For the next ~35 years it was treated as complete. The problem was that Ampere's Law, in its original form, only worked for closed steady currents — and a charging capacitor is the canonical case where it visibly fails.

In the early 1860s, '''James Clerk Maxwell''' was working to unify the existing laws of electricity and magnetism into a single mathematical framework. While studying Faraday's work on induction, Maxwell noticed the asymmetry: a changing magnetic field produced an electric field (Faraday's Law), but no equivalent law existed for a changing electric field producing a magnetic field. In his 1861 paper ''On Physical Lines of Force'' and more rigorously in his 1865 paper ''A Dynamical Theory of the Electromagnetic Field'', Maxwell introduced the displacement current term to fix Ampere's Law and restore symmetry.

The payoff was immediate and stunning. With the corrected equation, Maxwell could derive a wave equation directly from his four laws, and the predicted wave speed <math>1/\sqrt{\mu_0\epsilon_0}</math> came out to within experimental error of the measured speed of light. Maxwell concluded — correctly — that light itself is an electromagnetic wave. Heinrich Hertz experimentally confirmed the existence of these waves in 1887, and the entire modern world of radio, television, radar, and wireless communication followed.

== See also ==

The Ampere-Maxwell Law relates to the other Maxwell equations:

*[[Gauss's Law]]

*[[Magnetic Flux ]]

*[[Faraday's Law ]]

It might also help to read up on [[Maxwell's Electromagnetic Theory ]]

Also for continuity purposes, it can be helpful to look at the incomplete Ampere's Law:

*[[Ampere's Law]]

===Further Reading===

Matter and Interactions: Volume 2 by Ruth Chabay and Bruce Sherwood (4th Edition)

===External Links===

Maxwell's Equations:
http://study.com/academy/lesson/maxwells-equations-definition-application.html

James Maxwell Biography:
http://www.biography.com/people/james-c-maxwell-9403463

==References==

http://bulldog2.redlands.edu/fac/eric_hill/Phys232/Lectures/Ch%2023%20lect%201.pdf
http://ocw.mit.edu/high-school/physics/exam-prep/electromagnetism/maxwells-equations/
http://ocw.mit.edu/courses/physics/8-02-physics-ii-electricity-and-magnetism-spring-2007/readings/summary_w13d1.pdf
http://www.schoolphysics.co.uk/age16-19/Wave%20properties/Wave%20properties/text/Electromagnetic_radiation/index.html

[[Category:Which Category did you place this in?]]

Ampere-Maxwell Law

2026-04-27T03:32:38Z

Rohitnaras: /* Computational Model */

'''Claimed by Rohit Naras (Spring 2026)'''

The Ampere-Maxwell Law is one of the four [[Maxwell's Equations]] and represents a generalization of Ampere's Law. It states that magnetic fields are produced not only by electric currents but also by time-varying electric fields (the so-called "displacement current"). This addition by James Clerk Maxwell was the missing piece that unified electricity and magnetism, and it directly predicts the existence of electromagnetic waves traveling at the speed of light.

==The Main Idea==

Ampere's original law relates the magnetic field along a closed (Amperian) loop to the conduction current piercing any surface bounded by that loop. This works perfectly for steady currents in continuous wires, but it fails in a famous test case: a charging capacitor. If you draw an Amperian loop around the wire feeding the capacitor and choose a flat surface, you enclose a current ''I''. But if you bulge the surface so it passes between the capacitor plates instead, no conduction current pierces it at all — yet the magnetic field around the loop is the same. Ampere's Law gives two different answers for the same loop, which is a contradiction.

Maxwell resolved this by recognizing that between the plates, the electric field is changing in time, and this changing E-field must itself act as a source of magnetic field. He added a new term — the '''displacement current''' — proportional to the rate of change of electric flux. The corrected law is symmetric with [[Faraday's Law]]: just as a changing magnetic flux produces a curly electric field, a changing electric flux produces a curly magnetic field.

This symmetry is not just aesthetic. When you combine the Ampere-Maxwell Law with Faraday's Law, the equations support self-sustaining oscillations of E and B fields propagating through space at speed <math>c = 1/\sqrt{\mu_0 \epsilon_0} \approx 3 \times 10^8</math> m/s. Light is an electromagnetic wave, and the Ampere-Maxwell Law is half the reason why.

===A Mathematical Model===

The full Ampere-Maxwell Law is:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

where:

* <math>\oint \vec{B} \cdot d\vec{l}</math> is the line integral of the magnetic field around a closed Amperian loop
* <math>I_{enc}</math> is the conduction current passing through any surface bounded by that loop
* <math>\Phi_E = \int \vec{E} \cdot \hat{n}\, dA</math> is the electric flux through that same surface
* <math>\mu_0 = 4\pi \times 10^{-7}</math> T·m/A is the permeability of free space
* <math>\epsilon_0 = 8.854 \times 10^{-12}</math> C²/(N·m²) is the permittivity of free space

The second term, <math>\mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>, is Maxwell's contribution. The quantity <math>I_d = \epsilon_0 \frac{d\Phi_E}{dt}</math> is called the '''displacement current''', though it is not a true current of moving charges, it is a name given to a changing electric flux that has the same magnetic effect as a real current.

==Computational Model==

The simulation below uses VPython/GlowScript to visualize the Ampere-Maxwell Law for a charging parallel-plate capacitor. As the capacitor charges, the electric field between the plates grows, producing a changing electric flux. By the Ampere-Maxwell Law, this changing flux generates a curly magnetic field between the plates — even though no real current flows there.

The simulation shows:
* Two circular capacitor plates being charged by a current ''I''
* A growing electric field <math>\vec{E}</math> (red arrow) between the plates
* The induced curly magnetic field <math>\vec{B}</math> (blue arrows) circulating around the field axis
* Real-time graphs of <math>E(t)</math>, <math>dE/dt</math>, and the resulting <math>B(r)</math> at a fixed radius

You can run the simulation here: [Insert your GlowScript link after publishing — instructions below]

The full GlowScript code is also provided below for transparency and reproducibility.

Link: https://trinket.io/glowscript/faf9ad543e9a

==Examples==

===Simple===

'''Question:''' A 30 A current is charging a parallel-plate capacitor with circular plates of radius ''R'' = 2.5 mm. The plate separation is 4 mm. What is the magnitude of the magnetic field between the plates at a radial distance ''r'' = 1 mm from the central axis?

[[File:Example1.png]]

'''Solution:'''

Apply the Ampere-Maxwell Law to a circular Amperian loop of radius ''r'' centered on the axis between the plates. There is no conduction current between the plates, so:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

The electric field between the plates of a capacitor is <math>E = \frac{Q}{\epsilon_0 A_{plate}}</math>, where <math>A_{plate} = \pi R^2</math>. The flux through our Amperian loop (area <math>\pi r^2</math>) is:

:<math>\Phi_E = E \cdot \pi r^2 = \frac{Q}{\epsilon_0 \pi R^2}\cdot \pi r^2 = \frac{Q r^2}{\epsilon_0 R^2}</math>

Taking the time derivative and recognizing <math>dQ/dt = I</math>:

:<math>\frac{d\Phi_E}{dt} = \frac{I r^2}{\epsilon_0 R^2}</math>

By symmetry, B is constant in magnitude around the loop, so <math>\oint \vec{B}\cdot d\vec{l} = B(2\pi r)</math>. Solving:

:<math>B = \frac{\mu_0 I r}{2\pi R^2}</math>

Plugging in ''I'' = 30 A, ''r'' = 0.001 m, ''R'' = 0.0025 m:

:<math>B = \frac{(4\pi \times 10^{-7})(30)(0.001)}{2\pi (0.0025)^2} \approx 9.6 \times 10^{-4}\ \text{T} \approx 0.96\ \text{mT}</math>
===Middling===

Pick a closed, rectangular path in the xz plane with a height h and a width w. Calculate the speed v of the slab.

[[File: AMexample.png]]

At a time change in t, we can calculate the area as the height (h) times the speed over the change in time. ⩟ A = v (⩟ t) h
Because the electric field is constant in this region we can also calculate the change in electric flux over time as Evh(⩟ t) /(⩟ t) which is the same as Evh
Calculating the path integral for the magnetic field we get that ∮B . dl = Bh cos 0 = Bh
An important thing to notice is that there is no current I, so, using the Ampere-Maxwell Law we can see that

Bh = μ. [I+ ε. (vEh)]

but since there is no current,

B = μ. ε. (vE)

From Faraday's law we get that the emf equals the rate of change of the magnetic flux: Eh = Bvh

Substituting E = Bv into our previous equation we get that

B = μ. ε. (v(vB))

Solving for v we get that:

[[File: AMvelocity.png]]

The example shows that the speed of light relates the a time varying electric and magnetic field.

====What this implies====

In the example above we saw that the speed of light relates a time varying electric and magnetic field. This translates to the fact that an electromagnetic wave propagates at the speed of light.

[[File: AMwave.png]]

===Difficult===

[[File:Example3.jpg]]

[[File:Example3b.jpg]]

Question 1: First find the line integral of B around a loop of radius R located just outside the left capacitor plate.

Answer:

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I(t)</math>

Question 2: Now find an expression for the same line integral of B around the same loop located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the Ampère-Maxwell law above.

Answer:

:<math>\Phi_E = \int E \cdot dA = A E(t)</math>

:<math>I = ε_0 d \frac{(AE(t))}{dt} = A ε_0 \frac{dE(t)}{dt} </math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I</math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 A ε_0 \frac{dE(t)}{dt}</math>

Question 3: Express the normal current in terms of the charge on the capacitor plate.

Answer:

:<math>I = ε_0 \frac{d\frac{q}{ε_0}}{dt}</math>

the expression for the normal current is:

:<math>I = \frac{dq(t)}{dt}</math>

the current is directly related to charge on left capacitor plate

===Conceptual Questions===
[[File: AMconceptual.png|500px|Image: 500 pixels]]

The plot above shows a side and a top view of a capacitor with charge Q with electric and magnetic fields E and B at time t. The charge Q is:

1) Increasing in time

2) Decreasing in time

3) Constant in time

4) I don't know

Answer:

The unit vector N points out of the plane. Given that the magnetic field curls clockwise, the electric flux would be positive and decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q.

[[File: AMcapacitor.png|500px|Image: 500 pixels]]

Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is

1) Zero

2) Positive

3) Negative

4) Can’t tell

Answer:

One thing to notice is that there is no enclosed current through the disk. When integrating in the direction shown, the electric flux is positive. Because the plates are charging, the electric flux is increasing. Therefore the line integral is positive.

==Connectedness==

As a computer engineering student, the Ampere-Maxwell Law is foundational to almost every piece of hardware I work with. Every PCB trace carrying a high-frequency signal radiates because of changing electric fields between adjacent conductors, this is exactly the displacement current term in action. Signal integrity engineering, antenna design, and even the GHz-range coupling that causes crosstalk in dense chip layouts are all governed by this equation. Without Maxwell's correction, we would have no theoretical basis for wireless communication, RF circuits, or the high-speed serial links inside modern processors.

'''Interesting industrial application'''

A particularly elegant application is the '''wireless charging''' standard used in modern smartphones (Qi). Two flat coils: one in the charging pad, one in the phone, couple through a time-varying magnetic field. But at the higher-frequency variants (resonant inductive and especially capacitive wireless power transfer), the displacement current term becomes the dominant coupling mechanism. Capacitive wireless power transfer literally relies on the Ampere-Maxwell Law: there is no closed conductor between transmitter and receiver, only a changing electric flux that carries the energy across the gap.

==History==

Andre-Marie Ampere formulated his original law in 1826, relating magnetic fields to steady electric currents. For the next ~35 years it was treated as complete. The problem was that Ampere's Law, in its original form, only worked for closed steady currents — and a charging capacitor is the canonical case where it visibly fails.

In the early 1860s, '''James Clerk Maxwell''' was working to unify the existing laws of electricity and magnetism into a single mathematical framework. While studying Faraday's work on induction, Maxwell noticed the asymmetry: a changing magnetic field produced an electric field (Faraday's Law), but no equivalent law existed for a changing electric field producing a magnetic field. In his 1861 paper ''On Physical Lines of Force'' and more rigorously in his 1865 paper ''A Dynamical Theory of the Electromagnetic Field'', Maxwell introduced the displacement current term to fix Ampere's Law and restore symmetry.

The payoff was immediate and stunning. With the corrected equation, Maxwell could derive a wave equation directly from his four laws, and the predicted wave speed <math>1/\sqrt{\mu_0\epsilon_0}</math> came out to within experimental error of the measured speed of light. Maxwell concluded — correctly — that light itself is an electromagnetic wave. Heinrich Hertz experimentally confirmed the existence of these waves in 1887, and the entire modern world of radio, television, radar, and wireless communication followed.

== See also ==

The Ampere-Maxwell Law relates to the other Maxwell equations:

*[[Gauss's Law]]

*[[Magnetic Flux ]]

*[[Faraday's Law ]]

It might also help to read up on [[Maxwell's Electromagnetic Theory ]]

Also for continuity purposes, it can be helpful to look at the incomplete Ampere's Law:

*[[Ampere's Law]]

===Further Reading===

Matter and Interactions: Volume 2 by Ruth Chabay and Bruce Sherwood (4th Edition)

===External Links===

Maxwell's Equations:
http://study.com/academy/lesson/maxwells-equations-definition-application.html

James Maxwell Biography:
http://www.biography.com/people/james-c-maxwell-9403463

==References==

http://bulldog2.redlands.edu/fac/eric_hill/Phys232/Lectures/Ch%2023%20lect%201.pdf
http://ocw.mit.edu/high-school/physics/exam-prep/electromagnetism/maxwells-equations/
http://ocw.mit.edu/courses/physics/8-02-physics-ii-electricity-and-magnetism-spring-2007/readings/summary_w13d1.pdf
http://www.schoolphysics.co.uk/age16-19/Wave%20properties/Wave%20properties/text/Electromagnetic_radiation/index.html

[[Category:Which Category did you place this in?]]

Ampere-Maxwell Law

2026-04-27T03:31:58Z

Rohitnaras: /* Computational Model */

'''Claimed by Rohit Naras (Spring 2026)'''

The Ampere-Maxwell Law is one of the four [[Maxwell's Equations]] and represents a generalization of Ampere's Law. It states that magnetic fields are produced not only by electric currents but also by time-varying electric fields (the so-called "displacement current"). This addition by James Clerk Maxwell was the missing piece that unified electricity and magnetism, and it directly predicts the existence of electromagnetic waves traveling at the speed of light.

==The Main Idea==

Ampere's original law relates the magnetic field along a closed (Amperian) loop to the conduction current piercing any surface bounded by that loop. This works perfectly for steady currents in continuous wires, but it fails in a famous test case: a charging capacitor. If you draw an Amperian loop around the wire feeding the capacitor and choose a flat surface, you enclose a current ''I''. But if you bulge the surface so it passes between the capacitor plates instead, no conduction current pierces it at all — yet the magnetic field around the loop is the same. Ampere's Law gives two different answers for the same loop, which is a contradiction.

Maxwell resolved this by recognizing that between the plates, the electric field is changing in time, and this changing E-field must itself act as a source of magnetic field. He added a new term — the '''displacement current''' — proportional to the rate of change of electric flux. The corrected law is symmetric with [[Faraday's Law]]: just as a changing magnetic flux produces a curly electric field, a changing electric flux produces a curly magnetic field.

This symmetry is not just aesthetic. When you combine the Ampere-Maxwell Law with Faraday's Law, the equations support self-sustaining oscillations of E and B fields propagating through space at speed <math>c = 1/\sqrt{\mu_0 \epsilon_0} \approx 3 \times 10^8</math> m/s. Light is an electromagnetic wave, and the Ampere-Maxwell Law is half the reason why.

===A Mathematical Model===

The full Ampere-Maxwell Law is:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

where:

* <math>\oint \vec{B} \cdot d\vec{l}</math> is the line integral of the magnetic field around a closed Amperian loop
* <math>I_{enc}</math> is the conduction current passing through any surface bounded by that loop
* <math>\Phi_E = \int \vec{E} \cdot \hat{n}\, dA</math> is the electric flux through that same surface
* <math>\mu_0 = 4\pi \times 10^{-7}</math> T·m/A is the permeability of free space
* <math>\epsilon_0 = 8.854 \times 10^{-12}</math> C²/(N·m²) is the permittivity of free space

The second term, <math>\mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>, is Maxwell's contribution. The quantity <math>I_d = \epsilon_0 \frac{d\Phi_E}{dt}</math> is called the '''displacement current''', though it is not a true current of moving charges, it is a name given to a changing electric flux that has the same magnetic effect as a real current.

==Computational Model==

The simulation below uses VPython/GlowScript to visualize the Ampere-Maxwell Law for a charging parallel-plate capacitor. As the capacitor charges, the electric field between the plates grows, producing a changing electric flux. By the Ampere-Maxwell Law, this changing flux generates a curly magnetic field between the plates — even though no real current flows there.

The simulation shows:
* Two circular capacitor plates being charged by a current ''I''
* A growing electric field <math>\vec{E}</math> (red arrow) between the plates
* The induced curly magnetic field <math>\vec{B}</math> (blue arrows) circulating around the field axis
* Real-time graphs of <math>E(t)</math>, <math>dE/dt</math>, and the resulting <math>B(r)</math> at a fixed radius

You can run the simulation here: [Insert your GlowScript link after publishing — instructions below]

The full GlowScript code is also provided below for transparency and reproducibility.

<iframe src="https://trinket.io/embed/glowscript/faf9ad543e9a?start=result" width="100%" height="356" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe>

==Examples==

===Simple===

'''Question:''' A 30 A current is charging a parallel-plate capacitor with circular plates of radius ''R'' = 2.5 mm. The plate separation is 4 mm. What is the magnitude of the magnetic field between the plates at a radial distance ''r'' = 1 mm from the central axis?

[[File:Example1.png]]

'''Solution:'''

Apply the Ampere-Maxwell Law to a circular Amperian loop of radius ''r'' centered on the axis between the plates. There is no conduction current between the plates, so:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

The electric field between the plates of a capacitor is <math>E = \frac{Q}{\epsilon_0 A_{plate}}</math>, where <math>A_{plate} = \pi R^2</math>. The flux through our Amperian loop (area <math>\pi r^2</math>) is:

:<math>\Phi_E = E \cdot \pi r^2 = \frac{Q}{\epsilon_0 \pi R^2}\cdot \pi r^2 = \frac{Q r^2}{\epsilon_0 R^2}</math>

Taking the time derivative and recognizing <math>dQ/dt = I</math>:

:<math>\frac{d\Phi_E}{dt} = \frac{I r^2}{\epsilon_0 R^2}</math>

By symmetry, B is constant in magnitude around the loop, so <math>\oint \vec{B}\cdot d\vec{l} = B(2\pi r)</math>. Solving:

:<math>B = \frac{\mu_0 I r}{2\pi R^2}</math>

Plugging in ''I'' = 30 A, ''r'' = 0.001 m, ''R'' = 0.0025 m:

:<math>B = \frac{(4\pi \times 10^{-7})(30)(0.001)}{2\pi (0.0025)^2} \approx 9.6 \times 10^{-4}\ \text{T} \approx 0.96\ \text{mT}</math>
===Middling===

Pick a closed, rectangular path in the xz plane with a height h and a width w. Calculate the speed v of the slab.

[[File: AMexample.png]]

At a time change in t, we can calculate the area as the height (h) times the speed over the change in time. ⩟ A = v (⩟ t) h
Because the electric field is constant in this region we can also calculate the change in electric flux over time as Evh(⩟ t) /(⩟ t) which is the same as Evh
Calculating the path integral for the magnetic field we get that ∮B . dl = Bh cos 0 = Bh
An important thing to notice is that there is no current I, so, using the Ampere-Maxwell Law we can see that

Bh = μ. [I+ ε. (vEh)]

but since there is no current,

B = μ. ε. (vE)

From Faraday's law we get that the emf equals the rate of change of the magnetic flux: Eh = Bvh

Substituting E = Bv into our previous equation we get that

B = μ. ε. (v(vB))

Solving for v we get that:

[[File: AMvelocity.png]]

The example shows that the speed of light relates the a time varying electric and magnetic field.

====What this implies====

In the example above we saw that the speed of light relates a time varying electric and magnetic field. This translates to the fact that an electromagnetic wave propagates at the speed of light.

[[File: AMwave.png]]

===Difficult===

[[File:Example3.jpg]]

[[File:Example3b.jpg]]

Question 1: First find the line integral of B around a loop of radius R located just outside the left capacitor plate.

Answer:

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I(t)</math>

Question 2: Now find an expression for the same line integral of B around the same loop located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the Ampère-Maxwell law above.

Answer:

:<math>\Phi_E = \int E \cdot dA = A E(t)</math>

:<math>I = ε_0 d \frac{(AE(t))}{dt} = A ε_0 \frac{dE(t)}{dt} </math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I</math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 A ε_0 \frac{dE(t)}{dt}</math>

Question 3: Express the normal current in terms of the charge on the capacitor plate.

Answer:

:<math>I = ε_0 \frac{d\frac{q}{ε_0}}{dt}</math>

the expression for the normal current is:

:<math>I = \frac{dq(t)}{dt}</math>

the current is directly related to charge on left capacitor plate

===Conceptual Questions===
[[File: AMconceptual.png|500px|Image: 500 pixels]]

The plot above shows a side and a top view of a capacitor with charge Q with electric and magnetic fields E and B at time t. The charge Q is:

1) Increasing in time

2) Decreasing in time

3) Constant in time

4) I don't know

Answer:

The unit vector N points out of the plane. Given that the magnetic field curls clockwise, the electric flux would be positive and decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q.

[[File: AMcapacitor.png|500px|Image: 500 pixels]]

Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is

1) Zero

2) Positive

3) Negative

4) Can’t tell

Answer:

One thing to notice is that there is no enclosed current through the disk. When integrating in the direction shown, the electric flux is positive. Because the plates are charging, the electric flux is increasing. Therefore the line integral is positive.

==Connectedness==

As a computer engineering student, the Ampere-Maxwell Law is foundational to almost every piece of hardware I work with. Every PCB trace carrying a high-frequency signal radiates because of changing electric fields between adjacent conductors, this is exactly the displacement current term in action. Signal integrity engineering, antenna design, and even the GHz-range coupling that causes crosstalk in dense chip layouts are all governed by this equation. Without Maxwell's correction, we would have no theoretical basis for wireless communication, RF circuits, or the high-speed serial links inside modern processors.

'''Interesting industrial application'''

A particularly elegant application is the '''wireless charging''' standard used in modern smartphones (Qi). Two flat coils: one in the charging pad, one in the phone, couple through a time-varying magnetic field. But at the higher-frequency variants (resonant inductive and especially capacitive wireless power transfer), the displacement current term becomes the dominant coupling mechanism. Capacitive wireless power transfer literally relies on the Ampere-Maxwell Law: there is no closed conductor between transmitter and receiver, only a changing electric flux that carries the energy across the gap.

==History==

Andre-Marie Ampere formulated his original law in 1826, relating magnetic fields to steady electric currents. For the next ~35 years it was treated as complete. The problem was that Ampere's Law, in its original form, only worked for closed steady currents — and a charging capacitor is the canonical case where it visibly fails.

In the early 1860s, '''James Clerk Maxwell''' was working to unify the existing laws of electricity and magnetism into a single mathematical framework. While studying Faraday's work on induction, Maxwell noticed the asymmetry: a changing magnetic field produced an electric field (Faraday's Law), but no equivalent law existed for a changing electric field producing a magnetic field. In his 1861 paper ''On Physical Lines of Force'' and more rigorously in his 1865 paper ''A Dynamical Theory of the Electromagnetic Field'', Maxwell introduced the displacement current term to fix Ampere's Law and restore symmetry.

The payoff was immediate and stunning. With the corrected equation, Maxwell could derive a wave equation directly from his four laws, and the predicted wave speed <math>1/\sqrt{\mu_0\epsilon_0}</math> came out to within experimental error of the measured speed of light. Maxwell concluded — correctly — that light itself is an electromagnetic wave. Heinrich Hertz experimentally confirmed the existence of these waves in 1887, and the entire modern world of radio, television, radar, and wireless communication followed.

== See also ==

The Ampere-Maxwell Law relates to the other Maxwell equations:

*[[Gauss's Law]]

*[[Magnetic Flux ]]

*[[Faraday's Law ]]

It might also help to read up on [[Maxwell's Electromagnetic Theory ]]

Also for continuity purposes, it can be helpful to look at the incomplete Ampere's Law:

*[[Ampere's Law]]

===Further Reading===

Matter and Interactions: Volume 2 by Ruth Chabay and Bruce Sherwood (4th Edition)

===External Links===

Maxwell's Equations:
http://study.com/academy/lesson/maxwells-equations-definition-application.html

James Maxwell Biography:
http://www.biography.com/people/james-c-maxwell-9403463

==References==

http://bulldog2.redlands.edu/fac/eric_hill/Phys232/Lectures/Ch%2023%20lect%201.pdf
http://ocw.mit.edu/high-school/physics/exam-prep/electromagnetism/maxwells-equations/
http://ocw.mit.edu/courses/physics/8-02-physics-ii-electricity-and-magnetism-spring-2007/readings/summary_w13d1.pdf
http://www.schoolphysics.co.uk/age16-19/Wave%20properties/Wave%20properties/text/Electromagnetic_radiation/index.html

[[Category:Which Category did you place this in?]]

Ampere-Maxwell Law

2026-04-27T03:29:27Z

Rohitnaras:

'''Claimed by Rohit Naras (Spring 2026)'''

The Ampere-Maxwell Law is one of the four [[Maxwell's Equations]] and represents a generalization of Ampere's Law. It states that magnetic fields are produced not only by electric currents but also by time-varying electric fields (the so-called "displacement current"). This addition by James Clerk Maxwell was the missing piece that unified electricity and magnetism, and it directly predicts the existence of electromagnetic waves traveling at the speed of light.

==The Main Idea==

Ampere's original law relates the magnetic field along a closed (Amperian) loop to the conduction current piercing any surface bounded by that loop. This works perfectly for steady currents in continuous wires, but it fails in a famous test case: a charging capacitor. If you draw an Amperian loop around the wire feeding the capacitor and choose a flat surface, you enclose a current ''I''. But if you bulge the surface so it passes between the capacitor plates instead, no conduction current pierces it at all — yet the magnetic field around the loop is the same. Ampere's Law gives two different answers for the same loop, which is a contradiction.

Maxwell resolved this by recognizing that between the plates, the electric field is changing in time, and this changing E-field must itself act as a source of magnetic field. He added a new term — the '''displacement current''' — proportional to the rate of change of electric flux. The corrected law is symmetric with [[Faraday's Law]]: just as a changing magnetic flux produces a curly electric field, a changing electric flux produces a curly magnetic field.

This symmetry is not just aesthetic. When you combine the Ampere-Maxwell Law with Faraday's Law, the equations support self-sustaining oscillations of E and B fields propagating through space at speed <math>c = 1/\sqrt{\mu_0 \epsilon_0} \approx 3 \times 10^8</math> m/s. Light is an electromagnetic wave, and the Ampere-Maxwell Law is half the reason why.

===A Mathematical Model===

The full Ampere-Maxwell Law is:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

where:

* <math>\oint \vec{B} \cdot d\vec{l}</math> is the line integral of the magnetic field around a closed Amperian loop
* <math>I_{enc}</math> is the conduction current passing through any surface bounded by that loop
* <math>\Phi_E = \int \vec{E} \cdot \hat{n}\, dA</math> is the electric flux through that same surface
* <math>\mu_0 = 4\pi \times 10^{-7}</math> T·m/A is the permeability of free space
* <math>\epsilon_0 = 8.854 \times 10^{-12}</math> C²/(N·m²) is the permittivity of free space

The second term, <math>\mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>, is Maxwell's contribution. The quantity <math>I_d = \epsilon_0 \frac{d\Phi_E}{dt}</math> is called the '''displacement current''', though it is not a true current of moving charges, it is a name given to a changing electric flux that has the same magnetic effect as a real current.

==Computational Model==

The simulation below uses VPython/GlowScript to visualize the Ampere-Maxwell Law for a charging parallel-plate capacitor. As the capacitor charges, the electric field between the plates grows, producing a changing electric flux. By the Ampere-Maxwell Law, this changing flux generates a curly magnetic field between the plates — even though no real current flows there.

The simulation shows:
* Two circular capacitor plates being charged by a current ''I''
* A growing electric field <math>\vec{E}</math> (red arrow) between the plates
* The induced curly magnetic field <math>\vec{B}</math> (blue arrows) circulating around the field axis
* Real-time graphs of <math>E(t)</math>, <math>dE/dt</math>, and the resulting <math>B(r)</math> at a fixed radius

You can run the simulation here: [Insert your GlowScript link after publishing — instructions below]

The full GlowScript code is also provided below for transparency and reproducibility.

==Examples==

===Simple===

'''Question:''' A 30 A current is charging a parallel-plate capacitor with circular plates of radius ''R'' = 2.5 mm. The plate separation is 4 mm. What is the magnitude of the magnetic field between the plates at a radial distance ''r'' = 1 mm from the central axis?

[[File:Example1.png]]

'''Solution:'''

Apply the Ampere-Maxwell Law to a circular Amperian loop of radius ''r'' centered on the axis between the plates. There is no conduction current between the plates, so:

:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

The electric field between the plates of a capacitor is <math>E = \frac{Q}{\epsilon_0 A_{plate}}</math>, where <math>A_{plate} = \pi R^2</math>. The flux through our Amperian loop (area <math>\pi r^2</math>) is:

:<math>\Phi_E = E \cdot \pi r^2 = \frac{Q}{\epsilon_0 \pi R^2}\cdot \pi r^2 = \frac{Q r^2}{\epsilon_0 R^2}</math>

Taking the time derivative and recognizing <math>dQ/dt = I</math>:

:<math>\frac{d\Phi_E}{dt} = \frac{I r^2}{\epsilon_0 R^2}</math>

By symmetry, B is constant in magnitude around the loop, so <math>\oint \vec{B}\cdot d\vec{l} = B(2\pi r)</math>. Solving:

:<math>B = \frac{\mu_0 I r}{2\pi R^2}</math>

Plugging in ''I'' = 30 A, ''r'' = 0.001 m, ''R'' = 0.0025 m:

:<math>B = \frac{(4\pi \times 10^{-7})(30)(0.001)}{2\pi (0.0025)^2} \approx 9.6 \times 10^{-4}\ \text{T} \approx 0.96\ \text{mT}</math>
===Middling===

Pick a closed, rectangular path in the xz plane with a height h and a width w. Calculate the speed v of the slab.

[[File: AMexample.png]]

At a time change in t, we can calculate the area as the height (h) times the speed over the change in time. ⩟ A = v (⩟ t) h
Because the electric field is constant in this region we can also calculate the change in electric flux over time as Evh(⩟ t) /(⩟ t) which is the same as Evh
Calculating the path integral for the magnetic field we get that ∮B . dl = Bh cos 0 = Bh
An important thing to notice is that there is no current I, so, using the Ampere-Maxwell Law we can see that

Bh = μ. [I+ ε. (vEh)]

but since there is no current,

B = μ. ε. (vE)

From Faraday's law we get that the emf equals the rate of change of the magnetic flux: Eh = Bvh

Substituting E = Bv into our previous equation we get that

B = μ. ε. (v(vB))

Solving for v we get that:

[[File: AMvelocity.png]]

The example shows that the speed of light relates the a time varying electric and magnetic field.

====What this implies====

In the example above we saw that the speed of light relates a time varying electric and magnetic field. This translates to the fact that an electromagnetic wave propagates at the speed of light.

[[File: AMwave.png]]

===Difficult===

[[File:Example3.jpg]]

[[File:Example3b.jpg]]

Question 1: First find the line integral of B around a loop of radius R located just outside the left capacitor plate.

Answer:

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I(t)</math>

Question 2: Now find an expression for the same line integral of B around the same loop located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the Ampère-Maxwell law above.

Answer:

:<math>\Phi_E = \int E \cdot dA = A E(t)</math>

:<math>I = ε_0 d \frac{(AE(t))}{dt} = A ε_0 \frac{dE(t)}{dt} </math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I</math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 A ε_0 \frac{dE(t)}{dt}</math>

Question 3: Express the normal current in terms of the charge on the capacitor plate.

Answer:

:<math>I = ε_0 \frac{d\frac{q}{ε_0}}{dt}</math>

the expression for the normal current is:

:<math>I = \frac{dq(t)}{dt}</math>

the current is directly related to charge on left capacitor plate

===Conceptual Questions===
[[File: AMconceptual.png|500px|Image: 500 pixels]]

The plot above shows a side and a top view of a capacitor with charge Q with electric and magnetic fields E and B at time t. The charge Q is:

1) Increasing in time

2) Decreasing in time

3) Constant in time

4) I don't know

Answer:

The unit vector N points out of the plane. Given that the magnetic field curls clockwise, the electric flux would be positive and decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q.

[[File: AMcapacitor.png|500px|Image: 500 pixels]]

Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is

1) Zero

2) Positive

3) Negative

4) Can’t tell

Answer:

One thing to notice is that there is no enclosed current through the disk. When integrating in the direction shown, the electric flux is positive. Because the plates are charging, the electric flux is increasing. Therefore the line integral is positive.

==Connectedness==

As a computer engineering student, the Ampere-Maxwell Law is foundational to almost every piece of hardware I work with. Every PCB trace carrying a high-frequency signal radiates because of changing electric fields between adjacent conductors, this is exactly the displacement current term in action. Signal integrity engineering, antenna design, and even the GHz-range coupling that causes crosstalk in dense chip layouts are all governed by this equation. Without Maxwell's correction, we would have no theoretical basis for wireless communication, RF circuits, or the high-speed serial links inside modern processors.

'''Interesting industrial application'''

A particularly elegant application is the '''wireless charging''' standard used in modern smartphones (Qi). Two flat coils: one in the charging pad, one in the phone, couple through a time-varying magnetic field. But at the higher-frequency variants (resonant inductive and especially capacitive wireless power transfer), the displacement current term becomes the dominant coupling mechanism. Capacitive wireless power transfer literally relies on the Ampere-Maxwell Law: there is no closed conductor between transmitter and receiver, only a changing electric flux that carries the energy across the gap.

==History==

Andre-Marie Ampere formulated his original law in 1826, relating magnetic fields to steady electric currents. For the next ~35 years it was treated as complete. The problem was that Ampere's Law, in its original form, only worked for closed steady currents — and a charging capacitor is the canonical case where it visibly fails.

In the early 1860s, '''James Clerk Maxwell''' was working to unify the existing laws of electricity and magnetism into a single mathematical framework. While studying Faraday's work on induction, Maxwell noticed the asymmetry: a changing magnetic field produced an electric field (Faraday's Law), but no equivalent law existed for a changing electric field producing a magnetic field. In his 1861 paper ''On Physical Lines of Force'' and more rigorously in his 1865 paper ''A Dynamical Theory of the Electromagnetic Field'', Maxwell introduced the displacement current term to fix Ampere's Law and restore symmetry.

The payoff was immediate and stunning. With the corrected equation, Maxwell could derive a wave equation directly from his four laws, and the predicted wave speed <math>1/\sqrt{\mu_0\epsilon_0}</math> came out to within experimental error of the measured speed of light. Maxwell concluded — correctly — that light itself is an electromagnetic wave. Heinrich Hertz experimentally confirmed the existence of these waves in 1887, and the entire modern world of radio, television, radar, and wireless communication followed.

== See also ==

The Ampere-Maxwell Law relates to the other Maxwell equations:

*[[Gauss's Law]]

*[[Magnetic Flux ]]

*[[Faraday's Law ]]

It might also help to read up on [[Maxwell's Electromagnetic Theory ]]

Also for continuity purposes, it can be helpful to look at the incomplete Ampere's Law:

*[[Ampere's Law]]

===Further Reading===

Matter and Interactions: Volume 2 by Ruth Chabay and Bruce Sherwood (4th Edition)

===External Links===

Maxwell's Equations:
http://study.com/academy/lesson/maxwells-equations-definition-application.html

James Maxwell Biography:
http://www.biography.com/people/james-c-maxwell-9403463

==References==

http://bulldog2.redlands.edu/fac/eric_hill/Phys232/Lectures/Ch%2023%20lect%201.pdf
http://ocw.mit.edu/high-school/physics/exam-prep/electromagnetism/maxwells-equations/
http://ocw.mit.edu/courses/physics/8-02-physics-ii-electricity-and-magnetism-spring-2007/readings/summary_w13d1.pdf
http://www.schoolphysics.co.uk/age16-19/Wave%20properties/Wave%20properties/text/Electromagnetic_radiation/index.html

[[Category:Which Category did you place this in?]]

Ampere-Maxwell Law

2026-04-27T03:00:42Z

Rohitnaras:

'''Claimed by Rohit Naras (Spring 2026)'''

==Ampere-Maxwell Law==

The Ampere-Maxwell Law expands on Ampere's Law relating magnetic field and current on a closed loop. Ampere's Law:

[[File: Amperes.PNG]]

allowed us to explore the relationship between current and magnetic field on a chosen closed loop called an "Amperian path". However, for certain examples in which electric field varies with time, simply using Ampere's Law is not sufficient. The Ampere-Maxwell Law accounts for these situations and establishes that a time dependent electric field is associated with a corresponding magnetic field.

The Ampere-Maxwell Law is also known as the Ampere's Circuital Law, and should not be confused with Ampere's Force Law.

===Mathematical Model===

[[File: Ampere-Maxwell.png]]

Where

*B is the magnetic field,

*dl is the change in path, and

*the sum of I is the sum of the charges inside the path.

==Examples==

===Simple===

Question: A 30-A current is charging a capacitor that has circular plates 5 mm in diameter. The plate separation is 4 mm. What is the magnetic field between the plates 1 mm from the center?

[[File:Example1.png]]

Answer:
:<math>\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 ε_0 \frac{d\Phi_E}{dt}</math>

this is simply the Ampere-Maxwell Law

:<math>\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 ε_0 \frac{d}{dt}(\frac{q \pi r^2}{ε_0 A})</math>

this step breaks down the electric flux part of the equation

:<math>\mathbf{B} \cdot 2\pi r = \mu_0 ε_0 \frac{dq}{dt}\frac{\pi r^2}{ε_0 A}</math>

Since the path is a circle, replace dl with circumference of the circle

:<math>\mathbf{B} = \frac{\mu_0 I r}{2A} = \frac{\mu_0 I r}{2 \pi R^2}</math>

Continue to simplify and plug in I for dq/dt and plug in values (where I = 30 A, r = 1 mm, R = 2.5 mm)

:<math>\mathbf{B} = 1 mT</math>

===Middling===

Pick a closed, rectangular path in the xz plane with a height h and a width w. Calculate the speed v of the slab.

[[File: AMexample.png]]

At a time change in t, we can calculate the area as the height (h) times the speed over the change in time. ⩟ A = v (⩟ t) h
Because the electric field is constant in this region we can also calculate the change in electric flux over time as Evh(⩟ t) /(⩟ t) which is the same as Evh
Calculating the path integral for the magnetic field we get that ∮B . dl = Bh cos 0 = Bh
An important thing to notice is that there is no current I, so, using the Ampere-Maxwell Law we can see that

Bh = μ. [I+ ε. (vEh)]

but since there is no current,

B = μ. ε. (vE)

From Faraday's law we get that the emf equals the rate of change of the magnetic flux: Eh = Bvh

Substituting E = Bv into our previous equation we get that

B = μ. ε. (v(vB))

Solving for v we get that:

[[File: AMvelocity.png]]

The example shows that the speed of light relates the a time varying electric and magnetic field.

====What this implies====

In the example above we saw that the speed of light relates a time varying electric and magnetic field. This translates to the fact that an electromagnetic wave propagates at the speed of light.

[[File: AMwave.png]]

===Difficult===

[[File:Example3.jpg]]

[[File:Example3b.jpg]]

Question 1: First find the line integral of B around a loop of radius R located just outside the left capacitor plate.

Answer:

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I(t)</math>

Question 2: Now find an expression for the same line integral of B around the same loop located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the Ampère-Maxwell law above.

Answer:

:<math>\Phi_E = \int E \cdot dA = A E(t)</math>

:<math>I = ε_0 d \frac{(AE(t))}{dt} = A ε_0 \frac{dE(t)}{dt} </math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I</math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 A ε_0 \frac{dE(t)}{dt}</math>

Question 3: Express the normal current in terms of the charge on the capacitor plate.

Answer:

:<math>I = ε_0 \frac{d\frac{q}{ε_0}}{dt}</math>

the expression for the normal current is:

:<math>I = \frac{dq(t)}{dt}</math>

the current is directly related to charge on left capacitor plate

===Conceptual Questions===
[[File: AMconceptual.png|500px|Image: 500 pixels]]

The plot above shows a side and a top view of a capacitor with charge Q with electric and magnetic fields E and B at time t. The charge Q is:

1) Increasing in time

2) Decreasing in time

3) Constant in time

4) I don't know

Answer:

The unit vector N points out of the plane. Given that the magnetic field curls clockwise, the electric flux would be positive and decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q.

[[File: AMcapacitor.png|500px|Image: 500 pixels]]

Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is

1) Zero

2) Positive

3) Negative

4) Can’t tell

Answer:

One thing to notice is that there is no enclosed current through the disk. When integrating in the direction shown, the electric flux is positive. Because the plates are charging, the electric flux is increasing. Therefore the line integral is positive.

==Connectedness==

The Ampere-Maxwell Law is extremely useful in many aspects of life. It relates current and time-varying electric fields and allow us to derive a magnetic field from these situations. The Ampere-Maxwell Law can help with understanding and building generators and transformers.

==History==

You may remember Ampere's Law equation from chapter 21 of the textbook ([[Ampere's Law]]). This equation had to be modified to address time-varying electric fields and their associated magnetic fields. If Ampere's Law were applied to a situation with a time-varying electric field, at different points along the chosen Amperian path, different answers for the current enclosed would be obtained. In the same way that Faraday discovered that a time-varying magnetic field is accompanied by an electric field, James Clerk Maxwell resolved this issue when he discovered that an electric field that changed with time was accompanied by a resulting magnetic field. Specifically, when the plates of a capacitor grows as the capacitor is charged, the electric field changes (with time) which produces a magnetic field.

== See also ==

The Ampere-Maxwell Law relates to the other Maxwell equations:

*[[Gauss's Law]]

*[[Magnetic Flux ]]

*[[Faraday's Law ]]

It might also help to read up on [[Maxwell's Electromagnetic Theory ]]

Also for continuity purposes, it can be helpful to look at the incomplete Ampere's Law:

*[[Ampere's Law]]

===Further Reading===

Matter and Interactions: Volume 2 by Ruth Chabay and Bruce Sherwood (4th Edition)

===External Links===

Maxwell's Equations:
http://study.com/academy/lesson/maxwells-equations-definition-application.html

James Maxwell Biography:
http://www.biography.com/people/james-c-maxwell-9403463

==References==

http://bulldog2.redlands.edu/fac/eric_hill/Phys232/Lectures/Ch%2023%20lect%201.pdf
http://ocw.mit.edu/high-school/physics/exam-prep/electromagnetism/maxwells-equations/
http://ocw.mit.edu/courses/physics/8-02-physics-ii-electricity-and-magnetism-spring-2007/readings/summary_w13d1.pdf
http://www.schoolphysics.co.uk/age16-19/Wave%20properties/Wave%20properties/text/Electromagnetic_radiation/index.html

[[Category:Which Category did you place this in?]]

Ampere-Maxwell Law

2026-04-27T02:59:54Z

Rohitnaras:

Claimed by Maria Rivero
Claimed by Yeon Jae Cho (FALL 2016)

'''Claimed by Stella Chen (Fall 2017)'''
'''Claimed by Rohit Naras (Spring 2026)'''

==Ampere-Maxwell Law==

The Ampere-Maxwell Law expands on Ampere's Law relating magnetic field and current on a closed loop. Ampere's Law:

[[File: Amperes.PNG]]

allowed us to explore the relationship between current and magnetic field on a chosen closed loop called an "Amperian path". However, for certain examples in which electric field varies with time, simply using Ampere's Law is not sufficient. The Ampere-Maxwell Law accounts for these situations and establishes that a time dependent electric field is associated with a corresponding magnetic field.

The Ampere-Maxwell Law is also known as the Ampere's Circuital Law, and should not be confused with Ampere's Force Law.

===Mathematical Model===

[[File: Ampere-Maxwell.png]]

Where

*B is the magnetic field,

*dl is the change in path, and

*the sum of I is the sum of the charges inside the path.

==Examples==

===Simple===

Question: A 30-A current is charging a capacitor that has circular plates 5 mm in diameter. The plate separation is 4 mm. What is the magnetic field between the plates 1 mm from the center?

[[File:Example1.png]]

Answer:
:<math>\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 ε_0 \frac{d\Phi_E}{dt}</math>

this is simply the Ampere-Maxwell Law

:<math>\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 ε_0 \frac{d}{dt}(\frac{q \pi r^2}{ε_0 A})</math>

this step breaks down the electric flux part of the equation

:<math>\mathbf{B} \cdot 2\pi r = \mu_0 ε_0 \frac{dq}{dt}\frac{\pi r^2}{ε_0 A}</math>

Since the path is a circle, replace dl with circumference of the circle

:<math>\mathbf{B} = \frac{\mu_0 I r}{2A} = \frac{\mu_0 I r}{2 \pi R^2}</math>

Continue to simplify and plug in I for dq/dt and plug in values (where I = 30 A, r = 1 mm, R = 2.5 mm)

:<math>\mathbf{B} = 1 mT</math>

===Middling===

Pick a closed, rectangular path in the xz plane with a height h and a width w. Calculate the speed v of the slab.

[[File: AMexample.png]]

At a time change in t, we can calculate the area as the height (h) times the speed over the change in time. ⩟ A = v (⩟ t) h
Because the electric field is constant in this region we can also calculate the change in electric flux over time as Evh(⩟ t) /(⩟ t) which is the same as Evh
Calculating the path integral for the magnetic field we get that ∮B . dl = Bh cos 0 = Bh
An important thing to notice is that there is no current I, so, using the Ampere-Maxwell Law we can see that

Bh = μ. [I+ ε. (vEh)]

but since there is no current,

B = μ. ε. (vE)

From Faraday's law we get that the emf equals the rate of change of the magnetic flux: Eh = Bvh

Substituting E = Bv into our previous equation we get that

B = μ. ε. (v(vB))

Solving for v we get that:

[[File: AMvelocity.png]]

The example shows that the speed of light relates the a time varying electric and magnetic field.

====What this implies====

In the example above we saw that the speed of light relates a time varying electric and magnetic field. This translates to the fact that an electromagnetic wave propagates at the speed of light.

[[File: AMwave.png]]

===Difficult===

[[File:Example3.jpg]]

[[File:Example3b.jpg]]

Question 1: First find the line integral of B around a loop of radius R located just outside the left capacitor plate.

Answer:

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I(t)</math>

Question 2: Now find an expression for the same line integral of B around the same loop located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the Ampère-Maxwell law above.

Answer:

:<math>\Phi_E = \int E \cdot dA = A E(t)</math>

:<math>I = ε_0 d \frac{(AE(t))}{dt} = A ε_0 \frac{dE(t)}{dt} </math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I</math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 A ε_0 \frac{dE(t)}{dt}</math>

Question 3: Express the normal current in terms of the charge on the capacitor plate.

Answer:

:<math>I = ε_0 \frac{d\frac{q}{ε_0}}{dt}</math>

the expression for the normal current is:

:<math>I = \frac{dq(t)}{dt}</math>

the current is directly related to charge on left capacitor plate

===Conceptual Questions===
[[File: AMconceptual.png|500px|Image: 500 pixels]]

The plot above shows a side and a top view of a capacitor with charge Q with electric and magnetic fields E and B at time t. The charge Q is:

1) Increasing in time

2) Decreasing in time

3) Constant in time

4) I don't know

Answer:

The unit vector N points out of the plane. Given that the magnetic field curls clockwise, the electric flux would be positive and decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q.

[[File: AMcapacitor.png|500px|Image: 500 pixels]]

Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is

1) Zero

2) Positive

3) Negative

4) Can’t tell

Answer:

One thing to notice is that there is no enclosed current through the disk. When integrating in the direction shown, the electric flux is positive. Because the plates are charging, the electric flux is increasing. Therefore the line integral is positive.

==Connectedness==

The Ampere-Maxwell Law is extremely useful in many aspects of life. It relates current and time-varying electric fields and allow us to derive a magnetic field from these situations. The Ampere-Maxwell Law can help with understanding and building generators and transformers.

==History==

You may remember Ampere's Law equation from chapter 21 of the textbook ([[Ampere's Law]]). This equation had to be modified to address time-varying electric fields and their associated magnetic fields. If Ampere's Law were applied to a situation with a time-varying electric field, at different points along the chosen Amperian path, different answers for the current enclosed would be obtained. In the same way that Faraday discovered that a time-varying magnetic field is accompanied by an electric field, James Clerk Maxwell resolved this issue when he discovered that an electric field that changed with time was accompanied by a resulting magnetic field. Specifically, when the plates of a capacitor grows as the capacitor is charged, the electric field changes (with time) which produces a magnetic field.

== See also ==

The Ampere-Maxwell Law relates to the other Maxwell equations:

*[[Gauss's Law]]

*[[Magnetic Flux ]]

*[[Faraday's Law ]]

It might also help to read up on [[Maxwell's Electromagnetic Theory ]]

Also for continuity purposes, it can be helpful to look at the incomplete Ampere's Law:

*[[Ampere's Law]]

===Further Reading===

Matter and Interactions: Volume 2 by Ruth Chabay and Bruce Sherwood (4th Edition)

===External Links===

Maxwell's Equations:
http://study.com/academy/lesson/maxwells-equations-definition-application.html

James Maxwell Biography:
http://www.biography.com/people/james-c-maxwell-9403463

==References==

http://bulldog2.redlands.edu/fac/eric_hill/Phys232/Lectures/Ch%2023%20lect%201.pdf
http://ocw.mit.edu/high-school/physics/exam-prep/electromagnetism/maxwells-equations/
http://ocw.mit.edu/courses/physics/8-02-physics-ii-electricity-and-magnetism-spring-2007/readings/summary_w13d1.pdf
http://www.schoolphysics.co.uk/age16-19/Wave%20properties/Wave%20properties/text/Electromagnetic_radiation/index.html

[[Category:Which Category did you place this in?]]

Ampere-Maxwell Law

2026-04-27T02:59:42Z

Rohitnaras:

Claimed by Maria Rivero
Claimed by Yeon Jae Cho (FALL 2016)

'''Claimed by Stella Chen (Fall 2017)'''
'''Claimed by Rohit Naras (Spring 2026)'''/n

==Ampere-Maxwell Law==

The Ampere-Maxwell Law expands on Ampere's Law relating magnetic field and current on a closed loop. Ampere's Law:

[[File: Amperes.PNG]]

allowed us to explore the relationship between current and magnetic field on a chosen closed loop called an "Amperian path". However, for certain examples in which electric field varies with time, simply using Ampere's Law is not sufficient. The Ampere-Maxwell Law accounts for these situations and establishes that a time dependent electric field is associated with a corresponding magnetic field.

The Ampere-Maxwell Law is also known as the Ampere's Circuital Law, and should not be confused with Ampere's Force Law.

===Mathematical Model===

[[File: Ampere-Maxwell.png]]

Where

*B is the magnetic field,

*dl is the change in path, and

*the sum of I is the sum of the charges inside the path.

==Examples==

===Simple===

Question: A 30-A current is charging a capacitor that has circular plates 5 mm in diameter. The plate separation is 4 mm. What is the magnetic field between the plates 1 mm from the center?

[[File:Example1.png]]

Answer:
:<math>\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 ε_0 \frac{d\Phi_E}{dt}</math>

this is simply the Ampere-Maxwell Law

:<math>\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 ε_0 \frac{d}{dt}(\frac{q \pi r^2}{ε_0 A})</math>

this step breaks down the electric flux part of the equation

:<math>\mathbf{B} \cdot 2\pi r = \mu_0 ε_0 \frac{dq}{dt}\frac{\pi r^2}{ε_0 A}</math>

Since the path is a circle, replace dl with circumference of the circle

:<math>\mathbf{B} = \frac{\mu_0 I r}{2A} = \frac{\mu_0 I r}{2 \pi R^2}</math>

Continue to simplify and plug in I for dq/dt and plug in values (where I = 30 A, r = 1 mm, R = 2.5 mm)

:<math>\mathbf{B} = 1 mT</math>

===Middling===

Pick a closed, rectangular path in the xz plane with a height h and a width w. Calculate the speed v of the slab.

[[File: AMexample.png]]

At a time change in t, we can calculate the area as the height (h) times the speed over the change in time. ⩟ A = v (⩟ t) h
Because the electric field is constant in this region we can also calculate the change in electric flux over time as Evh(⩟ t) /(⩟ t) which is the same as Evh
Calculating the path integral for the magnetic field we get that ∮B . dl = Bh cos 0 = Bh
An important thing to notice is that there is no current I, so, using the Ampere-Maxwell Law we can see that

Bh = μ. [I+ ε. (vEh)]

but since there is no current,

B = μ. ε. (vE)

From Faraday's law we get that the emf equals the rate of change of the magnetic flux: Eh = Bvh

Substituting E = Bv into our previous equation we get that

B = μ. ε. (v(vB))

Solving for v we get that:

[[File: AMvelocity.png]]

The example shows that the speed of light relates the a time varying electric and magnetic field.

====What this implies====

In the example above we saw that the speed of light relates a time varying electric and magnetic field. This translates to the fact that an electromagnetic wave propagates at the speed of light.

[[File: AMwave.png]]

===Difficult===

[[File:Example3.jpg]]

[[File:Example3b.jpg]]

Question 1: First find the line integral of B around a loop of radius R located just outside the left capacitor plate.

Answer:

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I(t)</math>

Question 2: Now find an expression for the same line integral of B around the same loop located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the Ampère-Maxwell law above.

Answer:

:<math>\Phi_E = \int E \cdot dA = A E(t)</math>

:<math>I = ε_0 d \frac{(AE(t))}{dt} = A ε_0 \frac{dE(t)}{dt} </math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 I</math>

:<math>\int \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 A ε_0 \frac{dE(t)}{dt}</math>

Question 3: Express the normal current in terms of the charge on the capacitor plate.

Answer:

:<math>I = ε_0 \frac{d\frac{q}{ε_0}}{dt}</math>

the expression for the normal current is:

:<math>I = \frac{dq(t)}{dt}</math>

the current is directly related to charge on left capacitor plate

===Conceptual Questions===
[[File: AMconceptual.png|500px|Image: 500 pixels]]

The plot above shows a side and a top view of a capacitor with charge Q with electric and magnetic fields E and B at time t. The charge Q is:

1) Increasing in time

2) Decreasing in time

3) Constant in time

4) I don't know

Answer:

The unit vector N points out of the plane. Given that the magnetic field curls clockwise, the electric flux would be positive and decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q.

[[File: AMcapacitor.png|500px|Image: 500 pixels]]

Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is

1) Zero

2) Positive

3) Negative

4) Can’t tell

Answer:

One thing to notice is that there is no enclosed current through the disk. When integrating in the direction shown, the electric flux is positive. Because the plates are charging, the electric flux is increasing. Therefore the line integral is positive.

==Connectedness==

The Ampere-Maxwell Law is extremely useful in many aspects of life. It relates current and time-varying electric fields and allow us to derive a magnetic field from these situations. The Ampere-Maxwell Law can help with understanding and building generators and transformers.

==History==

You may remember Ampere's Law equation from chapter 21 of the textbook ([[Ampere's Law]]). This equation had to be modified to address time-varying electric fields and their associated magnetic fields. If Ampere's Law were applied to a situation with a time-varying electric field, at different points along the chosen Amperian path, different answers for the current enclosed would be obtained. In the same way that Faraday discovered that a time-varying magnetic field is accompanied by an electric field, James Clerk Maxwell resolved this issue when he discovered that an electric field that changed with time was accompanied by a resulting magnetic field. Specifically, when the plates of a capacitor grows as the capacitor is charged, the electric field changes (with time) which produces a magnetic field.

== See also ==

The Ampere-Maxwell Law relates to the other Maxwell equations:

*[[Gauss's Law]]

*[[Magnetic Flux ]]

*[[Faraday's Law ]]

It might also help to read up on [[Maxwell's Electromagnetic Theory ]]

Also for continuity purposes, it can be helpful to look at the incomplete Ampere's Law:

*[[Ampere's Law]]

===Further Reading===

Matter and Interactions: Volume 2 by Ruth Chabay and Bruce Sherwood (4th Edition)

===External Links===

Maxwell's Equations:
http://study.com/academy/lesson/maxwells-equations-definition-application.html

James Maxwell Biography:
http://www.biography.com/people/james-c-maxwell-9403463

==References==

http://bulldog2.redlands.edu/fac/eric_hill/Phys232/Lectures/Ch%2023%20lect%201.pdf
http://ocw.mit.edu/high-school/physics/exam-prep/electromagnetism/maxwells-equations/
http://ocw.mit.edu/courses/physics/8-02-physics-ii-electricity-and-magnetism-spring-2007/readings/summary_w13d1.pdf
http://www.schoolphysics.co.uk/age16-19/Wave%20properties/Wave%20properties/text/Electromagnetic_radiation/index.html

[[Category:Which Category did you place this in?]]

Ampere-Maxwell Law

2026-04-27T02:59:24Z

Rohitnaras:

Conservation of Momentum

2025-12-03T03:21:39Z

Rohitnaras: /* A Computational Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. Thus,

<math>\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}</math> 
<math>\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}</math> 
<math>\sum p_{z,\text{initial}} = \sum p_{z,\text{final}}</math>

This is useful when dealing with collisions in the 2nd and 3rd dimensions, where the components of momentum must be treated independently.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/glowscript/c2378d7b848e]

The link below shows a simulation of a perfectly inelastic collision where the two objects stick together after colliding and move with shared velocity.

[https://trinket.io/glowscript/56497389cdce]

Here is a fun one! The link below shows a simulation of an explosion, where the original object breaks into two fragments while conserving total momentum.

[https://trinket.io/glowscript/29e700a9bba8]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T03:20:59Z

Rohitnaras: /* A Computational Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. Thus,

<math>\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}</math> 
<math>\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}</math> 
<math>\sum p_{z,\text{initial}} = \sum p_{z,\text{final}}</math>

This is useful when dealing with collisions in the 2nd and 3rd dimensions, where the components of momentum must be treated independently.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/glowscript/c2378d7b848e]

The link below shows a simulation of a perfectly inelastic collision where the two objects stick together after colliding and move with shared velocity.

[https://trinket.io/glowscript/56497389cdce]

Here is a fun one! The link below shows a simulation of an explosion, where the original object breaks into two fragments while conserving total momentum.

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T03:13:56Z

Rohitnaras: /* A Computational Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. Thus,

<math>\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}</math> 
<math>\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}</math> 
<math>\sum p_{z,\text{initial}} = \sum p_{z,\text{final}}</math>

This is useful when dealing with collisions in the 2nd and 3rd dimensions, where the components of momentum must be treated independently.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/glowscript/c2378d7b848e]

The link below shows a simulation of a perfectly inelastic collision where the two objects stick together after colliding and move with shared velocity.

[https://trinket.io/glowscript/56497389cdce]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T03:09:54Z

Rohitnaras: /* A Computational Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. Thus,

<math>\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}</math> 
<math>\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}</math> 
<math>\sum p_{z,\text{initial}} = \sum p_{z,\text{final}}</math>

This is useful when dealing with collisions in the 2nd and 3rd dimensions, where the components of momentum must be treated independently.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

<iframe src="https://trinket.io/embed/glowscript/c2378d7b848e" width="100%" height="356" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe>

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T03:03:20Z

Rohitnaras: /* A Computational Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. Thus,

<math>\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}</math> 
<math>\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}</math> 
<math>\sum p_{z,\text{initial}} = \sum p_{z,\text{final}}</math>

This is useful when dealing with collisions in the 2nd and 3rd dimensions, where the components of momentum must be treated independently.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/glowscript/7bc4cb8e9a]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T03:02:49Z

Rohitnaras: /* A Computational Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. Thus,

<math>\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}</math> 
<math>\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}</math> 
<math>\sum p_{z,\text{initial}} = \sum p_{z,\text{final}}</math>

This is useful when dealing with collisions in the 2nd and 3rd dimensions, where the components of momentum must be treated independently.

===A Computational Model===

The program below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

<iframe src="https://trinket.io/embed/glowscript/7bc4cb8e9a?outputOnly=true" width="100%" height="356" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen> </iframe>

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T03:02:11Z

Rohitnaras: /* A Computational Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. Thus,

<math>\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}</math> 
<math>\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}</math> 
<math>\sum p_{z,\text{initial}} = \sum p_{z,\text{final}}</math>

This is useful when dealing with collisions in the 2nd and 3rd dimensions, where the components of momentum must be treated independently.

===A Computational Model===

The program below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

<iframe src="https://trinket.io/embed/glowscript/7bc4cb8e9a?outputOnly=true" width="100%" height="356" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe>

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T02:52:33Z

Rohitnaras: /* A Mathematical Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. Thus,

<math>\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}</math> 
<math>\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}</math> 
<math>\sum p_{z,\text{initial}} = \sum p_{z,\text{final}}</math>

This is useful when dealing with collisions in the 2nd and 3rd dimensions, where the components of momentum must be treated independently.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/embed/glowscript/7bc4cb8e9a]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T02:50:51Z

Rohitnaras: /* A Mathematical Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. So:

<math>\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}</math> 
<math>\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}</math> 
<math>\sum p_{z,\text{initial}} = \sum p_{z,\text{final}}</math>

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/embed/glowscript/7bc4cb8e9a]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T02:50:10Z

Rohitnaras: /* A Mathematical Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. So:

<math>\sum p_{x,\text{initial}} = \sum p_{x,\text{final}},</math>
<math>\sum p_{y,\text{initial}} = \sum p_{y,\text{final}},</math>
<math>\sum p_{z,\text{initial}} = \sum p_{z,\text{final}}.</math>

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/embed/glowscript/7bc4cb8e9a]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T02:49:34Z

Rohitnaras: /* A Mathematical Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

Because momentum is represented by a vector quantity, the conservation law applies separately in each direction. So:

<math>\sum p_{x,\mathrm{initial}} = \sum p_{x,\mathrm{final}},</math>
<math>\sum p_{y,\mathrm{initial}} = \sum p_{y,\mathrm{final}},</math>
<math>\sum p_{z,\mathrm{initial}} = \sum p_{z,\mathrm{final}},</math>

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/embed/glowscript/7bc4cb8e9a]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T02:44:36Z

Rohitnaras: /* A Mathematical Model */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = -\Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/embed/glowscript/7bc4cb8e9a]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T02:41:50Z

Rohitnaras: /* The Main Idea */

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta t =-F_2* \Delta t \end{align} </math>

We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

It is important to note the argument presented above can be extended to any number of interacting objects, showing that the total momentum of an isolated system remains constant.

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = \Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/embed/glowscript/7bc4cb8e9a]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T02:28:03Z

Rohitnaras:

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects separate after collision and do not stick together. A separate class of inelastic collisions exist called perfectly inelastic collisions, and this is the case where the objects stick together after colliding. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions. The bottom line is to always remember that regardless of whether the collision if elastic or inelastic, the momentum of the system is always conserved.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta 1 =-F_2* \Delta 1 \end{align} </math>
We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = \Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/embed/glowscript/7bc4cb8e9a]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T02:00:16Z

Rohitnaras:

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. In an isolated system, meaning one in which there is no external impulse, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. Individual momenta can be changed due to internal forces between objects, but because internal forces occur in equal and opposite pairs, the total momentum of the system is unchanged. Since momentum is a vector, momentum conservation applies separately in each spatial direction (x, y, and z). Mastering the conservation of momentum is a powerful problem solving tool in classical mechanics.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects collide and do not stick together, and an inelastic collision is when the objects that collided stick together in their final state. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta 1 =-F_2* \Delta 1 \end{align} </math>
We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = \Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/embed/glowscript/7bc4cb8e9a]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]

Conservation of Momentum

2025-12-03T01:23:49Z

Rohitnaras:

Claimed by Rohit Naras (Fall 2025)

The conservation of momentum is one of the fundamental laws of physics. Within the definitions of a problem, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. The momentum is changed through the action of forces as in Newton’s law of motion. This is a powerful idea in solving problems.

==The Main Idea==

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects collide and do not stick together, and an inelastic collision is when the objects that collided stick together in their final state. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 (<math> \begin{align} F_1 \end{align} </math>) is equal in magnitude and opposite in direction to the force of object 2 on object 1 (<math> \begin{align} F_1 \end{align} </math>). So: <math> \begin{align} F_1=-F_2 \end{align} </math>

The objects collide during a certain time period (<math> \begin{align} \Delta t \end{align} </math>) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: <math> \begin{align} F_1* \Delta 1 =-F_2* \Delta 1 \end{align} </math>
We know that <math> \begin{align} F*t \end{align} </math> is the formula for impulse and since the change in impulse is equal to the change in momentum: <math> \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} </math> (The Law of Conservation of Momentum).

===A Mathematical Model===

<math> \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = \Delta p_2 \,. \end{align} </math>

<math> \begin{align} p_1 \end{align} </math> is the momentum of the first object, <math> \begin{align} m_1 \end{align} </math> is the mass of the first object, <math> \begin{align} v_1 \end{align} </math> is the velocity of the first object. <math> \begin{align} p_2 \end{align} </math> is the momentum of the second object, <math> \begin{align} m_2 \end{align} </math> is the mass of the second object, <math> \begin{align} v_2 \end{align} </math> is the velocity of the second object.

===A Computational Model===

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[https://trinket.io/embed/glowscript/7bc4cb8e9a]

==Examples==

===Simple===

'''Question'''

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

'''Solution'''

<math> \begin{align}
m_1 = m_2 = m = 0.165 \,. \end{align} </math>

<math> \begin{align}
v_1i = 9.8 \,. \end{align} </math>

<math> \begin{align}
v_1f = -2 \,. \end{align} </math>

<math> \begin{align}
V_2i = 0 \,. \end{align} </math>

<math> \begin{align}
m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} </math>

<math> \begin{align}
v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} </math>
The velocity of the second ball is 11.8 meters per second in the positive x direction.

'''Question'''

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

'''Solution'''

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision.
<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} </math>

<math> \begin{align}
5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} </math>
The final velocity of the big and little fish system is 0.909 m/s

===Middling===

'''Question'''

[[File: Middle Problem2.jpg|400px|thumb|right|Figure 1]]You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity <math> \begin{align} v_f \, \end{align} </math>. Find <math> \begin{align} v_f \, \end{align} </math>.

'''Solution'''

<math> \begin{align}
v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} </math>
<math> \begin{align}
v_B=[-5,0,0] \, \end{align} </math>

<math> \begin{align}
m_H = 60 kg \, \end{align} </math>
<math> \begin{align}
m_B = 50 kg \, \end{align} </math>

<math> \begin{align}
m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} </math>

<math> \begin{align}
v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} </math>

<math> \begin{align}
v_f = [-1.5155, -4.8503, 0]
\,. \end{align} </math> m/s

'''Question'''

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

'''Solution'''

This is an elastic collision, and the total momentum is conserved.

<math> \begin{align}
m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} </math> m/s

<math> \begin{align}
80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} </math> m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

===Difficult===

'''Question'''

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

'''Solution'''

<math> \begin{align}
m_b = 0.145 kg \, \end{align} </math>
<math> \begin{align}
m_E = 5.972 e 24 kg\, \end{align} </math>
<math> \begin{align}
v_b = 40 m/s \, \end{align} </math>

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

<math> \begin{align}
m_b*v_b=-m_E*v_E .\, \end{align} </math>

<math> \begin{align}
v_e= - (m_b*v_b)/m_E .\, \end{align} </math>

<math> \begin{align}
v_e=- 9.7120e-25 m/s .\, \end{align} </math>

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

'''Question'''

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

'''Solution'''

The initial velocity of the cannon and the projectile are both 0 m/s.
<math> \begin{align}
0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} </math>

v_x = -3.54 m/s

==Application==

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum. [[File:Soyuz TMA-5 launch.jpg|thumb|Soyuz TMA-5 launch]]

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

==History==

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in ''Philosophiæ Naturalis Principia Mathematica'' (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his ''La Methode Dex Fluxions'' until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

== See also ==

===Further reading===
For extra concept quetions see: [http://homepage.smc.edu/morse_peter/phy14/MotionForces/MOMENTUM%20ANSWERS.pdf]

===External links===
Newton's Philosophiæ Naturalis Principia Mathematica: [https://archive.org/stream/100878576#page/84/mode/2up]

==References==
1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.

[[Category:Momentum]]