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momentum

 Definition/Summary Momentum is mass times velocity. It is a vector. $$\mathbf{p} = m\mathbf{v}$$ Angular momentum is the cross-product of position and momentum. Being the cross-product of two vectors, it is technically a pseudovector, although it is normally referred to simply as a vector. $$\mathbf{l} = \mathbf{r} \times m\mathbf{v}$$ For a rigid body, angular momentum is also the ordinary product of moment of inertia (a scalar) and angular velocity (a pseudovector). $$\mathbf{l} = I\mathbf{\omega}$$ Momentum appears directly in Newton's second law, and angular momentum appears on taking the cross-product of Newton's second law with displacement.

 Equations By Newton's second law, Net Force = Rate of change of Momentum: $$\mathbf{F} = \frac{d \mathbf{p}}{dt}$$ By taking the cross-product of the above with the position vector from any fixed point, we find that net Torque about that point = Rate of change of Angular Momentum about that point: $$\mathbf{\tau} = \frac{d \mathbf{l}}{dt}$$

 Scientists Sir Isaac Newton (1643-1727)

 Breakdown Physics > Classical Mechanics >> Newtonian Dynamics

 Images

 Extended explanation 1. Informal introduction to momentum Supposing a mosquito approaches you with the velocity of 40 km/h. Even a collision would hardly affect you. However, if a truck was to approach you with the same velocity, it could be fatal. It naturally, is because of the truck's mass. However, it is not only mass that plays a role here. If it were, a truck standing still would've scared you too. Hence, an important property here is the product of mass and velocity. This product is known as momentum. Momentum is a very important property of closed systems, as it allows us to predict the behavior of such systems. 2. Law of conservation of momentum A closed system, is a system in which no external force applies. In such a system, the linear momentum does not change. This can be followed from the alternative definition of force i.e. rate of change of momentum. When force is zero, there is no change in momentum. Internal forces may act, as in a collision, but they cancel out each other. This follows from Newton's Third Law of motion. Not only a collision, but even in the case where two charged particles attract each other and brought towards/away from each other, the motion is such that the linear momentum is always the same. If the system was initially at rest, the linear momentum will continue to be zero. In the case of angular momentum, when no external Torque is applied on a system, there is no change in angular momentum. 2A. Vector components Momentum is a vector, and so any equation involving momentum applies to the components in each direction separately. In particular, even in a system with external forces, conservation of components of momentum will still apply in any direction in which there is no external force: for example, it will apply to the horizontal components of momentum of balls on a horizontal pool table, even if the pool table is accelerating vertically (because all the external forces are vertical). 3. Applications of conservation of momentum Supposing, a sphere is moving with the velocity $\overrightarrow{v}$ in the east direction. It explodes into two parts of equal masses, A and B. If it is found that A moves in the southwest direction with speed $\overrightarrow{v}$, we can use the conservation of momentum to find out the velocity of B. Let the mass of the sphere be M. Then, the initial momentum of the sphere was: $$\overrightarrow{p} = Mv \hat{i}$$ Now, the momentum of A can be given as: $$\overrightarrow{p_A} = -\frac{Mv}{2}(\cos(45^o) \hat{i} + \sin(45^o) \hat{j})$$ i.e.: $$\overrightarrow{p_A} = -\frac{Mv}{2\sqrt{2}}(\hat{i} + \hat{j})$$ Since they break in two parts, and A lies in the XY plane, B has to be in the XY plane as well. If it were not, then there would be an additional component along the $\hat{k}$ which would not be canceled, and hence the conservation law cannot be satisfied. Let, the momentum of B be $\overrightarrow{p_B} = \frac{M}{2}(a\hat{i} + b\hat{j})$. The momentum of a system is the same of the momentum of it's particles. Hence, the momentum of this system at this instant is given as: $$\overrightarrow{p} = \overrightarrow{p_A} + \overrightarrow{p_B} = -\frac{Mv}{2\sqrt{2}}(\hat{i} + \hat{j}) + \frac{M}{2}(a\hat{i} + b\hat{j})$$ $$\overrightarrow{p} = \frac{M}{2}\left(\left(a - \frac{v}{\sqrt{2}}\right)\hat{i} + \left(b - \frac{v}{\sqrt{2}}\right)\hat{j}\right)$$ This must equal the initial momentum, since no external force acted on the system. Hence, on equation the two momenta, we get: $$\overrightarrow{p_B} = \frac{Mv}{2}\left(\left(2 + \frac{1}{\sqrt{2}}\right)\hat{i} + \frac{1}{\sqrt{2}}\hat{j}\right)$$ Hence, we were able to arrive at the velocity vector for the particle B. In collision systems, the Kinetic Energy may or may not conserved, since it is converted to other forms of energy like heat, sound and light. However, both linear and angular momentum is always conserved when no external force and torque act on the system. 4. Momentum and system of particles If there are a number of particles and their linear momentum is needed, then the linear momentum is defined as the product of the mass of the system and the velocity of the center of mass of the system. In previous examples of collisions and explosions, when we say that the momentum of a system is conserved, it implies that the velocity of the center of mass of the system does not change. 5. Impulse Another important quantity is Impulse. Impulse has the same units as that of Momentum. Impulse is defined as a change in momentum, which is brought on by an acting force. Impulse is commonly denoted by 'j'. It is a vector quantity, since it is the difference of two vector quantities. Momentum can be defined as the time integral of Force. When on a system having momentum $\overrightarrow{p}_1$, a force $\overrightarrow{F}$ acts, for a time 't' from $t = 0$ to $t = t$, and changes the system of the momentum to $\overrightarrow{F}_2$, then the change in momentum of the system $\Delta \overrightarrow{p} = \overrightarrow{p}_2 - \overrightarrow{p}_1$ is defined as the 'Impulse' that acted on the system. This is written as: $$\int_{p_1}^{p_2} d\overrightarrow{p} = \overrightarrow{p}_2 - \overrightarrow{p}_1 = \Delta \overrightarrow{p} = \int_{0}^{t} \overrightarrow{F}{dt}$$ This definition is sometimes helpful in calculating the force in same cases. For example, consider the attached image: if a ball hits a wall at $45^o$ and rebounds at $45^o$ on the other side of the axis with the same speed $v$, then the impulse is defined as: $$\overrightarrow{J} = \Delta \overrightarrow{p} = -2Mv \hat{j}$$ If we assume that in a small time 't', a constant force $\overrightarrow{F}_2$ acted on the ball, by the definition above, we can say that: $$\overrightarrow{J} = \overrightarrow{F}t$$ or: $$\overrightarrow{F} = \frac{-2Mv}{t} \hat{j}$$ 6. Momentum in modern physics A change in momentum of a particle, under classical physics is brought about only by a change in it's velocity. However, under Relativity, a change in momentum also occurs due to a change in mass due to Mass dilation. Hence, momentum is now defined as: $$\overrightarrow{p} = \gamma m_o\overrightarrow{v}$$ where, $\gamma$ is the Lorentz factor and $m_o$ is the rest mass of the particle. This article however, is focused on the classical physics definition of momentum, and hence we shall not discuss it in further detail. Even massless particles, like the photon have momentum. The significance of momentum in this case again is that the momentum is conserved when a photon collides with another particle or is absorbed by it. The magnitude of momentum for such particles is defined as: $$p = \frac{h}{\lambda}$$

Commentary

 Govind.A.S @ 07:58 AM Oct21-12 Section 3 and 5 OK now. Question 2- not answered 2) I section6, p=h/wavelength, is this momentum a vector? If so, what is the direction? If not, why not? ~EDIT (tiny-tim): p= h/λ is the magnitude of the momentum

 Govind.A.S @ 06:30 AM Sep5-12 1) In section3 and 5, p, delta(p), J etc are vectors. Pls show arrow symbol on top. 2) I section6, p=h/wavelength, is this momentum a vector? If so, what is the direction? If not, why not? ~EDIT (tiny-tim): corrected (have i missed anything?): thanks, Govind.A.S.!

 -edward @ 03:04 AM Jul24-12 perfect

 PeterPumpkin @ 03:49 AM Jun25-09 Under Conservation of Momentum above this appears: "A closed system, is a system in which no external force applies". Shouldn't this be "... no NET external force"? Otherwise how could we apply the Conservation of Momentum to collisions between billiard balls etc --- eg gravity, an external force, is always present. ~EDIT (tiny-tim): that's because gravity, and the normal reaction, are vertical, so we can still apply conservation of momentum to billiard balls in any horizontal direction … see new section 2A.

 turin @ 11:48 AM Mar24-09 alternative "Breakdown" suggestions: physics > motion > momentum physics > momentum additional "See Also" suggestions: canonical variable, conjugate variable, deBroglie relation, conservation of, collision, elastic collision, inelastic collision ~EDIT (tiny-tim): the Breakdown has to follow the categories in the Defined browse drop-down list … it tells the server where to put the page

 tiny-tim @ 03:19 AM Feb2-09 Title changed to "momentum" from "momentum (non-relativistic)", because that was getting no autolinking at all. That didn't seemed to matter originally, when this entry used to be head of the top-viewed list, but it has gradually slipped to its present fifth place. And this entry has always contained a section on relativistic momentum ("Momentum in modern physics") anyway (a separate entry would presumably be titled "energy-momentum" or "four-momentum").

 freehuman79 @ 06:28 PM Jan30-09 is there any thing about the angular momentum in electric and magnetic field? ~EDIT(tiny-tim): this page is only about (linear) momentum: we don't have one yet on angular momentum.

 Hootenanny @ 06:50 AM May15-08 "In section 5, the integral [definite integral of dp] is written wrongly. p1 should be at bottom and p2 at the top of the integral sign. Please make the changes." -Done rohanprabhu: In future you might want to use the report function to detail your edits, there is no notification when a comment is posted.

 rohanprabhu @ 05:21 AM May15-08 In section 5, the integral [definite integral of dp] is written wrongly. p1 should be at bottom and p2 at the top of the integral sign. Please make the changes.

 Hootenanny @ 10:24 AM May14-08 Done

 rohanprabhu @ 10:02 AM May14-08 Section 5 in the extended explanation refers to the Image uploaded. Please make a mention. I tried to do it, but the article had been submitted already. thanks.

 Kurdt @ 08:05 AM May14-08 I changed some of the latex in the extended explanation as it wasn't displaying correctly. May I suggest a notation change from arrowed vectors to bold faced, and changing the variable for impulse as its too much like the unit vector in the y-direction.