In Newtonian mechanics, linear momentum, translational momentum, or simply momentum (pl. momenta) is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction. If m is an object's mass and v is its velocity (also a vector quantity), then the object's momentum is
p
=
m
v
.
{\displaystyle \mathbf {p} =m\mathbf {v} .}
In SI units, momentum is measured in kilogram meters per second (kg⋅m/s).
Newton's second law of motion states that the rate of change of a body's momentum is equal to the net force acting on it. Momentum depends on the frame of reference, but in any inertial frame it is a conserved quantity, meaning that if a closed system is not affected by external forces, its total linear momentum does not change. Momentum is also conserved in special relativity (with a modified formula) and, in a modified form, in electrodynamics, quantum mechanics, quantum field theory, and general relativity. It is an expression of one of the fundamental symmetries of space and time: translational symmetry.
Advanced formulations of classical mechanics, Lagrangian and Hamiltonian mechanics, allow one to choose coordinate systems that incorporate symmetries and constraints. In these systems the conserved quantity is generalized momentum, and in general this is different from the kinetic momentum defined above. The concept of generalized momentum is carried over into quantum mechanics, where it becomes an operator on a wave function. The momentum and position operators are related by the Heisenberg uncertainty principle.
In continuous systems such as electromagnetic fields, fluid dynamics and deformable bodies, a momentum density can be defined, and a continuum version of the conservation of momentum leads to equations such as the Navier–Stokes equations for fluids or the Cauchy momentum equation for deformable solids or fluids.
Linear Motion Equation to get the common velocity of the block and bullet just after collision:
v2=u2+2gs, I set v=0 at max height of s = 0.004 m and g = -9.81 m s-2
I got u = 0.28 m/s
Then I calculate the impulse of the block using formula J = mv - mu, where v=0 and u=0.28 m/s and I got J = 1.4...
IS my solution right? Comparing with the other solutions, the answer just exchange the signals, i don't know why,
THats what ifound.
And here is the three equations:
{i use the point which occurs the collision}
Lo = Lf >>
0 = Iw + M*Vcm(block)
Eg = ct>
mvo² = mvf² + MVcm² + Iw²
I = ml²/3...
The Schwarzschild metric seems to model, for example, the earth’s gravity field above the earth’s surface pretty well, even though the Earth is not really a golf-ball sized black hole down at the center. Can the same be said for the Kerr metric? Does it model a rotating extended body’s gravity...
Hi, I have just joined the forum. Thank you all for being a part of such places so that people like me can get answers to the questions on their minds!
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I have been trying to understand how a quadcopter yaws. Referring to the figure below which is bird's eye view of...
I have done question 1. But I'm struggling with the other one. So since the only thing I know about the rocket is the mass and the velocity, I guess I have to use momentum to solve this problem. From the first question, I found out that the x-velocity of the projectile is ##v_x=5...
On p.170 of French's book on special relativity there is this thougth experiment attributed to Lewis and Tolman (1909). It is about two individuals throwing identical balls of mass M at each other with identical speed. The balls bounce against each other and are caught again.
See attached...
Here is my calculation:
F = ma
50N = 1050kg * a
a = 0.0476m/s²
S = ut + ½at ²
1000m = 0t + ½(0.0476)t²
t = 204.980s
y = 204.980s (time to travel 1000m)
since impulse = momentum,
F * t = mv
F * x = m * distance covered/y
50N * x = 1050kg * 1000m/204.980s
50N * x = 5122.450N⋅s
x = 102.440s...
I'm reading the article https://www.researchgate.net/publication/267938119_ON_THE_GALILEAN_COVARIANCE_OF_CLASSICAL_MECHANICS (pdf link here), in which the authors want to establish the transformation rule for momentum, assuming only that ##\vec{F}=d\vec{p}/dt## and notwithstanding the relation...
Since the equations are, actually, the question, i will post the image with relevant equations here:
it seems strange, I'm almost sure that I didn't make a mistake in the differentiation, but differentiating 9.8b I found 9.7a with both positive terms
To my mind because one particle has momentum ##\vec{p}## and the other one ##\vec{0}##. It is for instance necessary to find reference frame where one momentum will be for instance ##\frac{1}{2}\vec{p}## and the momentum of other particle should be ##-\frac{1}{2}\vec{p}##. So it is necessary to...
I believe momentum conservation is to be used in this sum since there's no external force, but I am not sure how to write the equation.
Can someone please help me out:)
Angular momentum can be exchanged between objects in a closed system, but total angular momentum before and after an exchange remains constant (is conserved).
There is a proof about this conservation?
"A smooth horizontal disc rotates with a constant angular velocity ω about a stationary vertical axis passing through its centre, the point O. At a moment t=0 a disc is set in motion from that point with velocity v0. Find the angular momentum M(t) of the disc relative to the point O in the...
given z(0) = 0 as well as
˙z(0)=0
How would one find the angular momentum along the x-axis in terms of t.
Currently, I have formulated the following:
$${\ddot{z} = \frac{g}{1+(\frac{4R}{s})^2}}$$
Hi
With the 2-body problem relating to planetary orbits i have encountered the following ; the gravitational force on the reduced mass acts towards the large mass(Sun) and since it is a central force it exerts no torque about the fixed centre(Sun) so angular momentum is conserved.
Conservation...
Well I am pretty sure that the kinetic energy stays the same because in this case the velocity vector and energy make a ninety degree angle so no work is done, but I am lost about angular momentum. It could decrease maybe if the torque is clockwise while the ship is going in a counterclockwise...
So far I found the answer for a and b, but when I attempted to do the other ones I was completely lost.
A.) P= MV
M = 25g = .025kg
V = 18
.025 * 18 = .45kg*m/s
B.) KE= 1/2 mv^2
1/2 (.025)(18)^2
4.05 J
We have 4-tensor of second rank. For example energy-momentum tensor ##T_μν##
, which is symmetric and traceless. Then
##T_{μν}=x_μx_ν+x_νx_μ##
where ##x_μ##
is 4-vector. Every 4- vector transform under Lorentz transform as (12,12). If we act on ## T_{μν}##
, by representation( with...
A cylinder of radius R spins with angular velocity w_0 . When the cylinder is gently laid on a plane, it skids for a short time and eventually rolls without slipping. What is the final angular velocity, w_f?
The solution follows from angular momentum conservation. $$L_i = I \omega_0 = L_f =...
How is it that momentum is being preserved in a non elastic collision?
for example let's say that two balls are colliding head-on, not elastically and heat is produced, does that not reduce the momentum of the system?
I got the correct answer for the first part but I'm not sure why the answer for (b) is the same for (a). Wouldn't the rings falling off mean that I_f = \frac{1}{12}M_L L^2 only where I_F, M_L, L are the final moment of inertia, mass of the rod and length of the rod as opposed to I_f =...
One part of König's theorem states that ##\vec{L} = \vec{L}_{\text{COM}} + \vec{L}^{'}##. The term ##\vec{L}^{'}## simply refers to the angular momentum wrt. the centre of mass. This is just a point, and doesn't have an axis implicitly associated with it (we have infinitely many choices!).
The...
Recently I started wondering why there seems to be so few practical/engineering applications where you need to calculate the momentum of something. I must emphasize that I don't mean usage of the concept of momentum or the law of conservation of momentum, but the value of the quantity itself...
Hi ; I have a few question regarding the conservation of linear and angular momentum. Would appreciate any help.
1 - When no external forces act are both linear and angular momentum conserved in all 3 directions separately or just the total linear/angular momentum conserved ?
2 - if I approach...
Hi! This is a very simple question regarding terms of expressions.
One law of motion is: F=ma
Another, using L as the linear momentum, is: F = dL/dt
If the first equation can be characterized (ignoring reference frames) as a "coordinate-based equation" (since is concerned with the second...
Change in KE = Change in thermal energy
0.5 * (6)* vblock^2 = 0.4 * 6 * 9.81* 0.1
vblock = 0.885
By Conservation of Momentum,
(0.05)(854) = (0.05)*vbu + (6)(0.885)I am not sure whether Change in KE = Change in thermal energy is true coz there should be a change in internal energy of the block...
I want to know between which two quantities of energy, momentum, and mass there is an uncertainty principle going on. Can I measure any two of those at the same time? If yes, which ones?
Thank you in advance
QUESTION:
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For the purposes of this problem, we will define the direction of Vehicle A's initial velocity as the positive direction:
While driving on a road that is inclined at an angle of 10 degrees above the horizontal, Vehicle A and Vehicle B are in a head-on collision lasting...
why can't we know where electron goes after it was hit by light? Light has a travel direction, can't we assume that electron bounces to the same direction that the light was headed??
I am reading Tong's lecture notes and I found an example in which there are several aspects I do not understand.
This example is aimed at:
- Understanding what is the analogy in field theory to the fact that, in classical mechanics, rotational invariance gives rise to conservation of angular...
We know classical equations fail to follow conservation of momentum and energy when we are dealing with speeds closer to the speed of light. But does it fail in the center of mass reference frame of a system?
In quantum mechanics one sees what J^2 can offer but why do we even consider looking at the eigenstates and eigenvalues of J^2 and a component of J, say J_z? Why don't we just use J?
In inelastic collision their is loss of energy and according to my current knowledge energy can be transferred by either work out heat...Now The problem is that we use law of conservation of momentum in problems related with inelastic and if energy is transferred from our system (two masses)...
Hello! I am reading some papers and I often noticed that it is mentioned that a strong magnetic field is able to decouple certain angular momenta from each other. For example in this paper: https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.100.023003 they present a Hamiltonian (second column...
Consider the following experiment from the point-of-view of classical mechanics and classical electromagnetism: An originally free electron then passes through a magnetic field that is oriented so that it causes the electron to turn to, say, the right. During the “turning” of the electron (a...
Tell me if I'm right:
A) Angular momentum is conserved because there are no external torques. Linear momentum isn't conserved because gravity is acting on the spacecraft . Mechanical energy isn't conserved because it has to change between different orbits.
B) Parabolic orbit...
I have some doubts about ballistic pendulums.
First, we say that if a bullet hits the pendulum, the linear momentum is conserved. But when we consider a rod attached to a pivot at one of its ends instead of a pendulum we say that the linear momentum isn't conserved because the rod can't move...
Hello everyone!
It seems I can't solve this exercise and I don't know where I fail.
By inserting the metric on the lefthand side of I. and employing the chain rule, the equation eventually reads (confirmed by my notes from the tutorial):
$$m\frac{\mathrm{d}p_\delta}{\mathrm{d}t} =...
I solved a and b fine, I just don't know where to start c
a) i=Ft i=change in p and p=mv so i=90=45v so v=2m/s
b) same thing just using the other mass i=90=70v so v=1.3m/s
c) v=d/t so 1.3=d/1.5 but that would be d=1.95 and the answer key says d=.96m
The answer is required to be in terms of M,m and c only. But, I am not able to calculate the momentum of the m mass particle using the above two. Can anyone help me by telling me what I am missing?
In my lecture notes on paramagnetism and diamagnetism is stated that "according to van vleck critical considering an electron in motion around the nucleus diamagnetic and paramagnetic effect compensate (classical)"
The proof following takes the intrinsic momentum related to L as the momentum...
So if i had this problem where i am squaring a four momentum vector with itself which gives
P2 = (##\gamma mc## )2 - ##\gamma##2## m ##2##\vec v## *##\vec v##
I have been told that the gamma factor is not considered at all. why would the gamma factor drop off? Does this rule apply to any...
I assume KE is conserved therefore, the KE transferred will be the same. But intuitively, this doesn't seem correct. Seems like the object that breaks apart would transfer less energy than the object that doesn't. Any thoughts?
When I solved the problem using the conservation of angular momentum, I have got the correct result (ω = 0.006 rad/s). However, when I tried to find the answer using the conservation of energy the result was incorrect and I do not understand why.
a)
On the black part, all incoming light is absorbed. This means that the momentum of the left-light beam doesn't change (i.e. momentum before hitting the black screen is ##\vec p_0## and after hitting it is zero. Thus ##\Delta \vec p = \vec p_0##). If momentum doesn't change, we get no...