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.
A stationary observer sees a particle moving north at velocity v very close to the speed of light. Then the observer accelerates eastward to velocity v. What is its new total velocity of the particle toward the north-west relative to the observer?
I ask because while the particles total...
Question 7.6
Official solution
It seems that the solution uses the conservation of angular momentum to solve this question (τ=0). But the problem is that the frame is set on the centre of mass of the guy, which is non inertial. I would like to know why it is correct to do it this way. My...
I don't understand the reasoning of this question's answer. The answer is velocity = 0 (option A).
A while ago, I was told that, since the magnets were held at-rest (before being let go), they must have no velocity after the collision. What about the velocity which they had just before the...
So I've managed to confuse myself on this problem :)
Since the problem says we can assume ##m_p << m_b##, I'm assuming that the velocity of the bowling ball will be unchanged, such that ##\vec v_{b,i} = \vec v_{b,f} = -v_{b,0} \hat i##
I started out using the energy-momentum principle, ##(\vec...
List of relevant equations:
Angular Momentum = L (vector) = r(vector) x p(vector)
Angular velocity of rotating object = w(vector), direction found using right hand rule. Torque = T(vector) = dL(vector)/dt
I have a few questions about torque and angular momentum direction and...
Question: If we place the frame of reference on an accelerating point, does the total rotational momentum still remain the same?
I attempted to solve this question by manipulating the equations as shown below.
$$\text{Define that }\vec r_i=\vec R+\vec r_i'\text{, where r is the position vector...
First I calculated the momentum of m1. Since m2 was at rest after the collision, all its momentum was transferred, so m1 has a momentum of 158 i hat.
L=r x p, so its 916 k hat. This would also be the change in L because it was initially 0 when m1 had no velocity, so I know this is the net...
First I found the moment of inertia,
I=1.8(5.5^2+3.9^2+4.9^2)
=125.046
Then I tried to find the rotation rate using the equation L=rotation rate*I
rotation rate=3773/125.046=30.173
But the answer is suppose to be 21.263?
I know how to get to the answer but that's what is confusing me.
To find final velocity I multiply the acceleration by the time the object fell.
Then multiply the velocity by the mass to get momentum.
Now the angular momentum is r x p.
Since the initial angular momentum was 0, this was also...
So I first tried to find L using torque,
Torque=d/dt*L, and took the integral of this.
Ended up with 23.28484t
Now I square the equation L=rotation rate*I to get L^2=rotation acceleration *I^2
Angular acceleration=L^2/I^2
I feel like I am doing something wrong though, this doesn't give the...
m1v1+m2v2=m1vs'+m2v2' , if car hits small fluffy object m2, initially v2=0, and v1'=v2' ... so
m1v1=[m1+m2](v2')
but why not energy? Why is there a KElost?
.5m1v1^2+.5m2v2^2=.5m1v1'^2+.5m2v'2^2 +KElost , and again v2=0, v1'=v2'
.5m1v1^2=.5[m1+m2]v2'^2+KElost
using consv of momentum...
So to cut to the chase, I missed my class' lesson on momentum - have tried to catch up, quite successfully but am baffled about this question. I know the conservation of momentum etc. but after trying for ages it's just not happening this question so any help would be much appreciated,
Oscar.
Please scroll-sown to Question 52: https://www.undergraduate.study.cam.ac.uk/files/publications/engineering_s1_qp_2017.pdf
The correct answer is 'B'. This is the working I did:
F = (change in momentum) / (change in time)
change in momentum = mv - mu, where v = final velocity and u = initial...
So we know that the initial intertia of the merry go round is 250 kg m^2 and its angular speed is 10 rpm. MGRs angular momentum would be L=Iw=250(10)=2500kg m^2 rpm.
We know the mass if the child is 25kg, and the child's linear velocity is 6m/s. We convert linear to angular w= v/r = 6/2 =...
Hi, I have the following problem:
A homogeneous disc with M = 1.78 kg and R = 0.547 m is lying down at rest on a perfectly polished surface. The disc is kept in place by an axis O although it can turn freely around it.
A particle with m = 0.311 kg and v = 103 m/s, normal to the disc's surface at...
I know that the force must be a central force and that under central forces, angular momentum is conserved. But I am unable to mathematically show if the angular and linear momentum are constants.
Radial Momentum
##p=m\dot r = ma\dot \theta=ma\omega##
Angular Momentum
##L=mr^2\dot\theta =...
Let's say you start 2 rockets in the opposite direction from a platform that's close to two soon to be merging black holes, the first rocket starts way before the second, but the second will ultimately fly a bit further, before they stop and fly back to towards the black hole(s). When black...
We want to show that ##[\hat{ \vec H}, \hat{ \vec L}_T]=0##. I made a guess: we know that ##[\hat{ \vec H}, \hat{ \vec L}_T]=[\hat{ \vec H}, \hat{ \vec L}] + \frac 1 2 [\hat{ \vec H}, \vec \sigma]=0## must hold.
I have already shown that
$$[\hat{ \vec H}, -i \vec r \times \vec \nabla]= -...
I'm sure I've read somewhere that Jupiter has 99% of the solar system's angular momentum, which shouldn't be the case.
However, I can't find a source for this, and any search online for the topic doesn't bring up any science sites.
Did I mis-remember?
I am wondering why heavier bullets have a higher momentum than lighter bullets when using similar powder charges?
At the muzzle, a typical 150 grain bullet fired from a 30-'06 will travel at around 3000 ft/s.
A 200 grain bullet from the same rifle will travel around 2600 ft/s.
(The velocities...
1. When an object attached to a fixed point with a string, is given a velocity and the string goes taut.
So it says in this book (Applied Mathematics 1 by L. Bostock and S. Chandler) that when the string goes taut, the component of the velocity of the particle becomes zero in the direction...
I am confused with the following two questions:
1. A particle moves under the influence of a central force directed toward a fixed origin ##O##. Explain why the particle's angular momentum about ##O## is constant.
2. Consider a planet orbiting the fixed sun. Take the plane of the planet's...
A particle of mass m has kinetic energy E when it collides with a stationary particle of mass M. The two particles coalesce. Which of the following expressions gives the total kinetic energy transferred to other forms of energy in the collision?
I keep getting C as my answer when the correct...
Hi All,
I'm trying to get a better understanding of the momentum thrust given by an over-expanded rocket nozzle (I realize this case voids the isentropic flow assumption used for the 1D isentropic gas expansion equations typically used for rocket engine design since the normal shock is not an...
I think I get the approach. We first need to evaluate the term ##\dot A_{\mu} \nabla A^{\mu}## and then evaluate the 3D space integral; we may need to take the limit ##V \rightarrow \infty## (i.e ##\sum_{\vec k} (2 \pi)^3/V \rightarrow \int d^3 \vec k##) at some point.
The mode expansions of...
Wikipedia says that they are the equivalents of momentum and force in rotational motion but I don't understand why this comparison is possible. The torque's dimension is N*m it seems like energy. What is this energy? Why angular momentum is not mass times angular velocity?
Say we have a motor attached to the Earth with gear A, that drives identically sized gear B. Gear B spins on its own axis and but is also attached to ground. Torque between gears is equal.
Technically each gear has equal but opposite AM right, but If I take Earth into account, how is Ang...
I am trying to relate these equations to get the energy with respect to momentum based on particle wavelengths.
I did the following:
$$\lambda = h/p$$ so $$p= h/ \lambda $$
Then
$$p=mv$$ and $$E=1/2mv^2$$
So
$$E = pv/2 $$
But the answer was:
$$E = p^2/2m $$
I don't understand how they...
Is the following true if the momentum operator changes the direction in which it acts?
\langle \phi | p_\mu | \psi \rangle = -\langle \phi |\overleftarrow{p}_\mu| \psi \rangle
My reasoning:
\langle \phi | p_\mu | \psi \rangle = -i\hbar \langle \phi | \partial_\mu | \psi \rangle
\langle...
I am just solving the equation $$\frac{h}{2\pi i}\frac{\partial F}{\partial x} = pF$$, finding $$F = e^{\frac{ipx2\pi }{h}}C_{1}$$, and$$ \int_{-\infty }^{\infty }C_{1}^2 = 1$$, which gives me $$C_{1} = \frac{1}{(2\pi)^{1/2} }$$, so i am getting the answer without the h- in the denominator...
I don’t understand how energy is conserved here. The energy of the atom when n=5 is -.544eV. The energy of the photon is 1.14eV. After release, the energy of the atom is -.544 - 1.14 = 1.68eV. Using this value, I get n = 2.67, not an integer, so n = 3 and the atom has energy = -1.51 eV. I...
I looked in the instructor solutions, which are given by:
But I don't quite understand the solution, so I hope you can help me understand it.
First. Why do we even know we are working with wavefunctions with the quantum numbers n,l,m? Don't we only get these quantum numbers if the particles...
So if i have a photon of some energy and i want to find the magnitude of the momentum, i can get the right answer but the units don't make sense.
So i derive p = E/c since i know the energy of the photon and i used f=E/h and substituted this into p=hf/c
This means for units of the equation p...
Hi all,
If a body has a given initial momentum and then travels through a continuously less dense medium would it's velocity increase to conserve momentum?
Thanks
Jerry
First, introduce the energy – momentum equation E² = p²c² + (m0c²)².
Next, just think it in natural way.
If the energy – momentum equation reflects the stationary situation, then, momentum p naturally equals to zero. Then, we got E² = 0 + (m0c²)², namely: E = m0c². It can be denoted exactly...
i. Rearranging the equation for kinetic energy in terms of momentum;
1/2mv^2=(mv)^2/2m=p^2/2m
Inputting the values given KE=1^2/2*4kg = 1/8 J
Kinetic energy of a body with twice the momentum; KE=2^2/2*2kg =1 J
The ratio of the kinetic energy of the first body to the second body is therefore...
t is Torque
I is the inertia moment
P is the power
c is the constant light speed
r is the spot distance to the fiber
p is the torsional constant
theta is what we want
In the equilibrium $$t = 0$$
$$ F\Delta T = \frac{E}{c} = \frac{P\Delta T}{c} => *F* = \frac{P}{c} (1) $$
This will be the...
m1 + m2 = 8
COE
0.5(m1)(u1)^2 + (m1)(g)(30) + 0.5(m2)(u2)^2 + (m2)(g)(30) = 0.5(m1)(v1)^2 + 0.5(m2)(v2)^2 + (m2)(g)(16)
Can you check if my eqn is correct? And can you advise what to do after this?
I wanted to do COLM but i don't know what is the initial part.
When A hits B,
COLM
mV = -mVa + 2mVb
V = 2Vb - Va
COKE
0.5mv^2 = 0.5mVa^2 + 0.5(2m)Vb^2
V^2 = Va^2 + 2Vb^2
When B hits C
COLM
2mVb=4mVc
Vc = 0.5Vb
COE
0.5(2m)Vb^2 = 0.5kx^2 +0.5(4m)Vc^2
sub Vc = 0.5b
mVb^2 = KX^2
After that I am stuck, cause i can't find V in terms of Vb only
Hello!
Would it be possible to brake a magnet by means of short circuiting a coil placed at the end of a plastic tube where the magnet has been accelerated in ? Is there a conversion of magnet speed to electric current/energy inside the braking coil ? It is thaught for reusing the projectile...
Okay, i know that as a ball collides with a pivoting rod on an axis, the ball has angular momentum. Therefore after the collision, the ball is stopped or slowed, and the rod swings.
The ball provides a force and torque to the rod. But if I isolate the ball, isn't the only thing acting on the...
Question 1:
Since Force=Change in momentum = ∆P / ∆t
= mv / t
Momentum of the water coming out=mv
mv=ρVv=ρAv
Force d/dt (ρAv2t)=ρAv^2
Force = 1000*40*30ms^2
Force = 3.6*10^7 N
Question 2:
This is where I am confused because I understand in an elastic collision total kinetic energy and...