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.
Hello all,
I am an Engineering dropout turned Cable Splicer. In my job we do a lot of Heavy Duty underground cable pulling. Usually plastic jacketed cable through some type of ductwork (typically plastic as well). We use a winch truck and a heavy rope to pull this cable through the ducts...
d(ɣmv)/dt = qvB
(dɣ/dt)mv + ɣm(dv/dt) = qvB
Substituting gamma in and using the chain rule, it ends up simplifying to the following:
ɣ^3*m(dv/dt) = qvB
Now, I am confused on how to solve for v.
Parallel:
M1V1+M2v2=M1V1’+M2V2’
(0.5)(3)+0=(0.5)(cos60)(3)+V2’Cos(x)(0.5)
V2’cos(x)=
Perpendicular:
M1V1+M2v2=M1V1’+M2V2’
0=(0.5)(0.3)(sin60)+V2’sin(x)(0.5)
V2’sin(x)=
And the divide 2 by 1
Which is tan(x)=2/1
And then plug then back in to solve, but I don’t think we do it like this because...
My apologies if the prefix is too high of complexity. I don't know where this would fall, difficulty or academically speaking.
While it may be surprising to some given Hollywood's portrayal of it in movies, if a person in wearing hard bulletproof armor is struck by a projectile, the person is...
Suppose two objects, A and B, with large lengths LA and LB, and masses MA and MB, collide at time t0.
Both objects before collision are vertical and aligned concentrically, being object B positioned initially at a higher z coordinate than object A.
The bottom end of object A is rigidly...
I chose to set the upwards direction to be positive and dM/dt = R = 190 kg/s, so I can solve the problem in variable form and plug in. With the only external force being gravity, this gives
M(t) * dv/dt = -M(t) * g + v_rel * R
where M(t) is the remaining mass of the rocket. Rearranging this...
I'm not interested in the mathematical derivation, the mathematical derivation already is based on the assumption that momentum is a vector and kinetic energy is a scalar, thus it proves nothing.
Specifically, what happens if we discuss scalarized momentum? What happens if we discuss vectorized...
two moving and rotating, uniformly weighted disks perfectly inelastic collide. The disks are rotating in opposite directions (see the diagram) At the moment of their collision, the angles between their velocity and the line connecting their centers are 45 degrees. The velocities are therefore in...
Question 2a: It is really hard for me to get my head around this.
The solution of this question mentions the momentum of the ball after it rebounds is 12kgms. My attempt at this solution is as follows
Before collision
Momentum of ball= mv= 2x3= 6 kgms and momentum of wall= 0
Therefore Total...
1 = elephant
2 = fly
So I am trying to find v'2 which is the final velocity of the fly. I have v1 the initial velocity of the elephant 2.1m/s. So I plug it into the equation and have v'2=(2m1/(m1+m2))*2.1m/s. We are not given the masses so I just know m1>m2 but I don't understand how that will...
So I have a trolley of mass m that moves on a straight line.
A sphere of mass m, is attached on the trolley with a light string of length a and it is left to oscillate.
Just to give some idea of their positions:
r_trolley = xi
r_sphere = (x-asinθ)i - acosθj (θ is the angle between the string...
The goal I am trying to achieve is to determine the momentum (2D) in a quantum system from the wavefunction values and the eigenergies. How would I go about this in a general manner? Any pointers to resources would be helpfull.
The given lagrangian doesn't seem to correspond to any of the basic systems (like simple/ coupled harmonic oscillators, etc). So I calculated the momentum ##p## which is the partial derivative of ##L## with respect to generalized velocity ##\dot{q}##. Doing so I obtain
$$p =...
Hi everyone,
In my physics class, we are doing the Hollywood Physics Project. It's a project where you analyze the physics from a scene in a movie and talk about if it's accurate or not.
I chose the scene from the Avengers where Thor strikes Captain America's shield with his hammer. The...
Let's imagine an ideal scenario where you're lifting your own weight in its entirety. Let's say a woman weighing 100 lbs. Suppose she's doing an idealized handstand and pushup from that position. So she's lifting 100 lbs. Let's say ideally all of the forces are on her arms only. Do these forces...
Here we are talking about non-relativistic quantum physics. So we all know kinetic energy T = E - V = \frac{1}{2}mv^2 in classical physics. Here V is the potential energy of the particle and E is the total energy. Now what I am seeing is that this exact same relation is being used in quantum...
Maybe a silly question but on the above question using the conservation of momentum:
momentum before firing (0) = momentum after firing (55*35)+(M*2.5)
If I re-range the above it's M = -(55*35)/2.5 = -770kg. I can I reconcile that minus sign (basically get rid of it)?
Thanks
Textbook solution:
##v## is the instantaneous velocity,
$$P(t)=(M+b t) v$$
Then $$impulse = \Delta P = (M+b t) v = \int^{t}_{0} F dt'$$
Thus $$v=\frac{F t}{(M + bt)}$$
What I did instead was:
Let ##M## be the instantaneous mass, and ##M_0## be the initial mass, then $$M=M_{0} + b t$$...
For the sake of this question, I am primarily concerned with the position wave function. So, from my understanding, the wave function seems to 'collapse' to a few states apon measurement. We know this because, if the same particle is measured again shortly after this, it will generally remain in...
I have trouble solving this problem any help would be appreciated.
Problem statement
##J=\frac{mr^2}{2}##
a) Determine the motion of yoyos for ##n=1,2,3##
The case for ##n=1## is simple, however, I am having trouble with ##n=2## and ##n=3##.
for ##n=2## I started by drawing all the forces...
Why, in lagrangian mechanics, do we calculate: ##\frac{d}{dt}\frac{\partial T}{\partial \dot{q}}## to get the (generalised) momentum change in time
instead of ##\frac{d T}{dq}##?
(T - kinetic energy; q - generalised coordinate; p - generalised momentum; for simplicity I assumed that no external...
I honeslty don't quite know how to start. It seems like the Hooke's coefficent k is independent of the answer to this problem.
I would also appreciate any clue of expressing the condition when "balls will collide again". The fact that all balls can keep moving make this rather difficult.
It...
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...
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...
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.
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...
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 im stuck, cause i cant find V in terms of Vb only
So, what I did was suppose the mass of ramp is $ M_r$ and let velocity at B of block be v, then, after inellastic collsion both bodies v' velocity
at B ,
$$M\vec{v}= M_r \vec{v'}+ M \vec{v'}$$
or,
$$ \frac{M}{M +M_r} \vec{v}= \vec{v'}$$
Now,
Suppose I take the limit as mass of ramp goes to...
Let's say you have two masses on either side of a spring. Mass 1 is connected to the end of a spring. The spring itself has no mass. Mass 2 is free in space. So you have:
[M1]-[spring] [M2]
So it's more descriptive, I'll name the variables like you might in programming. Let's define...
I first got the velocity of the combined mass with conservation of momentum and as it was in the mean position the velocity can be written as v = wA ( w= angular frequency , A = amplitude ) as we have to take it back to natural lenght i put A as the initial extension but i am getting a wrong ans...
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...
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...
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
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...
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??
In A.P. French's Special Relativity, the author said the following,
As I understand, photons are massless, so I don't think the last equation above applies to photons, but then, when deriving it, he used an equation proper to photons (##E=pc##).
So in which context is ##m=p/c## valid?
Since Pi = Pf,
0 = MbVbg + McVcg
I just need to express Vbg in terms of Vbc and Vcg (that is, I need to express the velocity of the ball relative to the ground in terms that I know/want to solve for):
by reference frames:
Vbc = Vbg + Vcg
so Vbg = Vbc -Vcg
Now I can sub in and solve
0 =...
I really want to know which answer is correct. I don’t really know if I should include velocities to the left as negative velocities in the equation. Is it -1 or 4.33? Please help! Thanks!!!
Hints given:
-Start with free body diagram. Use the relationship between impulse and momentum to find the final velocity of the car after he has pushed for time t.
-Use a kinematic equation to relate the final velocity and time to the distance traveled.
-What is his initial velocity?
My...
I thought it would be a good idea to pretend that the walls are stationary and that each time the particle hits a wall, it gets a velocity addition of the velocity of the wall it’s hitting. Using this I ended up at the formula
V = initial velocity of particle + n(velocity of left wall) +...