What is Conservation of momentum: Definition and 756 Discussions
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.
Here is what Feynman says, "Suppose we have two equal masses, one moving with velocity v and the other standing still, and they collide and stick; what is going to happen? There is a mass 2m altogether when we are finished, drifting with an unknown velocity. What velocity? That is the problem...
Hi.
I had a question about railguns, but I think I can formulate the underlying problem more clearly and concisely, hence I'm opening a different thread.
Consider the following rigid arrangement of three pieces of wire and two parallel capacitor plates:
There's an open switch somewhere in the...
Alleyway kick
The kicker weighs approximately 64kg or 140lb and the gentlemen holding the air shield is approximately 73kg or 160lb.
How far does the kickee travel before hitting the chair and boxes and what speed is he travelling at?
The practice kick prior to the actual kick puts the men...
I've been working through Bernard Schutz's book on GR and have run into some confusion in chapter 4 problem 20 part b. In this chapter, the stress-energy tensor for a general fluid was introduced and was used to derive the general conservation law for energy/momentum, where we found that...
In Chapter 5.3, Ballentine uses geometrical arguments to obtain the initial magnitude of a hydrogen atom's bound electron momentum. How does equation (5.13) obtain? I tried to naively compute
$$p_e^2 \equiv \textbf{p}_e\cdot \textbf{p}_e = p_a^2+p_b^2+p_o^2 + 2\textbf{p}_a\cdot \textbf{p}_b -...
For this problem I was very confused whether conservation of angular momentum should be applied to the person, the swing or the person-swing system. It seems to me that there is no net torque on any of the three systems I listed above. However, it seems that the angular momentums of the three...
The rotating ball should push the vehicle first to the right and once it hits the airbag - to the left?? Even if this works, how are you going to automate it and repeat it?
-Solved for vf using equation 3 to get 20.0m/s (speed before explosion) then solved for the distance to reach the explosion using equation 4, to get 20.0m, which felt wrong having the same numbers but that may just be coincidence.
-Found the distance travelled of the lighter piece using 530m -...
So i am tried to conserve momentum and use conservation of mechanical energy but won't there be psuedo force acting on the block if i am solving from non inertial frame ?. If i ignore the pseudo force and simply use C.O.M.E and include the K.E of the wedge and solve normally i do get the...
I have a wrecking ball with a mass of .5kg traveling at 3.03 m/s that hits a stationary block .9 meters high, weighing .06kg. I calculated the ball's exit velocity after it hits the block to be -3.00 m/s .
I calculated the final velocity of th block to be 4.2 m/s
Vf = Sqrt 2(g)(h) = sqrt...
We know that if we take two particles and assume no external force is applied then by Newtons third law total momentum gets conserved after collision. If we take three particles and there is collision between them and no external force then the momentum is again conserved for each pair like in...
If we have a ball with mass m dropped from a height h down to the ground, how come we can't set the conservation of energy equation just as the velocity of the ball turns 0.
mgh = 0
If instead the ball were moving with an initial velocity v, would the equation be
##mgh + \frac{1}{2}mv^2 = 0##...
If I consider only the freight car's mass and the mass dm that's added to the freight car as part of the system, then I get this answer:
https://ibb.co/QfKSqQ5
But if I consider the freight car's mass, the mass dm, and the locomotive car as part of the system (maintaining the locomotive has...
I thought the answer is B because the angular momentum in conserved in all 3 pictures.
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In a system with no external force like gravity, the final and initial momentums are conserved.
If we have a firework that explodes radially (disregarding gravity), the momentum before the explosion should equal the momentum after explosion. But isn't the explosion caused by an external force?
By "DART will have a relative speed of 6250 ms-1 when it collides with the asteroid", I assume it is the relative speed of the DART with respect to the asteroid.
Using that assumption, I can answer question (a)
For question (b), I don't understand the solution from the teacher. He did it like...
I have a problem in understanding angular momentum equation (mrv), especially the part where radius is involved.
imagine an elastic collision occurred between sphere of mass (M) attached to a string forming a circle of radius (R) and moving with velocity (V) and another stationary sphere having...
So I am guessing the cannons final velocity will be 4 m/s to the left because there momentum before shot was 0 because of opposite and equal reaction so
50,000kg x -4 m/s + 20kg x 10,000 m/s = 0 ?
The classic way to go about this problem would be to use Kepler's laws and thus find the new time period of earth.
However I encountered this question in a test on rotational motion which deals with conservation of angular momentum.
The equation used here would be I1ω1= I2ω2
Replacing I with MR2...
Consider a rocket with mass ##m## in space is going to move forward. In order to do so, it needs to eject mass backwards. Let the mass that is ejected has velocity ##u## relative to the rocket. What is the equation for the final velocity?
It is said that after ##dt## second, the rocket will...
In a closed system consisting of a set of particles not at rest relative to each other and acting on each other only by classical mechanical collision (i.e. billiard balls model, not including gravity or other long-range interactions), does conservation of momentum imply that the system will...
I got curious about firearm ballistics and googled something similar to "bullet momentum vs kinetic energy".
IIRC, momentum P = mv (checked); and kE = (mv^2)/2 (also checked).
So I essentially wondered if it's worse to get hit by a bullet with greater kE than by one with lesser kE, presuming...
Assuming no friction anywhere, no drag and perfect inelastic collision
Using conservation of mechanical energy i can determine the rotational speed of the rod right before collision occurs.
mgh=1/2*i*w^2
center of mass falls 1/2*L so we have:
M*g*1/2*L = 1/2*(1/3*M*L^2)*w^2
Solving for w...
1) By conservation of mechanical energy we have ##m_2gl(1-\cos(\alpha))+m_1gl=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+m_1gl## and by conservation of linear momentum along the x-axis we have ##m_1v_1+m_2v_2=0## which gives us ##v_2=\sqrt{\frac{2m_1gl(1-\cos(\theta))}{m_1+m_2}}## and...
(a) ##u_{min}=\big(1+\frac{m_2}{m_1}\big)\sqrt{2\mu_k g d}##
(b) ##x_f=\sqrt{\frac{2h}{g}\Big(\big(\frac{m_1}{m_1+m_2}u\big)^2-2\mu_k g d\Big)}##
Can someone check please?
1) By conservation of linear momentum: ##m_1 v_1-m_2v_2=(m+m_1+m_2)v_{cm}\Rightarrow v_{cm}=\frac{m_1}{m+m_1+m_2}v_1-\frac{m_2}{m+m_1+m_2}v_2=\frac{3}{8}\frac{m}{s}##;
2) By conservation of angular momentum: ##-Rm_1v_1-Rm_2v_2=I_{total}\omega=(I_{disk}+m_1R^2+m_2R^2)\omega## so...
Now, deriving relativistic momentum isn't terribly difficult, but that's not the same as understanding it. I'm trying to figure out why conservation of momentum in special relativity requires the gamma factor.
When I looked at conservation of momentum in elementary physics, we basically just...
This problem I already solved using another resource (just get the coordinate of the center of mass reach and from it, get to the larger mass. R = (3v02) / (4g)). But I'm having some trouble calculating using moment conservation. Here what I've done so far:
$$ 3\vec v_0 = \vec v_1 +2\vec v_2 $$...
My proposed solution:
When the student stops at the end, suppose the carriage is moving at speed u.
0 = (M+2m)u - m(v - u)
==> u = mv/ M+3m
After jumping out, the total momentum of the Carriage + collector system is 0 - mu = -m^2v/ M+3m.
By conservation of momentum for the Carriage +...
Suppose a rocket is moving at radial velocity vr and tangential velocity vt in the Sun's gravitational field. At some time, the rocket enters the gravitational field of Mars (with the above mentioned velocities), and gravitation effects due to the Sun can be ignored. After more time, the rocket...
At time t = 0, the mass of the cart is ##M_0## and velocity is ##v_0## in a time interval ##dt## let a mass of ##dm## be added to the cart due to the pouring water and let the reduction in speed be ##dv##
##\lambda = dm/dt##
applying conservation of momentum from the ground frame gives $$M_0...
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
In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling in the same direction. The initial speed of the leading bumper car is 5.60 m/s and that of the trailing car is 6.00 m/s. Assuming that the mass of the drivers is...
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...
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...
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?
I am unable to find any angle for which the horizontal and vertical components of the linear momentum are conserved.
I have added an image of my attempt
I think we can.Although the wall is not moving, it is just because the wall has a huge mass.As rhe law of the conservation of momentum states(suppose the ball hits the wall from the left), when the momentum decrease by J, the momentum of the wall increase by J, which means the momentum of the...
I tried solving it using this method and I got 12.5m/s, and assumed the collision was elastic.
The answer is actually 6.32m/s [41.5 degrees counterclockwise from the original direction of the first ball]; the collision is not elastic: Ek = 12.1J Ek`= 10.2J
I have absolutely no idea how the...
According to the first equation, the final potential energy is equal to the initial kinetic energy of the block. So that means that Vblok is the instantaneous speed of the block right before it moves to the right and compress the spring, right? But doesn't the second equation (The initial total...
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!
I encountered a weird conflict between my thought process and that of author's solution in book:
The common viewpoint of both of us were invoking conservation of energy of this SHM system
But the author proceeds to solve it using conservation of momentum, taking the new mass added to system as...
So to start off,
the piece that hits the ground first is the smaller piece.
So I can form the equations where:
where
##8(u_{8kg})= m_{1}v_{1}+m_{2}v_{2}##
##m_{1}+m_{2}= 8##
After 2 seconds,
##30 = v_{1}(2)+\frac{1}{2}at^{2}##
##v_{1}= 5.2m/s##
##(30-16) = v_{2}(2)+\frac{1}{2}at^{2}##
##v_{2}=...
Problem Statement: Are this 3 topics comes under laws of conservation of momentum?
Relevant Equations: Are this 3 topics comes under laws of conservation of momentum?
Are this 3 topics comes under laws of conservation of momentum: energy lost due to impact, inelastic impact, purely elastic...
A cyclist coasts along a road, he drives across a small puddle of water, after which the wheels leave wet lines on the road.
Now we concentrate our attention to the linear momentum of the water on a wheel. It decreases. Momentum is conserved, so what got the momentum that the water had?
Part (iii) is the part I am stuck on and is a 5 mark question. I have some idea of how to attempt it shown below
momentum is conserved so mux = mvy + mvz
(where ux is the initial velocity before the collision of ball x, vy is the velocity after the collision of ball y and vz is the velocity...
Hello,
I am an undergrad and am in an introductory level astrophysics course. I have a bit of confusion that I didn't know where to get help from so I made an account here. Please let me know if I miss some common etiquette or something... I don't understand how the slingshot maneuver...