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"Conservation of momentum is probably a universal law, but that does not imply that action must be always equal to reaction, which as a matter of fact does not happen." Sure it does, at least in Newtonian physics where forces act instantaneously. Force is just the first derivative of momentum $ dp/dt $, so if $ dp_1 /dt = - dp_2 /dt $ for any two systems ...


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Why do physicists strive to make that statement universally applicable? They did do just that from the 17th century to the 19th century, but that is no longer the case. Electromagnetism, relativity theory, and quantum theory put an end to thinking that Newton's mechanics was universal. It's a trivial matter to deduce Newton's third law by assuming that ...


1

The earth also gains the positive work the same way the rough floor does, only it's not apparent at first. To understand how, let's go to the microscopic level to see what actually happens when you rub an object against a rough surface. You have the molecules and atoms of the object above, and those of the surface below as in the image shown. The ...


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I'll try to explain it like I would to my kid, as soon as he gets there. If you make the same pendulum swing in a horizontal circle, and look at it from the side, you see the same harmonic motion. The only difference is, it stays at the same height all the time, but since we are dealing only with small angles, that's not much of a difference. Now comes, ...


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Newton's 3rd law is nothing but the conservation of momentum. Recall that force can be expressed as the time rate change of momentum, $$\vec{F} = \frac{d\vec{p}}{dt}$$ In order for the conservation of energy to hold, when Object 1 pushes against Object 2, $$m_1d\vec{v}_1 = -m_2d\vec{v}_2$$ Since the mass does not change with time (conservation of mass), ...


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My Interpretation of your question : Why don't agents of the conservative force gain internal energy (i.e. heat) as in the case of friction, instead of gaining just potential energy. In any Conservative force, Mechanical Energy is conserved. $$\therefore~~ E_{k~(\text{initial})} + E_{p~(\text{initial})} = E_{k~(\text{final})} + ...


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It will turn out that there is no gravitational field inside a hollow sphere. I'm not going to do the integral to prove this for you, but here are some tips. $dV = r^2 sin \theta dr d\theta d\phi$, not with an $R$. You are integrating over the entire volume of the sphere. You are approaching the integral in the wrong way by converting to (x,y,z) space, ...


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Other answers being good, i'll add 2 more cents here. Inertia as quoted in question (btw, this is nice, to see actual historical sources), is (defined as) the tendency of matter to continue in its current state, unless changed. This is inertia, the term "force of inertia" is just another way to state the same thing but using Newton's second law, $F=ma$, so ...


4

The formula for the time period of a pendulum (for small angles of displacement from mean position) is $$T = 2\pi\sqrt{\frac{L}{g}}$$ Now, $L$ here is the length from the point of suspension to the center of mass of the bob. For illustration, assuming the bob is spherical, as the water leaks out, the center of mass will shift downwards, increasing $L$ and ...


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What acts here is called impulse (of a Force) Suppose balls A, B are made of stainless steel and (m = 0.1 Kg r = 0.03 m) they collide. B is at rest (v = 0): Ball A will exert on b the Impulse of a Force $J$ and its velocity, momentum and KE will increase: $$J = [F . t] = \Delta p$$ If you know exactly of what steel the balls are made you can calculate the ...


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Sure. An elastic collision is merely an idealized process which never occurs in nature. On the microscopic level, the two bodies are made up out of atoms. The electron shells of these atoms contain electrons. Since negative charges repel each other, the two bodys will repel each other in a smooth fashion (an inverse square force is acting).


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The question of mass has arguably been one of the two most important issues in physics (the other being the electromagnetism). Physics has tried to uncover the true nature of mass for hundreds of years, to no avail so far. Not surprisingly, its description is somewhat circular: “In physics, mass is a property of a physical body which determines the body's ...


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The equation for period of a pendulum is; $$T=2π\sqrt{\frac{L}{g}}$$ This equation holds for constant lengths, constant gravitation and small angles(such that the string is not horizontal and there is tension in the sting). Assuming these are true, you are not doing anything wrong concerning the experiment. Note that there is no mention of mass in the ...


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There are two issues at play here. If you lift something upward at constant speed, then the acceleration $\vec a$ is zero. This means that the net force $\vec F_\text{net}$ is zero by Newton's second law ($\vec{F}_\text{net}=m\vec{a}$). As long as something moves with constant velocity, all of the forces add up to zero (i.e., they cancel out). Yes, that is ...


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Does it have something to do with the curvature of the Earth which is assumed to be spherical You'll probably groan when you read this answer since it isn't nearly as complicated as you might think. Essentially, there is factor of $\pi$ since the angular frequency $\omega = 2\pi f = \frac{2\pi}{T}$ A well know result from the linearized pendulum ...


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First notice that simply by considering the dimension of the parameters involved, one can deduce that the time period of oscillations should go like $$T\propto\sqrt{\frac{\ell}{g}}. $$ This is because $g$ is acceleration hence has the dimensions of Length over Time squared and so the only way the quotient can have the dimension of time is to have the ...


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You are basically asking how to describe the equations of motion when viewed from an rotating frame of reference (the planets surface). This can be done by including the Coriolis effect. For example if you simplify the problem to just the equator of the planet the equations of motion would then become: $$ \ddot{x}=\frac{F_x}{m}-\frac{v_xv_y}{r}, $$ $$ ...


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So my question is: is this formula, that I frequently encounter in syllabi and books, correct? Where lies my mistake. The procedure $$\mathbf F = \frac{d\mathbf p}{dt} = \frac{d}{dt}(m\mathbf v) = \frac{dm}{dt}\mathbf v + m\mathbf a,$$ is based on the erroneous idea that the equation $$ \mathbf F = \frac{d\mathbf p}{dt} $$ is valid for systems with ...


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Firstly we divide the work up into internal and external components: \begin{equation} \sum _i\int ^{r_2}_{r_1}\vec{F_i}^{(e)}\cdot d\vec s _i+\sum _{i,j}\int ^{r_2}_{r_1}\vec {F_{ij}}\cdot d\vec s _i \end{equation} The factor of a half comes in since we are summing over both $i$ and $j$, and (I think) we can assume that $V_{ij}=V_{ji}$ (why?). They ...


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I don't agree that the problem would be under-determined. If the materials are not frictionless, then the friction coefficients must be given as well as the coefficient of elasticity of the materials. This additional information will allow the calculation of the forces!


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This can be answered in two different frames of reference. If we look at this problem from the perspective of an outside observer, then as the wheel moves forward, the bottom of the wheel doesn't move at all and the top of the wheel moves at twice the speed of the wheel itself. In this frame, as the wheel moves forward, the centripetal force provided by the ...


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This solution uses multiple integrals but you don't need to compute them. The final computation is a single integral. The ball has spherical symmetry so its momentum of inertia is the same with respect to the $x$-axis, the $y$-axis and the $z$-axis $I=I_x=I_y=I_z$ with $$I_z=\iiint \rho(r)\left(x^2+y^2\right)\mathrm dx\mathrm dy\mathrm dz$$ now sum up the ...


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The bottom of the wheel is a different part of it at every moment. If you follow a particular point on the wheel, you'll see it moves down and slows in forward motion until it touches the surface at zero speed and immediately starts to move up and accelerate forward again. Up to twice as fast at the top to catch up and get on the forward side again and then ...


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The problem is that "the bottom of the wheel" is not a specific physical part of the wheel. It is a role or description that applies to each part of the wheel as it moves around the axle. You could just as well wonder how the top of the wheel can travel twice as fast as the wheel, and still stay connected. The answer is that the double speed "top of the ...


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As you can see, $\rho$ only depends on a single variable: $r$. Thus, it should be intuitive that one can do this problem by integrating only over the variable $r$. To see what you are supposed to do, consider what happens if you fix $r$: You obtain a spherical shell (as was pointed out in the comments). The moment of inertia of a spherical shell is quite ...


0

Let us simplify and assume that the pivot on the blade moves in circular motion because of the hand. Then if you look at the forces acting over the blade, and from the reference frame of the blade, you will see a centrifugal force because you are in a non-inertial rotating system. In this frame of reference you experience a pseudo force (the centrifugal ...


0

Yes, but note that $U_i = K_f$ because $U_f$ is 0 and $K_i$ is 0. The rest of the working seems fine. $ {1 \over 2 }kx^2 = {1 \over 2}mv^2 $ $ {1 \over 2} (1290.9{N \over M})(.275m)^2 = {1 \over 2}(.435kg)v^2$ but we know $N/m = kg/s^2$ $97.62{kg\cdot{m^2} \over s^2} \over (.435kg)$ = $ v^2$ You may have multiplied wrongly. $N/m * m^2 = N\cdot{m}$ and ...


1

Suppose a body(very big but not bigger than Earth) moves against gravitational force of Earth. Let us take a real situation: a large rocket is fired straight up. It takes kinetic energy from the chemical explosions; from momentum conservation part of the available energy from the explosion moves the earth in the opposite direction. The gravitational ...


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The comments seem to provide the answer. $$ L = p \times r $$ $$ L = m ( v \times r ) $$ Thus if r (the distance to the focus point of the orbit) changes, the velocity can change without the angular momentum changing. This is not possible for circular orbits where v is always perpendicular to r and the magnitude of r is constant. However, in elliptical ...


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It comes from the work made by the gravitational force of the big object that attracts the Earth. I updated my answer to a previous question from you. But remember that the description in terms of conservation of energy and the use of potential energies is an alternative description (usually simpler) than that using all the forces involved (specially ...


0

If you assume linear force displacement relationship (small displacements) with stiffness $k$ then $F(t) = k\, x(t)$ where $x(t)$ is the separation of the two objects. Also initially the impact speed is $V = \dot{x}(0)$. The impact time is characterized by the natural frequency of the system $$\omega = \sqrt{\frac{k}{m}}$$ where $m = \left( \frac{1}{m_1} + ...


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(make comment as an answer) As mentioned in the comment here, the impact forces that are active during the time-frame of the actual impact are 1) unknown 2) difficult to put in analytic form That is why results like the conservation of momentum theorem are used. One can do estimations or approximations of these impact forces but would have to use more ...


1

In the equation $F_{net}=ma$, normally we would assume that $F_{net}=0$ implies $a=0$ on the right-hand side. However, for a massless object, we can satisfy the equation by having $F_{net}=0$, $m=0$, and $a\ne0$. In reality, of course, the pulley is not massless, so $m$ is small, $a$ is some nonzero number, and $F_{net}$ is small. The above reasoning is the ...


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Ok so first things first. 1) I presume that by "not attached to a ceiling " you mean to say that the entire system is being pulled in the upward direction by a constant force F . 2)The force equations you provided are wrong.If the system is being pulled upwards then we can neglect tension as it is an internal force. 3)acceleration A of the system is F/M ...


1

what is the force that the first marble applied one the second marble? The collision is almost instantaneous. Wouldn't that make the force in ΣF=Δmv/Δt insanely large because Δt is so small? Suppose two steel balls A, B of equal mass (m = 0.1 r = 0.03 m) collide and B is at rest: Ball A will exert on b the Impulse of a Force $J$ and its velocity, ...


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As stated in other answers, $\Delta t$, although small than times you're used to perceiving, is not 0. Because the marbles do not deform very much, the collision does not appear to take much time. However, in the grand scheme of things, the change in momentum for each marble during the collision is also small. Although answers vary, a quick google search ...


1

The force can be surprisingly large, but $\Delta t$ is not zero, and the force is not infinite. Make some estimates: the duration of the collision is so short that our eyes and brain cannot perceive it. Make an estimate for an upper limit for the duration. (There's no right answer, but a lot of wrong answers. For example, I would think that a duration ...


0

You need to tell us more about the problem. Is, for example, the position $y(t)$ of the mass $M$ is a system "input" i.e. is its functional form is given in the problem? If so, then $M$ is simply extraneous data: $y(t)$ is what it is and independent of $M$ (or at least, whatever is forcing the $y(t)$ to have the given functional form imparts the right ...


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In this type of collision where you have what amounts to a very quick change in velocity, the force is called an impulse force and it is best to think of the equation a little differently. For example, instead of: $$ \sum F = \frac{\Delta mv}{\Delta t} $$ Think of $\int F \mathrm{d}t$ being equal to the change in momentum, that is: $$ \Delta mv = \int ...


0

Make the y-axis height and the x- velocity. This will give you a parabolic shape indicating that it is a quadratic variation.


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It is an engineering question because it depends on the building, a building codes one, if the building collapses because of a bomb, an earthquake or an airplane falling on it as in the towers. In earthquakes whole apartment buildings end squashed, if the reinforcements in the lower columns are bad. So the roof is safer. In old apartment houses we are ...


1

I reminded you elsewhere that you know very well when 3rd law is applicable, and that you even teach it to others for example here Newton's third law of motion is not applied on a single body. - user36790 but then, inexplicably, you forget to use it properly in your own posts: Let a body move and a conservative force oppose its motion. ...


0

There will be no friction if both the cylinder and the walls are perfectly rigid. Imagine the cylinder is on wheels that have friction with the planes, but no internal resistance. If there is friction, it would produce torque on the wheels causing them to turn. The wheels will not turn when in equilibrium, so there is no friction acting on them then. ...


1

The problem with your solution is that the inelastic collision and assumption that kinetic energy is conserved are mutually exclusive. You can see that in your math when you try to solve for $v_2$. Rewriting equation $(1)$ gives $v_1=\left(m+M\right)v_2/m$ which inserted into $(2)$ yields $$ m \left(\frac{m+M}mv_2\right)^2 = \left(m+M\right)v_2^2.$$ This ...


0

though momentum as well as energy is conserved but definitely the sum of individual momentum of particles is not equal to sum of individual K.E. of the particles. also there may be different value of K.E. for same momentum. So can not make any result by manipulating the equations.


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Both energy and momentum are conserved as always. But to understand why this statement is true you have to look at the system as you described it a little more closely: In order for blocks A and B to stick together after the collision, the force between them should be zero when the velocity difference is zero -otherwise that force would continue to ...


3

If you have another model, then use it to generate a set of predictions, and let's compare those to the predictions we get in the world we see. If you think there's some ether filling the universe, then write down some properties it has, and make some predictions with those properties. Talking in vague generalities will only lead you down a rabbit hole of ...


2

The first one is correct. The problem is that in general the force acting on the particle will no longer be conservative in the moving reference frame and we can no longer associate the potential energy $U$ to it. To see why this is so, realize that a force $\vec{F}$ is conservative if and only if: $$\oint_C\vec{F}\cdot d\vec{r}=0$$ to any closed curve $C$, ...


1

I do not think the question ask what happens if the bobs collide, rather, it is a small amplitude so the pendulums interact through the horizontal cord until they reach the same average kinetic energy at equilibrium (by virial's theorem). If they have the same kinetic energy they each will rise the same relative height from their vertical position, so the ...


1

B will move faster, the reason is that the acceleration, $a$, of A is smaller for two reasons (remember that $F_{applied}-F_{drag}=ma$) : 1) the same force forward is applied so the contribution to the acceleration on the smaller ball will be larger 2)the drag force on the larger ball will be larger (see Rennie's comment) on A because the cross sectional ...



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