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Here is a general figure of an hard spheres collision drawn in the center of mass of the mass $m_2$ before the collision. The black dot is attached to this frame. To solve the problem, you need to observe Conservation of energy: $m_1v_1^2=m_1(v'_1)^2+m_2(v'_2)^2$. Conservation of momentum: $m_1\vec v_1=m_1\vec v'_1+m_2\vec v'_2$ Conservation of torque ...

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I didn't see anyone mention the practical reason to use an approximation for energy. It is that in most problems you will be computing differences in energy. In that case, for small velocities, you can not only go the approximation ${m v^2\over 2}+m c^2$, but if you are also not converting mass to energy or vice-versa, you can drop the $m c^2$ as well ...

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Unless I made some mistakes, impulse is equal to momentum and not to change of momentum. Where did I go wrong, or, what is the final word? I'm not sure I completely understand your notation in your reasoning preceding the statements quoted above but 3rd line certainly doesn't look correct: Acceleration = a=[1]Δv change of velocity (m) v/s The ...

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A direct answer: Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given. The equation $E^2 = (mc^2)^2 + (pc)^2$ does work for all values. As other answers have explained, at low speeds (or equivalently, low momenta), $E = \frac{1}{2}mv^2 + mc^2$ gives approximately the same ...

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Your problem is that acceleration isn't the change of velocity, it's the change of velocity divided by the change in time. I still agree with one of the answers you link in saying that impulse and change in momentum are different concepts. Suppose we didn't know anything about Newton's second law. Then we could (maybe) imagine the concept of force and ...

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If the body's speed $v$ is much less than $c$, then the equation reduces to That is an imprecise wording, and the cause of your confusion, I believe. The correct description is that the second equation is an approximation of the first, and it will be closer to accurate the lower $v$ is. The approximation is important for understand the connection ...

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Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given. Yes, it would be nice if we could find an equation that was "universal". In the case of mass m, energy E, and momentum p, there is such a universal equation: $$E^2 = (mc^2)^2 + (pc)^2$$ Einstein, always wanted to ...

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The above answers are all good, but I want to add something else. You can derive the energy-momentum relation from at least principle, the action is $A=-m\int \sqrt{1-v^2}{\rm d}t$ . Lagrangian is $\cal L$$=-m\sqrt{1-v^2}, then you can get momentum p from derivative with respect to v (this is the real momentum, not mv). Energy H=E=pv-\cal L , just ... 5 If you put p = \gamma mv where \gamma = \frac{1}{\sqrt{1 - \beta^2}} and \beta = \frac{v}{c} in Einstein's equation, you get E^2 = (pc)^2 + (mc^2)^2 = (\gamma mvc)^2+ (mc^2)^2 = (\frac{c^2}{c^2 - v^2})v^2(mc)^2 + (mc^2)^2= (\frac{v^2}{c^2 - v^2})(mc^2)^2 + (mc^2)^2 = (\frac{c^2}{c^2 - v^2})(mc^2)^2 = (\gamma mc^2)^2 \therefore E = ... 2 Special equations of the kind you mention are useful as they elucidate limiting behaviors. Your example is no different from a statement like: the hypotenuse c of a right triangle with legs a and b with b \ll a is given by c = a + \frac12 b^2/a. The Pythagorean relation c^2 = a^2 + b^2 is valid for any right triangle, but the special case b ... 3 First, your findings are correct and are found by Einstein. However the fact that a new term appears is only showing that our previous knowledge was incomplete. The new term expresses the Energy related to the rest mass of the body which was not considered before. And it makes sense because before, in equating the Energy of a body only were included those ... 11 First, the non-relativistic equation$$ E= mc^2 + \frac{mv^2}{2} $$is equivalent to its second power,$$ E^2 = (mc^2)^2 + m^2 c^2 v^2+ \frac{m^2v^4}{4} $$If v/c\ll 1, then the last term is much smaller than the previous two, and the first two terms on the right hand side are equivalent to the correct relativistic$$ E^2 = (mc^2)^2+ (pc)^2 $$which ... 0 momentum is not conserved if there is friction, gravity, or net force (net force just means the total amount of force) what it means is that if you act on an object, its momentum will change. This should be obvious, since you are adding to or taking away from the object's velocity and therefore changing its momentum. Bonus: Momentum actually is conserved ... 0 If you have a system A with non-conserved momentum {\bf p}_A,$$ \frac{d}{dt}{\bf p}_A = {\bf F} \ne 0, $$then you have simply chosen not to consider the larger system A+B with conserved momentum {\bf p}_A + {\bf p}_B,$$ \frac{d}{dt}\left({\bf p}_A + {\bf p}_B\right) = 0, $$where$$ \frac{d}{dt} {\bf p}_B = -{\bf F}. $$For example, if a ball ... 1 Momentum. It's all momentum, you want to use your legs, shoulder and hip all to be "behind" your punch so that it has more momentum. You also want your hand to be continuously accelerating from the start to the end of the punch, as this will create more momentum by increasing velocity. Increasing mass also helps with impact. So, increasing the size of your ... 1 No. There are more conservation laws: conservation of angular momentum, electric charge, color charge, weak isospin. Without the charge conservation you cannot explain why electrons do not decay into neutrinos. 0 Not actually just energy conservation and momentum conservation but there are some other rules also that are fundamental in physics. If all the rules that govern the universe are given and the data of universe at any one instant of time is given then a computer that calculates all the simulation at a very good speed can tell all the events that will occur in ... 1 I don't think energy conservation and momentum conservation can explain charge conservation in particle physics. 0 No. Consider the radioactive decay of an atom. The energy and momentum is conserved but the direction of emission is not known. We do know that on average, i.e. if we take many identical atoms and observe their decay, that the direction of emission is isotropic. However, for a single event we just can't seem to predict this direction beforehand! These kinds ... 5 As lemon says, momentum is always conserved. Nevertheless, there are situations where we do not want to take the full system into account. We instead write an effective description in terms of a reduced set of variables. If the neglected degrees of freedom can absorb momentum, then the effective theory for the interesting variables looks like it does not ... 2 This is discussed in detail in A Relativists Toolkit by Eric Poisson. The original ADM paper is on the arxiv, and while the notation is old, the arguments are clear. 1 There's a discussion of this stuff in Wald, pp. 291ff. It must be a pretty accessible treatment, because I was able to understand some of it. The style is Wald's usual dry, mathematical, concise one, but he also gives some simple interpretations. 7 Calculus method The Taylor series of a function of d variables is as follows:$$f(\mathbf{x} + \mathbf{y}) = \sum_{n_1=0}^{\infty}\ldots\sum_{n_d=0}^{\infty} \frac{(y_1-x_1)^{n_1} \ldots (y_d - x_d)^{n_d}}{n_1!\ldots n_d!} \left. \left( \partial_1^{n_1}\ldots \partial_d^{n_d} \right) f \right|_{\mathbf{x}}.$$where \partial_i^{n_i} f means "the ... 1 No, the electrons are not moving like the balls on Newton's cradle. In Newton's cradle, the special thing is that the intermediate balls do not move at all, and even the moving ones don't half of the time. On the other hand, the electrons pictured show a wave movement, where all electrons move continuously. Now if you look at the Newton cradle ... 1 It isn't possible to comment on your viewpoint that relativistic mass is "an effect of the potential difference between reference frames", because that statement is too vague to be either true or false. In particular, I can't even tell whether you think that the relativistic equations involving mass are correct but can be expressed in an illuminating way ... 3 It is, in principle, measurable by a balance scale. In practice, this is difficult to do, because how do you user your scale when the thing your trying to weigh with it is flying away at close to lightspeed? If, however, you confine the object to a circular track so that you can keep it positioned over the scale for an arbitrary length of time to make the ... 2 I) The (Lagrangian) canonical conjugate momentum$$\tag{1} p_i ~:=~\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}^i} $$transforms as a one-form/co-vector$$\tag{2} p_i~=~ p^{\prime}_j \frac{\partial f^j(q,t)}{\partial q^i} $$under (possibly time-dependent) position coordinate transformations$$\tag{3} q^i~\longrightarrow~ q^{\prime j}~=~f^j(q,t)$$... 2 The objects that we call vectors in Euclidean space are, in more general situations, actually two possible different types of objects. If you change coordinates (q_i\to q_i'), the components change with different rules, according to how the Jacobian \Lambda_{ij} = \frac{\partial q'_i}{\partial q_j} (using your notation) appears. 1) Vectors (or tangent ... 5 This may be a bit more abstract than what you're looking for, but this is just a special case of a more general construction; determining the infinitesimal generators of a representation of a Lie Group acting on a vector space. In the case at hand, the Lie group is G=\mathbb R^3, the group of spatial translations, and the vector space is the Hilbert space ... 1 ... if the momentum of the system is conserved, the velocity of the centre of mass of the system should remain the same. True. ... the velocity of the centre of mass of the system has to be different after the collision for the kinetic energy to be different. False. So, how can there be a change in kinetic energy of the ... 1 if mass is assumed to be constant, the velocity of the centre of mass of the system has to be different after the collision for the kinetic energy to be different. However, if the momentum of the system is conserved, the velocity of the centre of mass of the system should remain the same. 1) mass is not constant and velocity is different: in ... 9 A simple counterexample: Imagine two particles with opposite direction and equal speed. The center of mass does not move, yet the kinetic energy of the system is non-zero. Now let both particles come to rest (by friction, hitting a wall, whatever). The kinetic energy is now zero, and total momentum has been conserved, while energy is not. The crucial ... 1 When you push on the ball, it pushes back on you with exactly the same force for the same amount of time. Both of you therefore always experience the same impulse. However, the speed imparted to a object by a specific impulse is inversely proportional to its mass. Both you and the tennis ball will move away from the point of contact, but the tennis ball ... 0 "imagine a spherical cow, in vacuum...." Newton's Third Law explains the question. Regarding the follow-up in the comment, the trajectory of your spaceship will not be altered, same reason. You cannot push against the back wall without reacting against something else, and that something else is part of the same solid body (or will be when it hits the front ... 0 It looks like in this problem the rain moves at whatever v the train is currently at, giving an infinite amount of energy as lim_{t\rightarrow \infty}. However it doesn't look like your using energy anyway, just be careful if you do. As BMS said, use the product rule. This gives you F= {\Phi}{v(t)}+\frac{dv}{dt}{(m+\Phi)} then subtract \Phi v(t) to ... 1 You should be careful, since you need to take account of the force that the rain exerts on the trolley, or the momentum of the rain. Your second approach does this rather nicely, (with the assumption that the rain falls vertically, and hence doesn't contribute to the initial momentum). In the first approach, you could redo it to add the force that the ... 2 Newton's 2nd law in differential form (ignoring vectors) is$$F_\text{net}=\frac{dp}{dt}=\frac{d}{dt}\left(p_\text{train} + p_\text{water in trolley} + p_\text{rain just hitting trolley}\right). \tag{1}$$You must take into account the change in momentum of the rain that occurs when it falls into the trolley and accelerates up to the speed of the train. ... -1 L(angular momentum) = (moment of Inertia) x (angular velocity) 0 This is because in deep space everything is being attracted equally in all directions since the universe is homogeneous. Hence, all of these tiny gravitational forces acting on the objects in your ship cancel out. Therefore you are left with a net force equal to zero on these objects in your spacecraft. On reflection, forget deep space for the time being ... 0 Your question is actually quite a complicated one as there are lots of different factors at play. However if you're asking why the objects inside the rocket don't attract each other then the main reason is that gravity is actually a very weak force. If a book is floating a metre away from me then the gravitational acceleration of the book due to my mass ... 0 Your answer of gravity being weak is exactly correct. The forces involved can be calculated with the following equation:$$ F = \frac{GMm}{d^2} $$A fueled Saturn V was about 3.0 \times 10^6 kg. Most of it doesn't reach space but we'll be conservative and say your spaceship is that massive. If you had a 1kg book that could somehow be only 1 meter away ... 1 The rocket is in free fall along with the book. The nearest gravitating bodies are very far away, so whatever meager acceleration they cause will be almost exactly the same on the rocket and on the book. Suppose you instead went on a very close flyby of a neutron star. Now the book will fall rapidly, and away from the rocket's center of mass. The only way ... 0 The entire system will be balanced and not move, as the momentum is conserved. There however will be an small shake back and fourth. First when it fires so the momentum move the rocket and the next one where the gas will hit the other side and conserve momentum and move it back to original location of starting point. 1 The angular momentum L_{A/B} of a rigid body A/B about its center of mass is$$L_{A/B} = I_{A/B} \omega_{A/B},$$where$I_{A/B}$is the inertia matrix of$A/B$about its center of mass in the world frame and$\omega_{A/B}$is the angular velocity of$A/B$. The angular momentum$L_{A/B}^0$of a rigid body$A/B$about the origin of the world frame is ... 3 I'm with Floris that the correct answer is "neither", but I'd like to to take the explanation in a slightly different direction. The notion that something is "moving an object" carries an unstated assumption that the natural state of an object is to be at rest. That is one of those things that is both obvious and wrong.1 The natural state of an object is ... 0 To 'derive' conservation of Energy from$ \vec{F} = m \vec{a}$, we take a dot product ($\hat{i} \cdot \hat{i} = 1$) which means that we have one (scalar multi-variable non-linear differential) equation with potentially many unknowns. Energy is nice because it provides a common language with all the physical sciences, but in classical mechanics, it's mostly ... 6 The answer to your title question is "neither". An object that moves has both momentum and kinetic energy, but it acquired these because a force was applied for a certain length of time. Now$F \cdot \Delta t$has the same dimensions as momentum (and in fact,$F \Delta t = m \Delta v$so there is a direct relationship between impulse which is$force * ...

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Maximum penetration will be achieved if there is a small (on the order of several centimeter) gap between the muzzle and wall. This is because, with muzzle against wall, the wall impedes the escape of air and combustion gases. Somewhat analogous to a muffler lowering the effective horsepower of an engine. Beyond a "small gap for exhaust gases to escape," ...

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Here are two separate ways to address the issue you bring up. One is more mathematical---comparing the relations $mv$ and $\frac{1}{2}mv^2$. The other has more to do with force and energy, which I'm calling physical. Mathematical Let's imagine two objects that are moving in the same direction collide with each other. Just to keep things simple, let's also ...

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Let's take an example with simple numbers : 1+2=3 3+0=3 This can represent the momentum conservation. Now look at the sum of squares : 1*1+2*2=5 3*3+0*0=9 The sum is not conserved because the momentum that was transferred changed differently the result of the squares. In a word, kinetic energy doesn't change linearly with speed (which is obvious since ...

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