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2

There is, effectively, only gravitation and friction acting on the pack of gum. However, the friction is not that strong (it is mostly independent of the velocity of the book, and dynamic friction is weaker than static friction) and it doesn't have that much time to act. Hence it doesn't affect the momentum of the gum noticeably. This is very related to the ...


0

The article is a little wordy relative to what you're asking, so I'll offer a summary which encompasses the definition of the problem being solved. The equation of motion at work is a version of conservation of momentum. $$ F = m a = m\frac{dv}{dt}=-v_e\frac{dm}{dt} $$ You could really say this is a parametric differential equation, and for a known thrust ...


2

$V_e=V-v_e $ Why does this occur? This is just saying that the exhaust velocity is measured with respect to the engine. If the rocket is moving forward ($V$), then the observed exhaust velocity ($V_e$) with respect to the ground (or other specified frame) is reduced by the engine's velocity. And since we are concerned with the changes in velocity with ...


1

The continuity equation (without sources) is usually written as follows $$\partial_t \rho + \nabla \cdot \mathbf{j} = 0$$ If you identify $\rho$ as the mass density, integrate over some volume $V$ and use the divergence theorem you get the result that you mention in your question. Namely, the change in mass in $V$ equals the amount of mass flowing through ...


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From a mathematical standpoint, any collision in which no mass is lost is described by two equations: Conservation of energy: $ m_1v_1^2 + m_2v_2^2 = m_1v_1'^2 + m_2v_2'^2 + E $ Conservation of momentum: $ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' $ You know the masses and initial velocities, so this reduces to a system of two equations in three unknowns ($ ...


1

To parameterize the degree of inelasticity you use the "coefficient of restitution" which is 1 for elastic processes and 0 for inelastic processes. This is described by $$ \text{coef. of restitution} = c_R = \frac{\text{final relative speed}}{\text{initial relative speed}} = \frac{v_2 - v_1}{u_1 - u_2} \,. \tag{*} $$ This also tells you how to compute the ...


1

Perfectly elastic and perfectly inelastic collisions are just limiting cases on a scale of how much kinetic energy is retained. As noted in @Nathan's answer, if you work in the center-of-mass frame, a perfectly inelastic collision results in 0% of the kinetic energy retained, while perfectly elastic collisions have 100% of kinetic energy retained. So, you ...


3

There is not such thing as a "partially elastic" collision. Classical collisions between particles can be separated into two categories: elastic and inelastic. Elastic collisions are defined as collisions in which no energy leaves the system (i.e. $E_i = E_f$). All other collisions are inelastic, as some energy is lost ($E_i > E_f$). A perfectly inelastic ...


0

Neither, however... If you are standing on such a board, there is a simple way to propel yourself, assuming you can change direction (i.e. steer) the board. You cannot propel yourself forward, but you can propel yourself sideways, by pushing the board to one side. This gives you some sideways velocity. Then (before you fall over) turn the board so it is ...


0

You can't actually propel yourself forwards or backwards in this way (unless you are taking advantage of significant friction in the bearings). Moving your body forward or backward would cause the platform to move in the opposite direction but only so much as to leave the person+platform system's center of mass unchanged. Another consideration. Where would ...


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(Not an answer as such, more an extended comment and suggestion). You wrote "On the other hand, the 1-D elastic collision problem can't be solved with Conservation of Momentum alone - Conservation of Energy is also required. " Actually the 1D elastic collision problem can be solved with (i) Conservation of Momentum (COM) (ii) application of symmetry (iii) ...


1

The Heisenberg inequalities reads : $$\Delta x \Delta p_x \geq \frac{\hbar}{2},$$ where $$\hbar = \frac{h}{2\pi}. $$ Therefore, for a free particle, your add the expressions $$ p_x=mv_x$$ so you get : $$\Delta x \Delta v_x \geq \frac {\hbar}{2m}$$ you have the uncertainty on the velocity : $\Delta v_x \geq \frac{\hbar}{2m\Delta x}$ the Heisenberg ...


-1

The bullets time in the air, having been fired horizontally, depends on it's velocity. In general that varies from about 900 fps to near 3000 fps for a rifle, so obviously the time it takes to hit the ground varies as well.


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The collision of a rod with a point mass is the similar to the collision of two masses but with the effective mass of the rod being $$ m' = m_{rod} \frac{I_{rod}}{I_{rod}+m_{rod} r^2} $$ where $r$ is the distance between the point of impact and the center of mass, and $I_{rod}$ is the mass moment of inertia about the center of mass. If the rod is slender ...


1

Try showing that the kinetic energy of a particle of mass $m$ and momentum $\vec { p } $ can be written as $$K=\frac{p^2}{2m}$$ The solution to your problem should be clear from there. This expression for kinetic energy is actually a pretty handy formula to have memorized.


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Your particles have the same momentum, so one has twice the velocity of the other. Their KE is in the ratio $$\frac{\left( \frac{m}{2}v^{2} \right)}{m\left( \frac{v}{2} \right)^{2}}=2$$ It gets confusing because energy is not conserved as a particular form. Once you get a handle on the work-energy theorem or the relationship of work and energy many things ...


2

You can write kinetic energy as $$K=\frac{1}{2}mv^2=\frac{mv^2}{2}=\frac{(mv)^2}{2m}.$$ You can take the next few steps of rewriting this in terms of momentum $p=mv$.


1

Here's a parallel answer to LuboŇ°'s but purely classical. Start by noting that the momentum vector of a plane wave with wavelength $\lambda$ is: $$ \vec{p} = \frac{2\pi}{\lambda} $$ In some elastic scattering experiment, e.g. X-ray or some other diffraction measurement, we have something like: where $\vec{p}_{in}$ is the momentum of the incoming wave ...


5

Microscopically, i.e. in the quantum theory the scattering with radiation is a collision of particles with photons such as $$ e^- + \gamma \to e^- + \gamma$$ The momentum vectors of the particles above are $$ \vec p_1+\vec p_2= \vec p_3 + \vec p_4$$ where the identity holds due to momentum conservation. But in general $\vec p_1\neq \vec p_3$ and $\vec ...


1

Recall the Heisenberg principle which tells us that a we cannot know both position and momentum at the same time $$\Delta x \Delta p \geq \hbar/2$$ For a particle, we have actually at most $\Delta x = L$, so that we cannot in principle have a sharp momentum state. Fortunately, there are eigenstates which are eigenstates of $\hat{P}^2$, that is, states ...


1

Taking the standard $[0,L]$ problem, eigenfunction and energy eigenvalues are: $$ \varphi_n=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}, \qquad E_n=\frac{\hbar^2\pi^2n^2}{2mL^2}. $$ This means that stationary 1D box systems (e.g. insulated ones) only admit states with a discrete set of possible energies, as above. Now, that as far energy is concerned. What about ...


3

I understand that the inner product of two 4-vectors is conserved under the Lorentz transformations Yes, $p_1.p_2$ is a Lorentz invariant So that the absolute value of the four momentum is the same in any reference frame. It is not correct to speak about the "absolute value" of a (quadri)vector. Which is conserved in a Lorentz transformation ...


2

In special relativity, if you add two velocities, you have to use the formula $$v = (v_1+v_2)\left(1+\frac{v_1v_2}{c^2}\right)^{-1} \text{ .}$$ So you cannot simply add two velocities together. Usually, velocity is not a good variable to work with in special relativity. It's much easier to use four-momentum conservation, which is simply given by $$p = p_1 ...


-2

You can just verify each component and they are just momentum conservation in 3-momentum. There is no velocity conservation so you cannot add them together.


0

It would be safer to swing closer to the hill if I am understanding your question. I've never seen the movie you are talking about but, in general, a lesser height would provide a smaller time frame in which the jumpers would accelerate. A lower impact velocity would result in a higher chance of withstanding the crash.


2

I don't understand why the time component of the 4-vector [ $m~\gamma(|\vec v|)~(c, \vec v)$ ] is being denoted as $E/c$. So the underlying question is two-fold: Why is "energy" considered the time component of some 4-vector at all?, and Why this specific time component expression, among time components of all different 4-vectors imaginable? (Where ...


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Your book may be treating things a little backwards from the way they are usually done. The usual way is to define the momentum four-vector as the combination $(E/c, \vec{p})$, where $E$ is already known to be the total energy (the thing that reduces to $mc^2 + \frac{1}{2}mv^2$ for $v\ll c$) and then go on to show that it satisfies the properties expected of ...


1

In general, the elasticity of a collision is dependent on the properties of the colliding objects. In a perfectly elastic collision, no kinetic energy is dissipated, which means the collision creates no heat, no sound, etc. In a perfectly inelastic collision, the maximum possible amount of kinetic energy is dissipated as heat, sound, etc. This corresponds ...


0

Yeah, definitely. One example is an inelastic collision, where both masses will have the same velocity after colliding. In this case, let's say a bullet of mass $m$ and speed $v_0$ hits a stationary rock of mass $M$ and they stick together and move with a final speed $v$. Intuitively, you can already tell that their velocity must be in the same direction as ...


1

theoretically that might be possible but practically it is IMPOSSIBLE you just cant apply same force on every atom of every single molecule of the car. and you should also see at the general formula of momentum in which force is inversely proportional to the time taken and directly proportional to the change in momentum according to your fictional scenario ...


0

No, there is not. Force is, by its very nature, the change of momentum over time: $$ F = \frac{\Delta p}{\Delta t} $$ If you want to stop something (which means $\Delta p = p$, where $p$ is the momentum of the object) "instantaneously", i.e with very small $\Delta t$, this will inevitably lead to a large force on the object.


-1

You would need to apply appropriate force to all elements of the object (including those inside it) at the same time, similar to gravity force. That seems very difficult.


0

If the masses are the same, then as stated above, the initially moving mass stops and the other one acquires the velocity of the initially moving one. This is the mechaism behind the so called "Newton's cradle". BTW, the collision has to be head on, ie take place in one dimension.


2

Normally you solve an elastic collision with just momentum and energy conservation, because you really don't know what happens at impact. The formulas are given in this answer. For equal masses in one dimension the velocities are exchanged. It turns out your interaction time and acceleration multiply to get the correct velocity change, but since you ...


0

For a collision to be inelastic, all energy stored in the two (or more) objects are contained in those objects before and after the collision. This is in reality impossible as some of the energy will always be converted into some other form. In a car accident for example, some (allot) of the energy contained in the two cars will go into mainly the ...


0

The elastic and inelastic collision are models used in point mechanics of what really happens in the continuum mechanics world. In reality, the objects both deform when they collide. As contact is made, the object become coupled by non-interpenetration forces in the direction normal to their contact area, and friction forces in the direction tangential to ...



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