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0

The most important equation to remember to use is: $F\Delta t = m\Delta v$ Where "F" is the magnitude of the thrust and "t" is for how long (in this case) constant force is being applied. If the thrust changes based on time, you have to use: $\int F dt = m \int dv$ EDIT: Kyle pointed out a flaw in my answer, so I will try to remedy it. Mass depends on ...


1

It does follow from the conservation of momentum. Consider the diagram (from Wikipedia) of a rocket expelling gas of mass $\Delta m$: At $t=0$, the initial momentum is $$ p(t=0)=\left(m+\Delta m\right)V\tag{1} $$ but at $t=\Delta t$, we've lost some mass and gained some velocity, $$ p(t=\Delta t)=m\left(V+\Delta V\right)+\Delta m \left(V-v_e\right) ...


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In an inelastic collision, some of the energy is absorbed by the colliding bodies - this is why you cannot use conservation of energy to calculate the resultant velocities of the bodies involved - you don't know how much is absorbed. But you do know that momentum is conserved, and assuming that the bodies remain intact (no pieces are separated from the ...


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Solving this particular set of equations would yield two values for $u1$ and as such two values for $u2$ - 10 and 5 ,that is, if the value of $u1$ were 5 the value of $u2$ would be 10 and vice-versa.You may arrive at this by solving a quadratic equation in $u1$ (or for that matter,$u2$) Now,you may have, while solving quadratics seen we,sometimes acquire ...


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The average of $v$, i.e. the expectation value $\langle v \rangle$ must be zero, as $\langle p \rangle = 0$ for the eigenstate $n$ (here $n = 1$). For $\langle v \rangle \neq 0$, the particle couldn't stay inside the box. But $\langle p^2 \rangle \neq 0$: $\langle p^2 \rangle = \frac{\hbar ^2\pi ^2}{a^2}$, as noted by the poster. And with $\langle p^2 ...


1

Suppose someone suggests that following a perfectly elastic collision, two billiard balls are each traveling twice as fast as they were before (and opposite to their original directions). You can't prove him wrong using conservation of momentum, but you can prove him wrong using conservation of energy. Therefore conservation of energy has implications that ...


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The math is almost trivial for someone beyond algebra 1. Write the kinetic energy of each particle as $p_n^2/2m_n$. Then converse momentum and kinetic energy in the center-of-momentum. You will see that the magnitude of the momentum each particle does not change.


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Observe that $v\dot v=\frac12\frac{\text d}{\text dt}v^2 = \frac12\frac{\text d}{\text dt}\mathbf v\cdot\mathbf v = \mathbf v\cdot\dot{\mathbf v}$ and therefore the integral of the OP becomes (assuming a constant mass $m$, e.g. a point particle) $$\Delta\left(\frac12mv^2\right)=\int_a^b mv\ \text dv = \int_a^b \frac{\text d}{\text dt}(m\mathbf v)\cdot\mathbf ...


0

You are talking about two derivatives, one derivative to time and the other to velocity. If you would derive the kinetic energy two times to the velocity, you would end up with m, not F.


10

Felt recoil is partly a matter of momentum, partly a matter of force. When a bullet with mass m leaves a gun with a velocity v, the gun must have an equal-but-opposed momentum MV, where M is the mass of the gun and V is the recoil velocity, or $$mv + MV = 0$$. If there are two possible gun sizes, $M_1$ and $M_2$, each will have a recoil velocity $V_1$ and ...


4

Newtons 3rd Law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body. Momentum is product of mass and velocity. The heavier gun has more mass, so, for the same momentum, it must have less "backwards" velocity, so less felt recoil.


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The larger firearm has more mass, and therefore more inertia for the recoil momentum of the bullet to overcome. Also, small firearms may be more difficult to secure a good grip on.


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You know that $$m\vec a=q\vec v \times \vec B,$$ So in particular, since $\vec B = B\hat z,$ we have $$ma_x=qv_yB,\text{ and } ma_y=-qv_xB. $$ And in our case $B=-\beta x$ so we have $$m\ddot x=-q\dot y\beta x\text{ and } m\ddot y=q\dot x\beta x. $$ You can take the time derivative of the left equation and get $$m\dddot x=-q\ddot y\beta x-q\dot y\beta ...


2

We can write total energy $E$ two ways: \begin{equation} E^2=p^2c^2+m^2c^4 \\ E=T+mc^2, \end{equation} where $T$ is kinetic energy. Eliminating $E$ from those two equations will give you the desired result.


2

Hint: $T = E - E_0 = m\gamma c^2 - mc^2 = mc^2(\gamma -1)$ and $p = |\vec p| = m\gamma |\vec v| = m\gamma v$


1

If the slope is truly frictionless, then the ball will never stop moving. You are assuming that after passing through the V at the bottom, it will roll back up the other slope. But It is not rolling but sliding (no friction - no torque to make the ball roll) When it hits the V, it will bounce - so it will lose contact with the surface Exactly how it ...


0

If the pencil were at absolute zero it would necessarily assume its lowest energy state, which is not the vertical state. If the pencil were modeled as a quantum rotor with an infinite potential barrier covering half its solid angle space (i.e. the table) then there are certainly excited but stable states where the pencil remains in a more or less vertical ...


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GIVE IT A SPIN, with the max possible rpm for such massive object. Do the experiment at the ISS, well above Earth surface. Unconstrain the problem from the 'surface' issue. Remove the 'surface' below the atom tip -- where is bellow without gravity? -- or approximate the surface as much as you can but without contact (the atom's electronic cloud prevent this, ...


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Yes, the temperature of the gas would decrease quite fast, given that the molecules in the container are still, which implies zero temperature for container. However, if the container's temperature is non-zero, it sometimes happens that gas molecules will instead gain energy because the molecule it collides with is moving fast enough in the opposite ...


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If you assume that the gas is ideal then each collision of a molecule of gas with the wall conserves the kinetic energy. Hence the temperature will stay the same.


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The temperature of the gas will eventually reach equilibrium with the walls of the container, and since a perfect insulator is not possible, the gas, walls and outside environment will, given enough time, be at the same temperature.


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Ok, I might have "solved it" although there are still grey areas. First the equation for the force experienced by a body colliding against a body with infinite mass: $$ \textbf{F} = \textbf{n} \delta e_r m k_0 $$ Explanation: $ \textbf{n} $: unit normal pointing outside the colliding wall. This is the direction of the force (ie, where we want the ball ...


1

Given the mass m of the ball, the incident normal speed v, and the coefficient of restitution $\rho$, Then the integral of F over the duration of the collision $\Delta t$ is $$\int_0^{\Delta t} F dt = \frac{m(1 + \rho)v }{\Delta t}$$ assuning no rotational effects are incurred. This follows from the fact that at any instant the acceleration of the ball away ...


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Simple momentum change will do: $$\vec F = \frac{d \vec p}{dt} $$ Your case is an elastic collision, so it simplifies to the difference between end states: $\vec F = \frac{\Delta \vec p}{\Delta t}$. Key thing: you need to know about the duration of the impact. Think about it... If your wall is very elastic - a vertical trampoline - the ball might still ...


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The second solution will mathematically satisfy the conservation equations, but corresponds the objects not actually colliding. Or they ``ghost'' and fly right through each other. :)


-1

I need the Latex environment to correct the answer provided by @Urgje, so forgive me for writing a new answer. As shown by Hitoshi, the commutator between the two components of kinetic momentum is: $$[P_x,P_y]=i\frac{e\hbar}{c}B_z$$ Let's now show that the kinetic momentum does not commute with the Hamiltonian. ...


1

The case for momentum of a system to be conserved is that no external force should be acting on the system. This comes from newtons second law. On alaysing the bullet rod system there are 2 forces, that acts on it:: Gravity- if we conserve momentum at the time just before the bullet hits the rod, to the momentum of the system just after the bullet collides ...


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If you made the graphic of forces of the system, you could see that the angular momentum is conservated, you can use that to find the final angular momentum of the system when the bullet hit the rod, then the final velocity of the bullet.


-2

This question goes back to the question that Feynman asked his father about why a ball in a wagon rolled to the back, when he pulled on the wagon. His father replied, "That, nobody knows." In other words, you're asking what relativistic momentum is. We don't even know what non-relativistic momentum is. Sure, we know the effects, we measure them, I've ...


0

Inertia is what we simply called 'quantity of material'. The word material has been used here to specify the matter of body. For example, a plastic chair, a wood chair and an iron chair. Among them, a plastic chair will have less inertia because it will apply less reaction force, so it is easy to lift it. And the word quantitative is used to define the ...


2

First, the chemical reaction that takes place in the cartridge of a gun produces high temperature, high-pressure gas, without the need for external oxygen. The gases then push the gun and the bullet apart. Secondly, the bullet will leave the barrel with a certain amount of momentum, found by multiplying the mass of the bullet by the velocity of the bullet. ...


1

Your question is not all that clear to me, but I assume you are puzzled as to how momentum can be conserved whenever forces act on the individual bodies to accelerate such bodies. First things first: if a force is exerted on a particular body, A, then it must be caused by a particular other body, B. By Newton's 3rd Law, A must also exert a force on B, a ...


1

So the problem here is that you are confusing quantum mechanics of a single harmonic oscillator with quantum field theory, which is quantum mechanics of a field and can also be considered as quantum mechanics of an infinite number of harmonic oscillators. In quantum mechanics of a single oscillator, the ground state $|0\rangle$ can be represented as: ...


15

Neutrinos are weakly interacting quantum mechanical point particles, with very small mass. Refraction is a classical mechanics phenomenon, happens to waves traveling in a medium and it is a collective synergy of many photons impinging on the field of the atoms and molecules of the medium. Individual photons are not refracted but are scattered. In synergy ...


2

Maybe the important step is to realise that the allowed range of momenta is $$ R=[-p_\text{max}:-p_\text{min}]\cup[p_\text{min}:p_\text{max}]. $$ Then the first two $\Theta$'s give one if $p$ is positive and in the allowed range, whereas the last two $\Theta$'s give a contribution that's only 1 if the $p$ is in the negative allowed range. The term out front ...



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