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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 ...


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.


3

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. :)


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 ...


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. ...


2

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


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.


1

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 ...


1

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.


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 ...


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: ...


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

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 ...


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 ...


1

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 ...



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