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6

Imagine that you have just two particles with the same mass and same speed, but going in opposite directions. They have opposite momenta, so the total momentum is zero. But they each have energy, and the total energy is not zero. The reason is because kinetic energy is just $\frac{1}{2} m v^2$. That square means that the kinetic energy can never be ...

4

Notice that \begin{align} i\frac{d(\psi^*\psi)}{dx} &=\frac{d\big[(-i\psi)^*\psi\big]}{dx} \\ &= \frac{d(-i\psi)^*}{dx}\psi + (-i\psi)^*\frac{d\psi}{dx} \\ &= \left(-i\frac{d\psi}{dx}\right)^*\psi + \psi^*\left(i\frac{d\psi}{dx}\right) \\ \end{align} Now subtract the second term on the right from both sides to get \begin{align} ...

3

A constant net force means: $$\Sigma\vec{F}=\frac{d\vec{p}}{dt}=C$$ where $C$ is some constant. This means that $$\int \ dp=p=C\int\ dt=Ct+p_0$$ where $p_0$ is the initial momentum. Now, you can easily verify that $$p_2-p_1=\Delta p=Ct_2+p_0-Ct_1-p_0=C(t_2-t_1)=C\Delta t$$ In particular, you see that $\Delta p \neq \frac{dp}{dt}$, unless $\Sigma ... 3 Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will". If you take$q|\vec v|$instead of$q\vec v$, you get something related to current (generally current times a factor). Current is conserved at a junction. Regarding your equal-and-opposite situation, the closest ... 2 This problem has a recursive flavor that we'll not try to avoid. Conservation of momentum tells us that $$m v_0 + (p+n-1)m v(n-1) = (p+n)m v(n).$$ Imposing the boundary condition$v(0)=0$we find $$v(n) = \frac{n}{n+p}v_0$$ as claimed. Let$a_n$be the time at which the$n$th bullet strike occurs. We have$a_1=x_0/v_0$and$$v_0 (a_n - T) = v_0 ... 1 Dan's answer is quite nice but I think the answer to this really depends on the interpretation one chooses to use. The straightforward answer I think is that you probably are confusing collimation with the creation of electrons. In classical mechanics one can independently specify the$\vec{x}$and$\vec{p}$very much indepently for the reason that Dan ... 1 The uncertainty principle's restriction on the minimum spreads in position and momentum is really small. An electron can be confined to a region in both position and momentum space that's extremely small compared to anything human-sized, but still have more than enough spread to obey the uncertainty principle. To give you an idea of how small this is, ... 1 You're not doing anything wrong, the objects will have different momenta in different reference frames. What should be the same in every reference frame is the forces acting on the objects during the collision. The laws of physics are the same in every reference frame, but not necessarily the numbers that go into the equations. By way of example, lets ... 1 energy is always positive or 0. And these are just numbers we associate with a body due to its motion according to different set of mathematical rules so that we can study these particles . As such they have no physical meaning . For example momentum is just$m$x$\vec v\$ . It is just a number we associated with a body by the quantities we defined ourselves ...

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