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Physics is not a discipline, but a way of looking at the world. See if you can use your sense of the world to explain everyday phenomena.

**paraxial** rays as shown in the figure below. If \( g=0.33~\mbox{mm}^{-1}\), determine the distance \(d\) **in millimeters** between two consecutive focal points.

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**in Watts** must develop the external forces to keep the ring rotating at constant angular speed \(\omega\) (on average).

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**in meters per second squared**? The radius of the balls is \(R=1~\text{m}\).

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An electron is accelerated by a potential difference of \( U_e= 1~\mbox{mV}\). It then enters a region with an inhomogeneous magnetic field \( \vec{B}(x,y,z)\) generated by a system of coils carrying currents \(I_{1}, I_{2}\ldots \).

We then perform a similar experiment with a proton. We first reverse the current in all the coils generating the magnetic field. We then accelerate the proton with a potential difference of \( U_p\) and it enters the region with the magnetic field. What must \(U_p\) be **in Volts** so that the trajectory of the proton is the same as that of the electron? Don't forget any sign changes!

Hint: Compute \[\frac{d}{ds} \frac{\vec{v}}{v},\] where \(s\) is the arc length along the trajectory:

\(s=\int_0^t v(t') dt'\).

**Details and assumptions**

The proton to electron mass ratio is approximately \(6 \pi^{5}\).

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In 1-d Newtonian mechanics the energy as a function of momentum for a particle is given by \(E=\frac{1}{2}mv^2\) or, in terms of momentum, \(E=p^2/(2m)\). In Einstein's theory of special relativity there is still energy and momentum, and they are still conserved, but the relationship between the two is different. In empty space, the energy of a particle as a function of its momentum \(\vec{p}\) and mass \(m\) is given by \(E^2=c^2 \vec{p} \cdot \vec{p}+m^2c^4\) where \(c\) is the speed of light. The photon, the "particle" of light, has no mass - its energy satisfies \(E=c|\vec{p}|\) in empty space. In a medium like water this can change - photons travel slower in a medium and their energy-momentum relation becomes \(E=c_m |\vec{p}|\), where \(c_m\) is the speed of light in the medium. Our question is the following: A very high energy proton with \(E_{prot}^2=c^2 \vec{p} \cdot \vec{p}+m_{prot}^2c^4\) enters a tank of water. What is the minimum magnitude of the proton's momentum, i.e. \(|\vec{p}|\), in **kg m/s**, such that the proton can emit a photon with non-zero momentum and still conserve energy and momentum?

**Details and assumptions**

- The speed of light in empty space is \(3 \times 10^8~\mbox{m/s}\).
- The mass of the proton is \(1.67 \times 10^{-27}~\mbox{kg}\).
- The speed of light in water is \(2.3 \times 10^8~\mbox{m/s}\).
- When photon emission begins to occur the initial proton, the final proton, and the emitted photon all travel in the same direction.

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