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3 years ago

7

A roller coaster car of mass m= 300 kg is released from rest at the top of a 60 m high hill (position A), and rolls with a negli

gible friction down the hill, through a circular loop of radius 20 m (Positions B,C,and D) and along a horizontal track (to position E). Describe the scientific principles involved in this scenario.

1 answer:

Andrews [41]3 years ago

4 0

Answer: The principle of conservation of energy, angular speed and centripetal force

Explanation:

At point A, the car experienced maximum of potential energy

As it moves down the hill, the potential energy decreases while the kinetic energy increases.

The maximum kinetic energy of the car is needed for the attainment of enough centripetal force to help the car move through the loop without falling .

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X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

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<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

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