2. An athlete with a mass of 78.75 kg is testing a net used by firefighters to save people. The net sags by 100 cm when the athlete jumps from a height of 15 m. Assuming the net behaves elastically like a spring, calculate how much it will sag when a person with a mass of 45 kg jumps from a height of 29 m.

Given:

$m_{1}=78.75$ kg;

$x_{1}=1 \mathrm{~m} ;$

$h_{1}=15 \mathrm{~m} ;$

$m_{2}=45 \mathbf{~ kg ; ~}$

$h_{2}=29 \mathrm{~m}:$
Solution. The mechanical system "Earth-athlete-net" $x_{2}-?$ can be considered closed. According to the law of conservation of energy, when the athlete jumps, his potential energy should completely transform into the energy of the elastic deformation of the net: $m_{2} g\left(h_{2}+x_{2}\right)=\frac{k x_{2}^{2}}{2} ;$
$m_{1} g\left(h_{1}+x_{1}\right)=\frac{k x_{1}^{2}}{2} ;$

$\frac{m_{2}}{m_{1}} \cdot \frac{\left(h_{2}+x_{2}\right)}{\left(h_{1}+x_{1}\right)}=\frac{x_{2}^{2}}{x_{1}^{2}} ;$

$\frac{45}{78.75} \cdot \frac{\left(29+x_{2}\right)}{(15+1)}=x_{2}^{2}$;

$\frac{\left(29+x_{2}\right)}{16}=\frac{x_{2}^{2} \cdot 78.75}{45} ;$

$29+x_{2}=x_{2}^{2} \cdot 1.75 \cdot 16$

$28 x_{2}^{2}-x_{2}-29=0$

$D=1+4 \cdot 28 \cdot 29=3249=57^{2}$

$x_{2}=\frac{1+57}{56} \approx 1.04 \mathrm{M}$.

3. Mass spectrometry is one of the most precise methods of substance identification, used in almost all areas of human activity: pharmaceuticals, medicine, ecology,

![](https://cdn.mathpix.com/cropped/2024_05_06_b0c760bc4b9ab1ab7326g-19.jpg?height=405&width=597&top_left_y=2202&top_left_x=1118)
oil and gas industry, nuclear energy. It is a kind of "weighing" of molecules: components are ionized, and then the mass-to-charge ratio of the ions is determined. It turned out that when separating ions with comparable mass, for example, two isotopes of argon ( ${ }^{36} \mathrm{Ar}$ and ${ }^{38} \mathrm{Ar}$ ), difficulties arise. Calculate the radius of curvature required for these particles to be 1.2 mm apart.

Ions, after passing through crossed electric and magnetic fields, and after the velocity selector, enter only the magnetic field and move along the arc of a circle.

$\left\{\begin{array}{c}v=\frac{E}{B} \\ q v B^{\prime}=\frac{m v^{2}}{R}\end{array} \rightarrow m=\frac{q B B^{\prime} R}{E}\right.$.

$x=D_{2}-D_{1}=2\left(R_{2}-R_{1}\right)$

All quantities except $R$ in formula (1) are determined by the parameters of the setup, therefore, for particles with the same charge, it is valid

$\frac{R_{2}}{R_{1}}=\frac{m_{2}}{m_{1}}=38 / 36=1.056$

$R_{2}-R_{1}=\frac{x}{2}=0.6(\mathrm{MM})$

$0.6+R_{1}=1.056 R_{1}$

$R_{1}=10.7$ mm

$R_{2}=11.3 \mathrm{Mm}$