Written exam in physics at MIPT in 1970

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Written exam
in physics at MIPT in 1970

V. E. Belonuchkin, S. M. Kozel

The written entrance exam in physics is not held in all universities. However, the tasks offered in the written work are also interesting for those who are only going to take the oral exam. Here we will analyze the tasks of one of the options for written work in physics for the entrance exam at the Moscow Institute of Physics and Technology.

Vilet No. 6

1. A particle of mass $m$, moving with a speed $v$, collides with a resting part of the mass $\frac{m}{2}$ and, after an urgent impact, rebounds at an angle $\alpha = 30^\circ$ to the direction of its initial motion. At what speed does the second part begin to move?

2. A standard cylinder contains 0.2 $e$ hydrogen and 3.2 $e$ oxygen at a temperature of 27°C. Hydrogen is contained with oxygen, and after the reaction is completed, the pressure inside the cylinder increases threefold. What will be the temperature inside the cylinder?

3. Attaching some additional resistance to the voltmeter increases the maximum voltage change by $n$ times. Another added resistance increases the maximum change by $m$ times. How many times will the maximum change in voltage by a voltmeter increase if these two resistances connected in parallel are connected in series with the voltmeter?

4. An astronomical camera for photographing the Sun has a lens with a focal length of $F = 10$ m. In front of the lens is a round circle with a diameter of $d = 5$ cm. The photographic film used has a sensitivity of $W = 10$ mox·sec. What should the exposure time t be under these conditions? It is known that direct solar rays create $E = 4 \cdot 10^4$ mox on a plane of illumination perpendicular to them. The angular diameter of the Sun $\alpha$ is taken equal to 0.01 rad.

Let’s consider solving these problems.

1. During an impact, the momentum and kinetic energy of the colliding bodies are conserved. However, momentum cannot be conserved in scalar form:

$$mv = \frac{m}{2} v_1 + mv_2.$$

This entry is correct if both particles, after impact, move along the same straight line along which the impacting part moved before the impact, that is, with a central impact. In our case, as can be seen from the conditions of the problem, the impact was non-central. Therefore, it must be taken into account that the amount of motion is a vector quantity and is added along the right vectors (Fig. 1).

Using the cosine theorem, we can write the law preserving momentum in the form

$$\left( \frac{m}{2} v_1 \right)^2 = (mv)^2 + (mv_2)^2 - 2(mv_2)(mv) \frac{\sqrt{3}}{2},$$

where $v_1$ denotes the speed of a particle with mass $\frac{m}{2}$, and $v_2$ denotes the speed of a particle with mass $m$. Besides,

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Let’s use the law of conservation of energy:

$$\frac{mv^2}{2} = \frac{mv_2^2}{2} + \frac{mv_1^2}{4}.$$

Eliminating $v_2$ from these equations, we find

$$v_1 = \frac{2}{\sqrt{3}} v.$$

The vector nature of momentum can also be taken into account. Let us introduce mutually perpendicular coordinate axes $x$ and $y$.

For projections of momentum onto such axes, scalar relations are satisfied

$$mv_{2y} + \frac{m}{2} v_{1y} = 0,$$

$$mv_{2x} + \frac{m}{2} v_{1x} = mv.$$

The law of conservation of energy will take the form

$$\frac{mv^2}{2} = m \frac{v_{2y}^2 + v_{2x}^2}{2} + \frac{m}{2} \frac{v_{1x}^2 + v_{1y}^2}{2}.$$

Having solved this system and taking into account the next relation $v_1 = \sqrt{v_{1x}^2 + v_{1y}^2}$, we get the same answer.

2. Many applicants believed that the words “the reaction has ended” meant that 3.4 $z$ of water had formed in the cylinder. However, from the conditions of the problem it follows that there is an excess of oxygen in the cylinder, since for the formation of 1 $z$-mol of water, 1 $z$-mol of hydrogen and only $1/2$ $z$-mol of oxygen are needed. Let $m_1$ and $m_2$ be masses

Рис. 1.

Let us denote the resistance of the voltmeter as $r$, the maximum voltage across the voltmeter as $U_0$, and the current flowing through the voltmeter at this voltage as $I_0$ (Fig. 2). When using additional resistance $R_g$, the variable voltage $U_s$ is distributed between $r$ and $R_g$ in proportion to their magnitude. The voltage on the voltmeter $U$, of course, still should not exceed $U_0$, and the variable voltage $U_s$ can be greater than $U_0$. Current

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Through a voltmeter and $R_g$ will not exceed $I_0$.

Based on these considerations, for the three servants described in the problem, we have:

$$I_0 (R_1 + r) = n U_0,$$
$$I_0 (R_2 + r) = m U_0,$$
$$I_0 \left( \frac{R_1 R_2}{R_1 + R_2} + r \right) = k U_0.$$

Here $R_1$ and $R_2$ are additional resistances, and $k$ is the desired value. Given that $I_0 r = U_0$, these equations can be transformed to the form:

$$R_1 + r = n r,$$
$$R_2 + r = m r,$$
$$\frac{R_1 R_2}{R_1 + R_2} + r = k r.$$

From the first two equations we express $R_1$ and $R_2$. Substituting them into the third, we get

$$k = \frac{mn - 1}{m + n - 2}.$$

4. The diameter of the image of the Sun on photographic film is $D = \alpha F$. Illuminated images can be found

Fig. 2.

Fig. 3.

taking into account that all light energy penetrating through the lens is distributed over the image area (we neglect light losses):

$$E_{\text{изозокр}} D^2 = Ed^2,$$

or

$$E_{\text{изозокр}} = E \frac{d^2}{\alpha^2 F^2} = 10^4 \text{ люкс.}$$

The exposure time will be found from the condition: $E_{\text{изозокр}} \tau = W$.

From here

$$\tau = \frac{W}{E_{\text{изозокр}}} = 10^{-3} \text{ сек.}$$

We invite you to decide for yourself the following option.

1. The jet boat takes in sea water and throws it out at the back with a speed of $V$ relative to the catarrh. At the same time, it moves with speed $v$. Three people aimed a towed vessel at the length of the cathar, the drag force of which, at the same speed of movement, is equal to the resistance of the catarr. Determine the speed of the tug if it is known that the drag forces of the boat and the towed vessel vary in proportion to their speeds.

2. A balloon with gel at a gel pressure of 65 am at a temperature of $-3^\circ\text{C}$ weighs 21 ke, and at a pressure of 20 am at the same temperature it weighs 20 ke. How much gel did the balloon contain at an initial pressure of 150 am and a temperature of $27^\circ\text{C}$?

3. What heat will be released in the circuit (see Fig. 3) when switching key $K$ from position 1 to position 2?

4. On a satellite flying in a circular orbit at an altitude $H = 100\text{ км}$, there is a camera whose lens has a focal length $F = 10\text{ см}$. The surface of the Earth under the satellite is photographed. Film resolution (that is, the minimum size of visible image details, determined by the grain structure of the photographic emulsion) $d = 10^{-2}\text{ мм}$. What is the minimum size for photographic items? What shutter speed should you choose (exposure time $\tau$) so that the orbital motion of the satellite does not affect the image quality?