Written exam in physics at NSU

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WRITTEN EXAM
IN PHYSICS AT NSU

G. V. Meledin

The three variants of problems that we publish in this issue of the journal are intended this year for those who entered the physics department of Novosibirsk State University. After the conditions of each task there are clear numbers. The first of them is the percentage of those who solved this problem from among those who entered, the second - from those who did not enter, the third - from the total number of applicants. The number of points for correct solution of the problem is indicated in parentheses.

In order to receive a satisfactory grade, it was necessary to score 5-9 points. An excellent score was given if the algorithm scored at least 17 points.

Option 1

1. A sports hammer is a cannonball on a cable that is thrown, spinning around itself with sufficient speed. Find the maximum distance that the hammer can fly if the holding force of the sportsman before throwing the shot is $n$ times the weight of the hammer. The distance from the axis of rotation to the core along the cable is $l$. The influence of sports and the growth of sports are in demand. 90, 53, 66 (5).

2. A rectangular wire frame with current on the sides of length $a$ and $b$ is fixed so that it can freely rotate around the horizontally located side $a$ (Fig. 1). The weight of side $a$ is $P_1$, and side $b—P_2$. The frame is in a vertical uniform magnetic field with induction $B$. Find the magnitude of the current in the frame if the angle of inclination of the frame to the horizon is $\alpha$.

3. The cosmic corpus moves towards the Moon under the influence of its gravity. At a great distance, the speed of the corps relative to the Moon was zero. The acceleration of gravity on the surface of the Moon is $n=6$ times less than on Earth ($g_{\pi} = \frac{g}{6}$). The radius of the Moon $R$ is about 1700 km. At what altitude $h$ should the braking engine be turned on to achieve a soft landing, if we assume that the engine creates a five-fold overload (5 $g$)? The change in the mass of the hull during braking due to the withdrawal of fuel and the dependence of the gravitational force of the Moon on the distance at the braking stage can be taken into account ($h$ is much less than $R$). 83, 14, 39 (7).

4. On a smooth wedge of mass $M$, which slides only horizontally, falls

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ball of mass $m$ (Fig. 3). The ball elastically hits a face forming an angle $\alpha$ with the horizon. The speed of the ball immediately before the strike is $v$ and is directed vertically downwards. Find the speed of the wedge after the blow. Trennium can be neglected. 7, 2, 4 (7).

Solutions

1. The cannonball’s flight range is maximum if its departure angle is $45^\circ$. Moreover, that $n \gg 1$. Then we can assume that when the nucleus unwinds, centripetal acceleration is imparted to it by the tension force of the thread. At the moment of throwing $\frac{mv^2}{l} = nmg$. This means that the speed that the athlete imparts to the core when throwing is equal to $v = \sqrt{lng}$.

Now it is not difficult to find the range of the cannonball. Let us write down the kinematic equations of its motion:

vertically $y = v_{\text{верт}} \cdot t - \frac{gt^2}{2}$, horizontally $x = v_{\text{гор}} \cdot t$.

At the moment of falling to the ground $y=0$. Substituting this value into the first equation, we find the flight time of the nucleus $t = \frac{2v_{\text{верт}}}{g} = \frac{v\sqrt{2}}{g}$. (the root $t=0$ corresponds to the moment of the throw).

During this time the cannonball will fly a horizontal distance

$$x = v_{\text{гор}} \frac{v\sqrt{2}}{g} = \frac{v^2}{g} = nl.$$

2. Let us write the condition that the sum of the moments of forces acting on the frame is equal to zero, taking into account that the moments of forces acting in the magnetic field on the sides of the frame balance each other (since the currents induce through them are opposite). $P_1b \cos \alpha + 2P_2 \frac{b}{2} \cos \alpha - BIa \sin \alpha = 0$.

Hence $I = (P_1 + P_2) \frac{\text{ctg} \alpha}{a}$.

3. Let us write down the law of conservation of energy for the stage of approaching the ship to the Moon. From the analogy of the gravitational field with the electric field, we can conclude that if the potential energy of the body is considered equal to zero indefinitely, further from the Moon, then at a distance of $H$ from its center the potential energy is equal to $-\gamma \frac{Mm}{H}$, where $\gamma$ is the gravitational constant, $M$ is the mass of the Moon and $m$ is the mass of the body. If there is no potential energy from the surface of the Moon, then $\Pi = \gamma \frac{Mm}{R} - \gamma \frac{Mm}{H}$.

Far from the Moon, the ship has only potential energy $G \frac{Mm}{R}$, and at an altitude $h$ from the Moon’s surface, it has kinetic energy $\frac{mv^2}{2}$ and potential energy $\Pi = \gamma \frac{Mm}{R} - \gamma \frac{Mm}{R+h}$.

Therefore $\gamma \frac{Mm}{R} = \frac{mv^2}{2} + \gamma \frac{Mm}{R} - \gamma \frac{Mm}{R+h}$.

Considering that on the surface of the Moon there is $\gamma \frac{Mm}{R^2} = mg_\pi$ and therefore $\gamma M = R^2g_\pi = \frac{1}{6} R^2g$, we can write that

$$\frac{v^2}{2} = \frac{1}{6} g \frac{R^2}{R+h}.$$

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The change in energy of the ship during braking is equal to the work of the engine. Since a five-fold overload is created during engine operation, its traction force should be equal to 5 mg. Because

$$5mgh = \frac{mv^2}{2} + mg_\pi h = \frac{mv^2}{2} + \frac{1}{6} mgh$$

(as $h \ll R$, then $\Pi = \gamma \frac{Mm}{R} - \gamma \frac{Mm}{R+h} = \gamma \frac{Mmh}{(R+h)R} \approx \gamma \frac{Mm}{R^2} h = mg_\pi h$).

Hence $\frac{v^2}{2} = \frac{29}{6} gh$.

Substituting this expression for $\frac{v^2}{2}$ into equation (1), we find $h \approx \frac{R}{29} \approx 59$ km.

4. Using the law of conservation of energy and the conditions for conservation of the components of the system’s momentum in the horizontal direction and along the wedge, we can write:

$$v_0^2 = v_x^2 + v_y^2, \quad mv_x = Mn,$$

$$v_0 \sin \alpha = v_x \cos \alpha - v_y \sin \alpha.$$

Hence $u = -\frac{v_0 \sin 2\alpha}{M/m} + \sin^2 \alpha$.

Option 2

1. The plane flies in a straight line from city $M$ to city $H$ and back. Find the ratio of the total flight times in the case when a wind blows from $M$ to $H$, the speed of which is equal to $u$, and when the wind blows with the same speed perpendicular to the line $M-H$.

The speed of the aircraft relative to the air in both cases is equal to $c$. 89, 74, 90 (5).

2. The apex point of the fixed ball is placed in the case when a wind blows from $M$ to $H$, the speed of which is equal to $u$, and the fixed ball; the center of gravity of the toy—point $A$—is located on a polyline of the radius of the hemisphere. Will the fork-stand fall off the ball? There is no slippage. 83, 22, 44 (6).

3. The circuit shown in figure 5 is connected to an alternating voltage source with an amplitude of 220 V and a period of $1/50$ s. Find the maximum voltage value across the resistance $R$ and must perniode, during which the current in the circuit is different from zero. Draw a graph of the voltage across resistance $R$ versus time.

Assume that the dependence of the current through the diode on the voltage applied to it has the form shown in figure 6. 83, 14, 39 (7).

4. When photographing a very distant object, a thin ruler of width $D$ was placed in front of the object camera at a distance equal to the focal length. The plane of the ruler is perpendicular to the optical axis of the object. The lens is a thin diameter $d$ ($d < D$). Determine the width of the shadow on the film (the width of the area in which all the rays coming from the photographed object will be shaded by the ruler). 29, 5, 14, (8).

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Solutions

1. The time ratio is
   $$\frac{L}{c+u} + \frac{L}{c-u} = \frac{c}{\sqrt{c^2-u^2}}$$
   ($L$ — distance between cities). If the wind blows perpendicular to line $M-H$, then the speed $c$ of the aircraft relative to the air must be directed so that its sum with the wind speed $u$ is directed along line $M-H$.

2. Let’s tilt the toy at a small angle $\alpha$ and see what happens to the center of gravity of the toy. If it goes down, the equilibrium position is unstable and the toy will fall; if it rises, then the equilibrium position is stable.

In our drawing (Fig. 7), lines $O'B$ and $A'C$ are vertical, and line segment $O'C$ is horizontal. $O'B=2R\cos \alpha, <A'O'D=\alpha, a <DO'C=90^\circ-\alpha$. Therefore $<A'O'C=\alpha-(90^\circ-\alpha)=2\alpha-90^\circ$ and
$$A'C = A'O'\sin(2\alpha-90^\circ) = -\frac{1}{2}R\cos 2\alpha.$$

Let’s find the displacement of the center of gravity.
$$\Delta h = AD - O'B + A'C = \frac{3}{2}R - 2R\cos \alpha + \frac{R}{2}\cos 2\alpha = R(1-\cos \alpha)^2 > 0.$$

Thus, we see that the center of gravity of the toy has dropped. The Vanka insert will fall from the hemisphere.

3. The graph of the current through a diode versus voltage can be deciphered simply: no current flows through the diode until the voltage applied to it reaches +10 $\theta$. As the voltage increases further, the diode is equivalent to a resistance of 10 $\omega$.

Let the alternating voltage change with time according to the law $u(t) = 220\cos \frac{2\pi t}{T}$. If we take into account the e.m.c. batteries, then it is clear that the diode opens when $u(t_1) = 210\theta$ and closes when the voltage drops again to $u(t_2) = 210\theta$. That is, it is open until $\cos \frac{2\pi t}{T} \geq \frac{210}{220}$.

In general, the fraction of the period during which the diode is open is equal to $\frac{t}{T} = \frac{1}{2\pi}\arccos \frac{21}{22}$.

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The maximum voltage across resistance $R$ is 5 $\theta$ (10 $\theta$ divided equally between the diode and resistance $R$). A qualitative graph of the voltage across resistance $R$ versus time is shown in figure 8.

4. Let’s construct an image of the shadow, taking into account that the film in the camera is located in the focal plane (Fig. 9). Since a distant object is being photographed, the lens collects a parallel beam of rays at the focus.

From the construction it is clear that $x = \frac{D - d}{3}$.

Now try to solve the problems of option 3 on your own.

Option 3

1. A constant voltage source $U$ is connected to the circuit of resistances $R_1, R_2, R_3$ and $R_4$ (Fig. 10) at points $a$ and $b$, and a high-resolution voltmeter is connected at points $c$ and $d$. What potential difference will the voltmeter show? 94, 55, 70 (3).

2. At the bottom of a vessel with liquid, the specific gravity of which is $d$, there is a $\Gamma$-shaped body with the dimensions indicated in figure 11. The liquid does not leak under the lower edge of the body. Specific body weight $d_0 = 2d$. At what height $h$ liquid level in the vessel will the equilibrium of the body be disturbed? 78, 22, 40 (6).

3. When the capacitor was always $P$ suspended externally, its length increased by $l$. How much more does the length of the spring change if a beam of electrons enters this capacitor parallel to the plates? Beam current $I$, particle speed $v$. Having passed the capacitor, the beam changes direction by an angle $\alpha$.

Electron charge $e$, mass $m$. 87, 29, 51 (7).

4. A lead sheet of weight $P$ rests on a roof with a pitch angle of $\varphi$. The friction coefficient of lead on the roof is $k > tg \varphi$. Lead linear expansion coefficient $\alpha$. Assuming that the temperature rises during the day, reaching a maximum value of $t_1^0$ C, and then drops to $t_2^0$ C, determine the position of the points that are motionless, respectively, when heating and leaving the sheet during the day, if the length of the sheet at a minimum temperature of $t_1^0$ C is equal to $l$. Find how far the leaf will slide in $N$ days according to the weather conditions. 42, 4, 19 (9).