### Разбалловка

C1  4.50 A weightless rod of a length $2R$ is placed perpendicular to a uniform magnetic field $\vec B$. Two identical small balls of mass $m$ and charge $q$ each are attached at the rod ends. Let us direct $z$-axis along the magnetic field and place the origin at the rod center. The balls are given the same initial velocity $v$ but in opposite directions so that one of the velocities is precisely in the $z$-direction. What are the maximum coordinates $z_\mathrm {max}$ of the balls? Express your answer in terms of $q, B, m, v$, and $R$. Find the magnitude of the ball accelerations at this moment and express your answer in terms of $q, B, m, v, R$, and $z_\mathrm{max}$.

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 Stated that the Lorentz force acts on bodies, $\vec{F} = q\left[ \vec{v} \times \vec{B} \right]$ 0.2 The 2nd Newton's law is written for the system of bodies $2m\dot{\vec{v}}_C = q\left[ \left( \vec{v}_1 + \vec{v}_2 \right) \times \vec{B} \right]$ Remark: The 2nd Newton's law written correctly for a single ball gives a full score 0.2 Stated that the center of mass doesn't move. (If this conclusion is made without showing math, the previous marks (0.2+0.2) are also given.) 0.4 The angular form of 2-nd Newton's law is written $\dfrac{d\vec{L}}{dt} = q \left[ \vec{R} \times \left[ \vec{v} \times \vec{B} \right] \right]$ Remark: this equation also may be written as $z$-axis projection 0.5 Calculated $L_z = \dfrac{qBz^2}{2}$ 0.8 Stated that $z$ is maximum when $v_z =0$ 0.2 The geometry is used correctly to find the direction of $\vec{L}$ when $z$ is maximum and written correct expression $L_z=\sqrt{R^2-z^2}mv$ 0.4 The kinetic energy is constant because the power of the Lorentz force is zero. 0.2 The velocity of the ball $v$ is constant 0.2 The biquadratic equation for $z$ is obtained: $\left( \dfrac{qB}{2mv} \right)^2 z^4 + z^2 - R^2=0$ 0.2 Calculated $z_\text{max} = \dfrac{\sqrt{2}mv}{qB} \sqrt{\sqrt{1+\left(\frac{qBR}{mv} \right)^2 }-1}$ 0.3 Tangential acceleration is $a_1=0$ because $v=\text{const}$ 0.2 Radial acceleration is $a_2 = v^2/R$ 0.3 Component of the acceleration in the 3-rd direction found from the 2-nd Newton's law: $ma_3=qvB\dfrac{z_\text{max}}{R}$ 0.2 Answer for $a$: $a=\sqrt{\left( \dfrac{v^2}{R} \right)^2 + \left( \dfrac{qvBz_\text{max}}{mR} \right)^2 }$ 0.2 Minor mistake: missing numerical factor or a non-dimensional typo when rewriting an equation -0.1 Major mistake: missing symbol or a dimensional typo when rewriting an equation -0.2