NBPhO 2023, Теория и експеримент — Задача 8. Connected charges

i) (2 т.)

As the ball $A$ enters the electric field, the force $qE$ is exerted on it, hence the acceleration of the system of two balls is $a_1 = qE/2m$; the acceleration remains constant until the ball $A$ exits the field. This will happen when $L = a_1 t_1^2/2$, hence $t_1 = \sqrt{2L/a} = 2\sqrt{Lm/qE}$ At that moment, the acceleration of the ball $A$ becomes zero, but at the very same moment, ball $B$ enters the field. The force $qE$ exerted on it will give it twice as big an acceleration as before, $a_2 = qE/m$, and the string becomes loose. So, the speed of the ball $B$ is now bigger than that of the ball $A$, and so they will eventually collide. The time $t_2$ spent by the ball $B$ in the field can be expressed as $t_2 = (v_f - v_i)/a_2$, where $v_i = \sqrt{qEL/m}$ is its initial speed, and $v_f = \sqrt{3qEL/m}$ — the final speed. So, $t_2 = \sqrt{Lm/qE}(\sqrt{3} - 1)$. At the moment when the ball $B$ exits the field, the distance between the balls is $s = L - (v_f - v_i)t_2/2 = v_i t_2 = L(\sqrt{3} - 1)$, and the difference between the velocities of the balls is $\Delta v = v_f - v_i = \sqrt{qEL/m}(\sqrt{3} - 1)$, hence the time remaining until the collision $t_3 = s/\Delta v = \sqrt{Lm/qE}$. Thus, the moment of collision $t = t_1 + t_2 + t_3 = \sqrt{Lm/qE}(\sqrt{3} + 2)$.

Grading:

• Initially both charges move with the same acceleration (0.2 pts)
• When A escapes the field region A moves with constant speed, B moves with higher acceleration (0.2 pts)
• When B escapes, both move with constant speed (0.1 pts)
• Correct $a_1$ (0.2 pts)
• Correct $t_1$ (0.2 pts)
• Correct $a_2$ (0.1 pts)
• Correct $v_i$ (0.2 pts)
• Correct $v_f$ (0.2 pts)
• Correct $t_2$ (0.2 pts)
• Correct $s$ (0.2 pts)
• Correct $t_3$ and $t$ (0.2 pts)

ii) (2 т.)

First solution: The centre of mass moves with constant acceleration $a_1 = qE/2m$, and has travelled distance $s_{CM} = L - l/2 + l/(2\sqrt{2}) = 0.8200L = a_1 T^2/2$, hence $T = 1.811\sqrt{\frac{mL}{qE}}$.

Grading:

• Use $s_{CM} = a_1 T^2/2$ for centre of mass (0.5 pts)
• Correct $a_1$ (0.5 pts)
• Correct $s_{CM}$ (0.6 pts)
• Correct $T$ (0.4 pts)
• If you assume small angle oscillation to calculate time necessary for the string to align with the $x$-axis. (up to 1.4 pts)

iii) (2 т.)

First solution: Due to the conservation of $y$-directional momentum, the $y$-components of the balls are always equal. At $t = T$, the string is parallel to the $x$-axis; since it is not stretchable, $x$-components of the balls are equal, too, at that moment. Hence, the speeds are equal, and energy conservation law states $EqL = mv^2$ so that $v = \sqrt{\frac{qEL}{m}}$.

Grading:

• Using energy conservation law (1.0 pts)
• Kinetic energy of both masses (0.6 pts)
• Work done by electric field (0.2 pts)
• Correct $v$ (0.2 pts)

Alternative solution: The time period $0 < t < T$ is most conveniently analysed in the centre-of-mass frame. In that frame, there is a force of inertia equal to $F_i = -ma_1 = -qE/2$. Hence, both balls perform non-linear oscillations around the centre of mass (it is nonlinear because the amplitude is not small). The speed of the ball at $t = T$ can be found from the energy conservation law. At $t = 0$, the speed is zero so that the full energy is due to the potential energy in the homogeneous force field $qE/2$: $\Delta U = \frac{qE}{2}\left(\frac{l}{2} - \frac{l}{2\sqrt{2}}\right) = \frac{qEL}{8}(2 - \sqrt{2}) = \frac{mv_y^2}{2}$, hence $v_y^2 = \frac{qEl}{4m}(2 - \sqrt{2}) = 0.1800\frac{qEL}{m}$. This gives us the velocity component in the $y$-direction; once we return to the laboratory frame, we need to add the $x$-component of the frame’s velocity $v_x = a_1 T = \frac{qE}{2m}1.811\sqrt{\frac{mL}{qE}} = 0.9055\sqrt{\frac{qEL}{m}}$ so that $v = \sqrt{v_x^2 + v_y^2} = \sqrt{\frac{qEL}{m}}$.

Grading:

• Horizontal speed $v_x$ (0.4 pts)
• Use energy balance or something similar to find rotational speed $v_y$ (0.4 pts)
• Correct distance travelled (0.6 pts)
• Rotational speed $v_y$ (0.4 pts)
• Pythagorean theorem (0.2 pts)

iv) (2 т.)

For the string to be parallel to the $x$-axis, the system must perform $n + \frac{1}{2}$ half-periods, where $n$ is an integer. Since now the amplitude is small, the oscillations are harmonic, and the circular frequency $\omega = \sqrt{\frac{qE}{ml}}$ (we can use the formula for a pendulum of length $l/2$ and equivalent gravity field $g = qE/2m$). So, the travel time must be $\tau = \pi\sqrt{\frac{ml}{qE}}(n + \frac{1}{2})$. On the other hand, we can use the result from task ii: $\tau = \sqrt{\frac{ml}{qE}}\sqrt{4\lambda - 2 + 2\cos\phi}$, where $\lambda = L/l < 1$ and $\cos\phi \approx 1$. So, $\pi(n + \frac{1}{2}) = \sqrt{4\lambda}$. Since $\lambda < 1$, this equality can only be satisfied with $n = 0$, hence $\lambda = \frac{\pi^2}{16} = 0.6169$ so that $l = L/\lambda = 1.6211L$.

Grading:

• Correct circular frequency (0.6 pts)
• If circular frequency deviates from the right by factor $\sqrt{2}$ (0.2 pts)
• Correct rotation time with $n$ (0.6 pts)
• Correct rotation time without $n$ (0.2 pts)
• Correct travel time from ii (0.4 pts)
• Derivation of $l$ (0.4 pts)