Goldilocks And The Three Spheres

Relevant wiki: Conservation of Momentum

The interactions between the balls occur in two stages. Initially, the string between balls A and B is taut, but goes slack as soon as A begins to move. It becomes taut again when ball A makes an angle of $\SI{60}{\degree} \equiv \pi/3$ to the horizontal with ball B, at which point B is pulled toward A along the line connecting them. This is because a section of rope can only pull along its tangent line (can you explain why?).

Just after B is pulled toward A (at $\theta=\SI{60}{\degree}$ to $\vec{v}_0$ ), C is pulled toward B by the taut rope connecting them (at an angle $\theta=\SI{60}{\degree}$ to the velocity of B). We have therefore argued that when balls B and C are pulled, their velocity is along the line connecting it to A or B, respectively.

Let's treat the first event. As soon as the string between A and B becomes taut, the momentum $m_Av_0$ is reallocated between the two balls in an energy-conserving process. Thus we have (dropping the factors of $\frac12$ ) $m_A v_0^2 = m_A \vec{v}_A^2 + m_B \vec{v}_B^2$ Momentum is conserved in each dimension. It's most convenient to treat the collision in a coordinate system $\left(x^\prime, y^\prime\right)$ in which one dimension lies along the taut string connecting A and B. We have $\begin{aligned} m_Av_0\cos\pi/3 &= m_Av_A^{x^\prime} + m_Bv_B^{x^\prime} \\ m_Av_0\sin\pi/3 &= m_Av_A^{y^\prime} \end{aligned}$ We have three equations with three unknowns, thus we can solve for the velocity of ball B $\vec{v}_B = \langle v_B^x, 0 \rangle$ . Writing the energy conservation relation in components, and applying conservation of momentum in the $y$ -dimension, we have $\begin{aligned} m_Av_0^2 &= m_A\left(v_A^{x^\prime}\right)^2 + m_A\left(v_A^{y^\prime}\right)^2 + m_B\left(v_B^{x^\prime}\right)^2 \\ &= m_A\left(v_A^{x^\prime}\right)^2 + m_B\left(v_B^{x^\prime}\right)^2 + m_Av_0^2\overbrace{\sin^2\pi/3}^{\frac34}\\ \frac14 m_Av_0^2 - m_B\left(v_B^{x^\prime}\right)^2 &= m_A\left(v_A^{x^\prime}\right)^2 \\ \end{aligned}$ Applying momentum conservation in the $x^\prime$ -dimension, we have $\begin{aligned} \frac14 m_Av_0^2 - m_B\left(m_B^{x^\prime}\right)^2 &= m_A\left(v_A^{x^\prime}\right)^2 \\ &= \frac{1}{m_A}\left(\frac12 m_Av_0 - m_Bv_B^{x^\prime}\right)^2 \end{aligned}$ Expanding, we have $\frac14 m_Av_0^2 - m_B\left(v_B^{x^\prime}\right)^2 = \frac14 m_Av_0^2 + \frac{m_B^2}{m_A}\left(v_B^{x^\prime}\right)^2 - v_0m_Bv_B^{x^\prime}$ which leads to $m_B\left(v_B^{x^\prime}\right)^2\left(1 + m_B/m_A\right) = v_0m_Bv_B^{x^\prime}$ and therefore the speed $v_b$ is given by $v_b = \left|v_B^{x^\prime}\right| = \frac{m_A}{m_A + m_B}v_0$ We argued above that the interaction between A and B is of the same nature as the one between B and C. Therefore, the initial speed of C is given by $\begin{aligned} v_c &= \frac{m_B}{m_B + m_C}v_b \\ &= \frac{m_B}{m_B + m_C}\frac{m_A}{m_A + m_B}v_0 \end{aligned}$ In the general case of $n$ items in an equilateral arrangement with $l(\overline{AB}) \ll l(\overline{BC})$ , $l(\overline{BC}) \ll l(\overline{CD})$ etc., the initial velocity of the last ball is given by $v_n = v_0\prod_{i=1}^{n-1}\frac{m_i}{m_i + m_{i+1}}$