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Newton's cradle geometry

Konstantinos Foutzopoulos

Based on PLANCKS 2014. Problem 2.

A Newton’s cradle is well-known physics demo where a swinging ball hits upon a number of other (same) balls at rest. The momentum is transferred (fully if elastic collisions are assumed) to the last ball letting the others at rest. Then that ball performs a pendulum-like motion and the motion is realized in reverse.

Let a Newton’s cradle of (total) NN balls and assume that the launched ball has velocity at the time of collision v0v_0 and after the collision the balls have a velocity viv_i. As one ball cannot move faster than the next one, the following constrain relationship holds.

0⩽vi−1<vi 0 \leqslant v_{i - 1} < v_i

Then, by the conservation of momentum

mv0=∑mvi⇒v0=∑vi m v_0 = \sum m v_i \Rightarrow v_0 = \sum v_i

and by the conservation of energy

E0=∑Ei⇒T0=∑Ti⇒v02=∑vi2 E_0 = \sum E_i \Rightarrow T_0 = \sum T_i \Rightarrow v_0^2 = \sum v_i^2

The first equation describes a hyperplane that contains the points Pi=(δijv0)P_i = (\delta_{i j} v_0) whereas the second describes a hypersphere centered at origin of radius r=v0r = v_0. The solutions are the intersection of the plane and the sphere.

For N=2N = 2

v0=v1+v2v02=v12+v22⇒v0=v1+v2v1v2=0⇒(v1,v2)={(0,v0),(v0,0)} \begin{array}{c} v_0 = v_1 + v_2\newline v_0^2 = v_1^2 + v_2^2 \end{array} \Rightarrow \begin{array}{c} v_0 = v_1 + v_2\newline v_1 v_2 = 0 \end{array} \Rightarrow (v_1, v_2) = \{ (0, v_0), (v_0, 0) \}

For this case the plane degenerates to a line and the sphere to a circle. Their intersection are two points. Utilizing the constrain, only one (1) unique solution is left. The (v1,v2)=(0,v0)(v_1, v_2) = (0, v_0).

For N=3N = 3

v0=v1+v2+v3v02=v12+v22+v32⇒v0=v1+v2+v3v2v3+v1v3+v1v2=0 \begin{array}{c} v_0 = v_1 + v_2 + v_3\newline v_0^2 = v_1^2 + v_2^2 + v_3^2 \end{array} \Rightarrow \begin{array}{c} v_0 = v_1 + v_2 + v_3\newline v_2 v_3 + v_1 v_3 + v_1 v_2 = 0 \end{array}

As there’re more unknowns than equations there’re infinite solutions. For this case those form a circle. Utilizing the constrain, the valid solutions form an arc of that circle.

Although for N=3N = 3 (and N>3N > 3) there’re infinite solutions only one is experimentally realized. This is explained by considering two-body (N=2N = 2) collisions. Splitting in steps at each ball and based on previous result, it’ll be (ui−1,ui)=(0,u0)(u_{i - 1}, u_i) = (0, u_0).

Therefore finally there’ll be one (1) unique solution, the υ=(vj=δjNv0)\mathbf{\upsilon}= (v_j = \delta_{j N} v_0).