# Colliding Hang Balls [closed] Two identical balls are attached at the ends of inelastic and very light string. Hanging on a fixed friction less pulley, one is pulled higher so that it is at right angle with the other ball. If then released the ball being held is ought to hit or touch the lower ball below. (see figure)

Mass of balls = 1 lb. (1" radius) Let L = length of string =10 ft. Let X = horizontal distance of pulled ball ( center to center of balls) Let Y = vertical distance of pulled ball ( center to center of balls)

How far is X so that when the higher ball is released it will collide with the lower ball?

• Hit on single swing only – TSLF Oct 28 '16 at 0:55
• Is this a puzzle, or physics? – greenturtle3141 Oct 28 '16 at 1:37
• This seems like a textbook physics problem to me, rather than a puzzle. – Dan Russell Oct 28 '16 at 3:15
• shouldn't this be on the physics SE? – MMAdams Oct 28 '16 at 9:40
• What a surprising result for such a simple setup! The cetrifugal force from ball X at the bottom of the swing pulls up ball Y very fast. – humn Oct 29 '16 at 3:56

(Now a community wiki, as the puzzle is closed, in case anyone wants to add or correct.)

Not a numerical solution but a couple of interesting, if true, conjectures.

Conjecture A. Ball X will always move downward more rapidly than ball Y.

Conjecture B. Although ball Y moves downward at first, it will be moving upward when the balls meet.

Really?

Conjecture A.   Ball X will always move downward more rapidly than ball Y.

This is because the overall downward force is almost always greater on X than on Y, and never less. While gravity pulls down on both balls with the same force, the tension along the cable always works more directly against gravity on ball Y than on ball X, except when they are both at vertical, where they experience identical forces.

Conjecture B.   Ball Y will be moving upward when the balls meet.

As the cable’s length is fixed, when the balls meet they must either have no up–down motion or have equally opposite up–down motions. (Ball X will also be moving sideways.) As ball X will be going downward more rapidly than ball Y (according to conjecture A), they cannot have 0 up–down motion and ball Y must be the one going upward.

## (from Rubio:)

• At starting position X=5.925' the balls will collide on the first swing.
• Both conjectures are true.

While I lack the math skill to solve this, I was able to write a simulation script to step through the kinematics in very small $\Delta$t and the results are incorporated in the following pretty graph (now with data included): Since the lower ball is a fixed distance away from the upper one, along the path of the string (i.e. up and over the pulley), and it's identical to the upper ball, we can actually disregard its existence entirely and treat it as a fixed magnitude force pulling the upper ball in the direction of the pulley. The only interesting part about it is that we care when its position and the upper ball's position coincide, which (ignoring the balls' diameter) happens when both are 5 feet below the pulley and directly under it (X=0,Y=-5). So now we have one mass to worry about, with two forces acting on it - one, downward force of gravity, and one, equal in magnitude but different in direction, the string pulling it in the direction of the pulley. Iterating over small steps in time and computing each acceleration, velocity, and position in both X and Y, gives the results shown in the graph.

Final conclusion:

• Initially both balls fall.
• Conjecture A holds; the upper ball never falls slower than the lower. This makes sense; at t=0, no part of the lower ball's force is counteracting the force of gravity on the upper ball - the tension of the string is directed entirely horizontally. As the direction of the tension force changes, the proportion of the tension that acts counter to gravity on the upper ball increases, eventually exactly countering it when the balls collide; but up to that point, the tension force of the string acts entirely against gravity for the lower ball, but only fractionally against gravity for the upper ball--the balance is what gives the upper ball its horizontal acceleration.
• At the outset, upper ball is being pulled equally downward and leftward, so it moves on a 45° heading downward/leftward.
• Around ⅓ second into its trip, the horizontal acceleration of the upper ball is largely dominating its motion, and continues to grow.
• Conjecture B holds; the upper ball accelerates leftward due to the horizontal component of the tension force pulling it toward the pulley, and for nearly the first ¾ of its trip that movement has been shortening the string length rightward of the pulley, allowing the lower ball to fall. From then on, however, the velocity vector of the upper ball causes it to increase the linear distance between itself and the pulley, so it begins pulling the string back over the pulley and shortening the length of string from which the lower ball hangs. The lower ball starts to move up.
• It is possible to arrange the relative movements of the balls so they will collide. The most direct hit I was able to make occurs when the upper ball starts at X=5.925' (roughly 71") to the right of the pulley, leaving the lower ball suspended roughly 49" below the pulley. Given these starting conditions, the balls collide at about 0.631 seconds after release, at a point where both balls are (of course) 5' directly below the pulley.

The code:

#! /usr/bin/perl -w
use strict;
my $dt=.000001; my$t=0; my $x1=0; my$y1=0; my $x2=5.925; # through trial and error my$y2=0; my $vx=0; my$vy=0; my $L=10; my$g=32.174;    # everything is in feet
while ($x2>$x1) {
$t+=$dt;
my $A=atan2(-$y2,$x2); my$Fx=-$g*cos($A); my $Fy=$g*sin($A)-$g;  # tension force, x & y components
$vx+=$Fx*$dt;$vy+=$Fy*$dt;   # apply force as acceleration to velocity (*)
$x2+=$vx*$dt;$y2+=$vy*$dt;   # apply velocity to position
$y1=sqrt($x2*$x2+$y2*$y2)-$L; # string left of the pulley
printf("%8.6f %11.8f %11.8f %11.8f %11.8f %11.8f $A\n",$t,$x2,$y2,$y1,$vx,\$vy);
}


(*)note we cheat here a little; since both masses are equal, there's no need to muck about with F=ma - force and mass are equal magnitude, just different units. it all works out (check if you don't believe it).

• A comment by OP on the question stipulates a collision after a single swing, which goes against conjecture C, I'm afraid. – LogicianWithAHat Oct 28 '16 at 10:11
• Thanks, @LogicianWithAHat, that saves a diagram. (First OP comment, at that, oooo.) – humn Oct 28 '16 at 10:14
• Posted as complete answer as I could fit in a couple of comments earlier - I'll add to the wiki when I'm back at a computer. – Rubio Oct 29 '16 at 3:42
• Hope I'm overlooking something, @Rubio, but keep getting that the leftward tension on X just after release is only half of gravitational force, so ball X's initial path might be steeper than 45 degrees. (Just before release, I do get that the tension equals gravity.) – humn Oct 29 '16 at 5:56
• I found something like the OP set up in wikipedia "Swinging Atwood Machine" that uses 2 pulleys, but i guess the lagrangian mechanics applies to both. – TSLF Oct 29 '16 at 13:52