Two identical balls roll across the top of a table on parallel paths. One of the balls has to roll down into and up out of a dip in the table. The other ball rolls on the flat all the time. Which ball gets to the far side of the table first, and why?.

The ball traversing the dips can be either slower or faster getting across the table, depending on the parameters of the dip.

Imagine a V-shaped dip and a ball small enough that the valley of the "V" doesn't matter.  The angle of the dip to the table top is theta, and horizontal length of the "V" equals the length of the table top.

Let V0=the initial velocity of the ball before the dip, y=the vertical distance below the table top at any time t, g=acceleration of gravity, and V= the speed of the ball at any time t.  Assume that the ball NOT traveling the dip reaches the other end at time t1

For the ball on the table top (no dip), the distance to the other side of the table is V0 * t1.

For the ball traversing the dip, the speed at any position y, below the table top is V=sqrt(v0^2 + 2*g*y). At the bottom of the dip then, the speed is V = sqrt (v0^2 + 2*g*h).  Since this speed change is due to gravity alone, and gravity is a constant acceleration, the AVERAGE speed of the ball during the time it is on the down side of the dip is  V0 + .5*sqrt(v0^2 + 2*g*h).  this is also the average speed for the up side of the dip., and therefore the average speed over the entire dip.  The distance traveled in time t1, at this average speed is [v0 + 0.5*sqrt( v0^2 + 2*g*h)] * t1, and the HORIZONTAL distance is this value  times cos(theta) or distance across table = [V0 +0.5* sqrt (v0^2 + 2*g*h)] * t1 * cos(theta). If this value is great than V0 * t1, then the ball on the dip will beat the other ball.  Or, rearranging, if  {1 + 0.5*sqrt(1+ 2gh/V0^2)}*cos(theta) > 1, then the ball in the dip will beat the other ball.  If one inspects the equation a bit, you can see that there are combinations of h, V0 and theta that will cause the expression on the left side to be less than, equal to or greater than 1.  QED

Edited on November 13, 2009, 8:04 pm

Posted by Kenny M on 2009-11-13 20:01:02