Controlling a Rolling Ball On a Tilting Plane - STEM2

where s 0 is the initial position. The change in kinetic energy is
m( v2
2 − v2 0
2 )=f(s − s 0)=ma(s − s 0 ),
where f is a constant force of acceleration. Dividing by m/2 weget
v 2 − v 2 0 =2a(s − s 0).
The velocity increases linearly with the time, so the average velocity multiplied
by the time is the distance traveled
The four equations
v + v 0
2
t = v 0 + at + v 0
t
2
=(v 0 + at
2 )t
= v 0 t + at2
2
=(s − s 0 ).
v = v 0 + at,
s = s 0 + v 0 t + a t2 2 ,
v 2 − v 2 0 =2a(s − s 0),
and
v + v 0
t =(s − s 0 ),
2
usually suffice to solve problems of this type.
Example 1. Suppose a ball is rolled off of the edge of a table top. Suppose
the table is at height height h and the ball lands on the floor at a distance
s from a point on the floor below the table edge. What was the horizontal
velocity of the ball as it left the table edge
Solution The ball is accelerated by the acceleration of gravity g downward.
The initial vertical velocity is zero, and the initial vertical position is zero.
So using the second formula
h = g t2 2 ,
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