So if you know this value you can determine the height to which the object will return back up to, and the speed at which it will leave the surface at.
the FIBA claim the mass of a basketball must between 567 grams (g) and 650 g. They also pass as a more stringent requirement "it must bounce at least 1300 mm when dropped from a height of 1800 mm on a hard surface with a mass of more than 1 ton"
express the coefficient of restitution as a value by substituting v in terms of s,
What is very important to realise is that if a strike an obstacle lying across its path it rebounds perpendicularly, there will be NO change in the parallel component of it velocity
i1 = i2, but j1 > -j2
so take note of the component of velocity that will change, do the calculations on this change, but LEAVE the other component ALONE!
I cannot emphasise this enough that only one component of the velocity will change, thats the one that is perpendicular to the surface. Think about it! A snooker ball rolls along a table, when it gets to the cushion it rebounds back, with no new velocity in the up or down directions. It only affects the i direction as the barrier lies in the j component.
Now lets go back to the book and work through example 7.7 on page 173
Then go and answer questions 4 & 5 on page 177
Remember that Tan = sin/cos
the sin is the j component of the velocity it is the one that will change so it is clear to see that e = -VsinB / UsinA
as the i direction wont change then the Tan B / Tan A will be ... this is yours to prove.
Enjoy it ...
In keeping with the ideas that we came up with in the previous section, we need to make the barrier the obstacle the focus when we orientate our axis. If we do not have the barrier on the i or the j plane, well then we make our calculations twice as difficult because now the rebound is not solely affecting the i component of velocity or the j component of velocity, but BOTH .... not good ...
When we have a object that strike a surface that is at an angle to our usual i & j axis it is more convienient to change the axis than to work out the traditional i & j values. In this way we can use what we took from the last page that the only change in speed occurs in the component at right angles to the surface.
So looking at Worked Example of 7.8, and 7.9 we will see that we need to alter our view and the problem becomes simple.
If we recalibrate the axis we also must recalculate the velocity in terms of i & j, then only one of these components will be affected by the e
Now go and do Qs 7,8,9 & 10
note that the sum to infinity for where -1< r < 1 is
sun=m to n objects is given by