Scope of coefficient of restitution

There are several several instances when we can use the fact that $e=1$ . to simplify collision problems which saves us from having to apply conservation of energy.

The result $e=1$ . is fairly obvious for head-on collisions, and for collisions of extended objects involving pure translation and for those involving pure rotation.

But my doubt is can I in general use the result for any collision among extended bodies too which might involve a combination of both rotation and translation ??

Can I say that if an object elastically collides with another,,, then the velocities of the points of contacts of both objects satisfy the relation $e=1$ . ??

example in the image,, if the collision between ball and rod is elastic,, can i say that the point of collision of the rod after collision necessarily moves at a speed v such that,, if the balls initial and final velocities are $u$ . and $f$ . respectively ,

then ,

$Tell$ $me$ whether I use :

1) $v-f=u$ . ??

2) $e=1$ ??

3) Do I have to use the usual way of energy conservation and momentum conservation and angular momentum conservation independently ??

Please any one who has the knowledge I seek help me ,, if possible also provide a qualitative or quantitative explanation

$Be$ $Careful$ . :

e is the coefficient of restitution, not to be confused with $2.7108$ .

$NOTE$ . :

I know it is applicable for pure rotating and pure translating objects whether extended or not because i can prove it)

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Comments

My question Challenges in mechanics by Ronak Agarwal (Part3) is based on this concept only.

Ronak Agarwal - 6 years, 8 months ago

gotta try it then,, i saw it before but it seemed to lengthy,, either way can you prove that e=1 for all sorts of elastic collisions? or is it just a general result you know and apply ? :)

Mvs Saketh - 6 years, 8 months ago

Yes it is a general result that $e=1 \Rightarrow$ Energy is conserved.

@Ronak Agarwal – got it,, and yes i got the correct solution to ur cricket answer,,, but why i cant i see the add answer button anymore,, either way applying conservation of angular momentum gives the same equation structure as my previous admittedly wrong manner,, what we need is that after the collision the velocity of ball should be directed towards a wicket,,, but the direction of angular momentum cannot be changed and hence the angle between the initial angular momenta by virtue of rotation and translation should be such that they match with the angle of line joining the extreme wicket and the line representing perpendicular distance from the wickets,, correct?

@Mvs Saketh – Now you are exactly right.

Yes you can in general use the result e=1 for any kind of collisions.

Why do you put two commas ,, where ever you have used them ?

Arya Samanta - 6 years, 8 months ago

I had edited Them For $Saketh$ .

Deepanshu Gupta - 6 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$