# A ball on a string

Show me how you get that 5/7 and 2/7 sharing

My notes: https://dl.dropboxusercontent.com/u/34732003/IMG_20150212_180625.jpg [Broken] (dropbox, because its too large)
Where did i go wrong?

Last edited by a moderator:
Any solid homogenous sphere that rolls without slipping will have the LInear KE and rotating KE in that ratio regardless of speed.

Take the problem from the start (work backwards)
I get the velocity of ball 2 at 120 ° (slack point) to be 2.236 m/s, giving a KE of 25 Joules
Its PE gained at that point = 10 * 10 * 1.5 = 150 Joules
So its KE at collision point must have been: 25 + 150 = 175 Joules
(velocity = sqrt ( 175 / ( ½ * m ) ) = 5.9161 m/s)
This is also the linear KE transferred from ball 1 at the collision
(equal mass, elastic collision)
This represents 5/7 of the total KE of ball 1
( 2/7 is tied up in rotational KE )
So ball 2 total KE ( linear + rotational ) at impact = 175 * ( 7 / 5 ) = 245 Joules
So if m*g*h = 245, then h = 2.45 metres

PS: Signing out until: 12.00 noon GMT Saturday

Right, got it.
https://dl.dropboxusercontent.com/u/34732003/IMG_20150213_220656.jpg [Broken]. Happened to do another exercise today that helped me a bit with this one.

Took a look at my earlier notes, damn i must've been tired. I think i need a break :F

Last edited by a moderator:
Glad to help, good for me also.
Dean