IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

This problem has been solved a lot of times here on this forum. Anyway, let me do it again:

Now, consider the instant just after the ball collides with the wall.

Using conservation of momentum, mv + mv' = 0, or v' = -v, which implies the velocity of the ball immediately after collision is v'.

Conserving angular momentum about the point of contact:

-I

= I

'

Now, MI of the sphere about the bottom point = 2MR²/5 + MR²

[I_cmw - mvr] = -I_cmw' - mv'r [using v = rw]

=3/7mvr = 7/5mv'r

which gives v' = 3v/7.

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