# Final speed of ball : collision with vertical plank at top

## Homework Statement ## The Attempt at a Solution

I have solved the first part already.
For the 2nd part, I am writing the relevant eqns.

Can I take motion of the plank as a pure rotation about the pivot ?
Quantities corresponding to pivot is subscripted by p.
Conservation of momentum gives,
## mv_0 =mv_f + MV_f ## (1)
Conservation of Angular momentum about the pivot gives,
##2mv_0 l = 2mv_f l+ [ I_p \omega_p = I_{cen} \omega_{cen} + MV_f R] ~~~~~~(2)
\\ \omega_p = \omega_{cen} = V_f/l ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)
\\ mv_0 = mv_f + \frac { 2MV_f } 3 ~~~~~~~~~~~~~~~~~~~~(4) ## which contradicts eqn. (1)

Conservation of kinetic energy,
## \frac 1 2 m v_0^2 = \frac 1 2 m v_f^2 + [\frac 1 2 I_{p} ω^2 = \frac 1 2 I_{cen} ω^2 + \frac 1 2 M V_f^2] ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(5)##

Are these equations correct?

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I think eq (1) is not correct as the pivot may exist force in horizontal direction,too.

haruspex
Homework Helper
Gold Member
I think eq (1) is not correct as the pivot may exist force in horizontal direction,too.
Quite so. Likewise, to be able to ignore this for conservation of angular momentum you must use the pivot as the axis.

(You mean "exert".)

• Pushoam
Likewise, to be able to ignore this for conservation of angular momentum you must use the pivot as the axis.
Thanks for giving a reason for this "must". I was looking for it.This adds to my intelligence of selecting proper axis of rotation for a given problem.

(You mean "exert".)
Yes.