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1)what is the velocity of the ball before the collision?

2)how much energy is lost during the collision.

what i did was

I=(M/3)L^2

u=omega*L

the momentum is conserved during the collision so

P1=P2+L2

mv=mu+(M/3)L^2(omega)

mv=mu+(M/3)Lu

v=u[1+(M/3m)L]

now using conservation of energy after th collision until the max height, saying that my gravitational potential energy is equal to 0 at the top of the pole, therefore my height, h, is negative from the top of the pole and is -L*cos(alpha) for the ball and -(L/2)*cos(alpha) for the pole[measured from the centre of mass, at the centre of the pole]

0.5mu^2+I(omega)^2-mgL-0.5MgL=-mgLcos(alpha)-Mg(L/2)cos(alpha)

0.5mu^2+(M/6)u^2=mgL+0.5MgL-mgLcos(alpha)-Mg(L/2)cos(alpha)

u^2[(m/2)+(M/6)=gL[m+(M/2)](1-cos(alpha))

u^2=2gl(m+(M/2))(1-cos(alpha))(3/(3m+M))

u^2=3gL[(2m+M)/(3m+M)](1-cos(alpha))

v^2=u^2*[1+(M/3m)L]^2

v^2=[3gL[(2m+M)/(3m+M)](1-cos(alpha))]*[1+(M/3m)L]^2

does this seem correct?? the answer in my textbook is

v^2=2gL(1-cos(alpha))(1+(M/2m))(1+(I/mL^2))

cant see where ive gone wrong and dont want to continue to the next part till i know this is right. please help