RBD

Ans not known

applying linear and angular impulse momentum theorem. linear : mVncm  (mVcm) =I Vncm= new velocity of COM sign convention: up=+ down= hence: mLw2/2 + mLw1/2 =I angular impulse theorem about the first end of rod: mVncmL/2 + mL^2w2/12 
(mL^2w1/3)=IL hence: mL^2w2/4 + mL^2w2/12 + mL^2w1/3=LI I= mLw2/6 + mLw1/3 equating with previous equation: mLw2/2 + mLw1/2 = mLw2/6 + mLw1/3 hence mLw1/6 = mLw2/3 hence : w2=w1/2