A bullet of mass m is fired with a horizontal velocity and at a height h = 1/2 R

Question

A bullet of mass m is fired with a horizontal velocity and at a height h = 1/2 R in to a wooden disk of much larger mass M and radius R. The disc rests on a horizontal plane and the coefficient of friction between disc and plane is finite bullet gets embedded in to the disc Linear velocity of the disc after the motion has become uniform is mv_o/3m If h = 3/2 R then disc will roll without sliding immediately after impact Angular velocity of the disc just after impact is mv_o/mR Angular velocity of the disc just after impact is 3mv_o/2MR

Explanation / Answer

conserve linear momentum about disc bottom most point

mv0 = (m+M)v

v = mv0/(m+M)

conserving angular momentum we get = mv0R/2 = (MR2/2 +mR2)w

since disc is large we can neglect mass of bullet =  mv0R/2 = (MR2/2 )w => w = mv0/MR

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