A uniform thin rod AB of mass M = 0.6 kg and length l = 60 cm stands at the edge of a frictionless table as shown in figure. A particle of mass m = 0.3 kg flying horizontally with velocity v0 = 24 m/s strikes the rod at point P at a height h = 45 cm from the base and sticks to it. The rod is immediately driven off the table. If the time after collision when the rod becomes horizontal for the first time is found to be =π∗∗sec, find the value of '**'(g = 10 m/s2).

The position of center the combined system with B as origin.y=0.6×30+0.3×450.6+0.3=35cm above the table If we take "rod + particle" as system we have no external force acting on the system in horizontal direction hence we can conserve the linear momentum of the system in horizontal direction just before and just after collision.Let vCM be the linear velocity of CM and ω is the angular velocity about an axis passing through CM immediately after collision. From conservation of linear momentum, we getmv0=(M+m)vCM or vCM=mv0M+m=(0.3)(24)0.6+0.3=8m/sThe system has no external impulse acting on the system. Hence we can conserve angular momentum of the system about any point. Let us conserve angular momentum about C.M of the system.Angular momentum of the system just before collision. L→initial =mv0(h−y)(−k^)=0.3×24×(0.35−0.30)(−k^)Just after collision L→initial =Icmω→Icm=Icmrod+Icmparticle =Ml212+My−l22+m(h−y)2⇒Icm=(0.6)(0.6)212+0.6(0.35−0.30)2+0.3(0.45−0.35)2=0.0225kgm2Hence L→final=0.0225ω(−k^)As L→initial =L→final ⇒0.75(−k^)=0.0225ω(−k^)Or ω=0.720.225=32rad/sInitially the rod was in vertical orientation then it become horizontal. It means angle rotated by the rod is π2.Hence time taken in this process, t=π/2ω=π2×32=π64sec