Two uniform thin rods A and B of length 0.6 m each and of masses 0.01 kg and 0.02kg respectively are rigidly joined end to end. The combination is pivoted at the lighter end, P as shown in fig. Such that it can freely rotate about point P in a vertical plane. A small object of mass 0.05kg, moving horizontally, hits the lower end of the combination and sticks to it what should be the velocity of the object so that the system could just be reised to the horizontal position.

Explanation

System is free to rotate but not free to translete. During collision net torque on the system (rod `A+rodB+mass` m ) about point P is zero
therefore, angular momentum of system before collision
`=` angular momentum of system just after collisio (about P)
Let `omega` be the angular velocity of system just after collision then
`L_(i)=L_(f)impliesmv(2l)=Iomega` ..(i)
`=m(2l)^(2)+m_(A)(l^(2)//3)+m_(B)[(l^(2))/(12)+((l)/(2)+l)^(2)]`
given `l=0.6m,m=0.05kg,m_(A)=0.01kg` and `m_(B)=0.02kg`
Substituting the values we get
`I=0.09kg-m^(2)`
therefore, from Eq. (i)
`omega=(2mvl)/(I)=((2)(0.05)(v)(0.6))/(0.09)`
`omega=0.67v` (ii)
Now after collision, mechanical energy will be conserved.
Therefore, decrease in rotational `KE=` increase in gravitational PE
or `(1)/(2)Iomega^(2)=mg(2l)+m_(A)g((l)/(2))+m_(B)g(l+(l)/(2))`
`omega^(2)=(gl(4m+m_(A)+3m_(B)))/(I)`
`=((9.8)(0.6)(4xx0.05+0.01+3xx0.02))/(0.09)`
`=17.64(rad//s)^(2)`
`thereforeomega=4.2rad//s` ..(iii)
Equating Eqs. (ii) and (iii) we get
`v=(4.2)/(0.67)m//s` or `v=6.3m//s`