A vertically oriented uniform rod of mass M and length l can rotate about its upper end. A horizontally flying bullet of mass m strikes the lower end of the rod and gets stuck in it, as a result, the rod swings through an angle `alpha`. Assuming that `m lt lt M`, find:
(a) the velocity of the flying bullet,
(b) the momentum increment in the system "bullet-rod" during the impact, what causes the change of that momentum,
(c) at what distance x from the upper end of the rod the bullet must strike for the momentum of the system "bullet-rod" to remain constant during the impact.

Explanation

(a) About the axis of rotation of the rod, the angular momentum of the system is conserved.
Thus if the velocity of the flying bullet is `v`.
`mvl=(ml^2+(Ml^2)/(3))omega`
`omega=(mv)/((m+M/3)l)~~(3mv)/(Ml)` as `m lt lt M` (1)
Now from the conservation of mechanical energy of the system (rod with bullet) in the uniform field of gravity)
`1/2(ml^2+(Ml^2)/(3))omega^2=(M+m)gl/2(1-cosalpha)` (2)
[because C.M. of rod raises by the height `l/2(1-cosalpha)`]
Solving (1) and (2), we get
`v=(M/m)sqrt(2/3gl)si nalpha/2` and `omega=sqrt((6g)/(l))si nalpha/2`
(b) Sought `Deltap=[m(omegal)+M(omegal/2)]-mv`
where `omegal` is the velocity of the bullet and `omegal/2` equals the velocity of C.M. of the rod after the impact. Putting the value of v and `omega` we get
`Deltap~~1/2mv=Msqrt((gl)/(6))si nalpha/2`
This is caused by the reaction at the hinge on the upper end.
(c) Let the rod starts swinging with angular velocity `omega^'`, in this case. Then, like part (a)
`mvx=((Ml^2)/(3)+mx^2)omega^'` or `omega^'~~(3mvx)/(Ml^2)`
Final momentum is
`P_f=mxomega^'+underset(0)overset(l)intyomega^'M/ldy~~M/2omega^'l~~3/2mvx/l`
So, `Deltap=p_f-p_i~~mv((3x)/(2l)-1)`
This vanishes for `x~~2/3l`