Q. A uniform rod AB of length `L` and mass `M` is lying on a smooth table.A small particle of mass `m` strikes the rod with a velocity `v_0` at point C at a distance `x` from the centre O. The particle comes to rest after collision. The value of `x` so that point A of the rod remains stationary just after collision is:-
The total linear momentum is the same before and after collision regardless the type of collision.
If you hit the rod in the middle, it will begin sliding only (figure 1 below). The total linear momentum is the same before and after collision regardless the type of collision.
`v = m/M*v_0`
If you hit the rod in one end, it will begin only to rotate about its center. We assume the collision is elastic. The angular momentum is the same before and after collision.
where `I =(ML^2)/12` and `omega=(v')/(L/2) =(2v')/L`
`mv_0*L/2 =(ML^2)/12 *(2v')/L`
`v' =3 (m/M)*v_0 = 3v`
Thus for A to be stationary mass `m` need to hit at a distance from the center O
`x =(1/3)*(L/2) =L/6`
The correct answer is B) `L/6`
More correctly would be to say that `Delta(v) =v -(-3v) =4v` so that the point x would lie at at distance from center O of
`x = (1/4)*(L/2) =L/8`
but this option is not on the list on answer choices.