A uniform rod of length `L` and mass `M` is lying on a frictionless horizontal plane and is pivoted at one of its ends as shown in Fig. There is no friction at the pivot. An inelastic ball of mass `m` is fixed with the rod at a distance `L//3` from `O`. A horizontal impulse `J` is given to the rod at a distance `2L//3` from `O` in a direction perpendicular to the rod. Assume that the ball remains in contact with the rod after the collision and impulse `J` acts for a small time interval `/_\t`. Now answer the following questions:

Find the magnitude of the impulse applied on the pivot during the time interval `/_\t`

Options

A. `(mJ)/((m+3M))`
B. `(mJ)/((3m+M))`
C. `(MJ)/((m+3M))`
D. `(MJ)/((3m+M))`

Answer

A

Explanation

Let the system starts with agular velocity `omega`. Angular velocity if the ball will also be `omega` as it remais struck to the rod.

Velocity of the ball `v=omegaL/3`
For the rod angular impulse= change in angular momentum:
`J(2L)/3-J_(b) L/3=(ML^(2))/3omega`.....i
For the ball, impulse `=` change in linear momentum
`J_(b)=mv=momega(L/3)`......ii
From eqn i and ii `omega=(6J)/((m+3M)L)`
and `J_(b)=(2mJ)/((m+3M))`

Let impulse on the pivot be `J_(1)`
For the rod and ball system total impulse `=` change in linear
mometum `J+J=Mv_(c)+mv`
Solve to get `J_(1)=(mJ)/(m+3M)`