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 figure. 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 ΔtNow answer the following questions:  Find the resulting instantaneous angular velocity of the rod after the impulse.Find the impulse acted on the ball during the time interval Δt.Find the magnitude of the impulse applied by the pivot during the time interval Δt.

Let the system starts with angular velocity ω. Angular velocity of the ball will also be ω as it remains struck to the rod. Velocity of the ball v=ωL3For the rod, angular impulse = change in angular momentum:J2L3−JbL3=ML23ω            …(i)For the ball, impulse = change in linear momentumJb=mv=mωL3               …(ii)From Eqs. (i) and (ii), ω=6J(m+3M)Land Jb=2mJ(m+3M)Let impulse on the pivot be J1.For the rod and ball system, total impulse = change in linear momentum : J+J1=Mvc+mvSolve to get : J1=mJm+3MLet the system starts with angular velocity ω. Angular velocity of the ball will also be ω as it remains struck to the rod. Velocity of the ball v=ωL3For the rod, angular impulse = change in angular momentum:J2L3−JbL3=ML23ω            …(i)For the ball, impulse = change in linear momentumJb=mv=mωL3               …(ii)From Eqs. (i) and (ii), ω=6J(m+3M)Land Jb=2mJ(m+3M)Let impulse on the pivot be J1.For the rod and ball system, total impulse = change in linear momentum : J+J1=Mvc+mvSolve to get : J1=mJm+3MLet the system starts with angular velocity ω. Angular velocity of the ball will also be ω as it remains struck to the rod. Velocity of the ball v=ωL3For the rod, angular impulse = change in angular momentum:J2L3−JbL3=ML23ω            …(i)For the ball, impulse = change in linear momentumJb=mv=mωL3               …(ii)From Eqs. (i) and (ii), ω=6J(m+3M)Land Jb=2mJ(m+3M)Let impulse on the pivot be J1.For the rod and ball system, total impulse = change in linear momentum : J+J1=Mvc+mvSolve to get : J1=mJm+3M