A rod AB of mass M and length L is lying on a
smooth horizontal surface. A particle of mass m
strikes to end A elastically with speed v in a
direction perpendicular to AB. After collision if
particle comes to rest, then the distance of point
from B which is in rest just after the collision is
Asked by dipanshusingla029
25th May 2018,
8:21 PM
Answered by Expert
Answer:
As shown in figure if mass m hits the rod at centre, there is only linear motion.
let m and v are mass are mass and speed of particle that is hitting the rod. Let M be the mass of rod and V be the speed after collision.
By conseravtion of momentum , we have m×v = M×V, hence V = (m/M)×v ..................(1)
If the mass hit the rod at end A, the rod will rotate about its center.
Angulr momentum of the rod is given by, 
where I is moment of inertia and V' be the linear speed at the end of rod. Solving for V' using 2 we get 
In the rotational motion, centre of the rod is in rest position. Hence the point which is L/2 from the end B is at rest after collision.
Answered by Expert
18th June 2018,
12:45 PM
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