A particle, moving horizontally, collides perpendicularly at one end of a rod having equal mass and placed on a smooth horizontal surface.

Options

A. Particle comes to rest if collision is perfectly elastic and centre of m rod starts to move with the same velocity.
B. Particle continues to move along the same direction, whatever is the value of `e`.
C. Particle may get rebound back.
D. Velocity of mid-point of the rod will be less than `v//2` if the particle gets stuck.

Answer

B, D

Explanation

Whenever two paticles having equal mass collide head on elastically, their velocities get interchanged. Therefore, if the particle collides at midpoint of the rod then velocities would get interchanged. In that case option a would be correct. But now the particle strikes at an end of the rod, hence particle head on collision does not take place. Therefore, option a is incorrect.
If the particle gets rebound back to its original path, then its final mometum will become negative. Since mass of the particle and rod is equal, therefore law of conservation of momenum will be satisfied only when velocity of centre of mass of the rod is greater than orginal velocity of te particle. Hence, kinetic energy of the system (just after collision) will become greater than that (just before the collision), which is impossible. Hence, the particle cannot rebound back or it will colntinue to move along the same direction. Hence, option b is correct but option c is incorrect. if the paticle gets stick to the rod, centre of mass o the system will lie at a distance `l//4` from theend at which the particle sticks. According to the law of conservation of momentum cenre of mass will start ot move wit velocity `v//2`. But the particle will simultaneously start to rotate about tecentre of mass m an anticlockwise direction. Angular velocity `omega` of that rotational motion can be calculated by applying law of mass `m` an anticlockwise direction. Angular velocity `omega` of that rotational motion can be calculated by applying law of conservation of angular momentum. Hence the resultant velocity of the midpoint of the rod will be equal to `[(v_(0))//2)-(lomega//4)]` which is obviously less than `v_(0)//2`. Therefore option d correct.