An uniform rod of mass `m=30 kg` and length `l=0.80 m` is free to rotate about a horizontal axis `O` passing through its centre. A particle `P` of mass `M=11.2 kg` falls vertically through a height `h=36/245 m` and collides elastically with the rod at a distance `l/4` from `O`. at the instant of collision the rod was stationary and was at angle `alpha=37^(@)` with horizontal as shown in figure

Velocity of particle `P` after collision `(g=10ms^(-2))`

Options

A. `7/9 ms^(-1)`
B. `7 ms^(-1)`
C. `9/7 ms^(-1)`
D. `1 ms^(-1)`

Answer

C

Explanation

Velcity component of particle, normal to rod(just after collision) is
`upsilon_(n)=(J-Mupsilon_(0)cosalpha)/(M)`
Since, the collision is elastic, therefore, there is normal loss of kinetic energy during collision. Hence kinetic energy of system of rod and particle just after collision=kinetic energy of particle just before collision.
`:. 1/2 I omega^(2)+1/2M(upsilon_(t)^(2)+upsilon_(n)^(2))=1/2mupsilon_(0)^(2)`
`J=24 Ns`

`:.` From equation `(1) omega=3 rad//sec`tangential component of velocity of particle `upsilon_(t)=upsilon_(0)sin alpha=sqrt(2gh) sin 37^(@)=36/35 ms^(-1)`