Question

Question No 2: A rod of length L and Mass M lies on a frictionless horizontal table. It is free to move in any way on the table. A small body of mass m moving with a velocity u collides in elastically with the rod.
(a) Find the velocity of the Centre of mass of the rod.
(b) Find the angular velocity of the rod about Centre of mass.

Answer 1

Answer:

). Suppose velocity of COM of the rod just after collision is v and angular velocity about is `omega` is `omega`. Applying following three laws

(1). External force on the system (rod `+` mass) in horizontal plane along x-axis is zero

`therefore` applying conservation of linear momentum in x-direction

`mv_(0)=Mv` ..(i)

(2). Net torque on the system COM of rod is zero

`therefore` Applying conservation of angular momentum about COM of rod, we get `mv_(0)((L)/(2))=Iomega`

or `mv_(0)(L)/(2)=(ML^(2))/(12)omega`

or `mv_(0)=(MLomega)/(6)` ...(ii)

(3) since, the collision is elastic, kinetic energy is also conserved

`therefore(1)/(2)mv_(0)^(2)=(1)/(2)Mv^(2)+(1)/(2)Iomega^(2)`

or `mv_(0)^(2)=Mv^(2)+(ML^(2))/(12)omega^(2)` ..(iii)

From Eqs (i), (ii) and (iii) we get the following result

`(m)/(M)=(1)/(4)`

`v=(mv_(0))/(M)` and `omega=(6mv_(0))/(ML)`

(b). point P will be rest if `xomega=v`

or `x=(v)/(omega)=(mv_(0)//M)/(6mv_(0)//ML)` or `x=L//6`

`thereforeAP=(L)/(2)+(L)/(6)` or `AP=(2)/(3)L)`

(c). After time `t=(piL)/(3v_(0))`

Angle rotated by rod `theta=omegat=(6mv_(0))/(ML).(piL)/(3v_(0))`

`=2pi((m)/(M))=2pi((1)/(4))`

`thereforetheta=(pi)/(2)`

Therefore, situation is as shown in figure

`therefore `resultant velocity of point `P` will be

`|V_(P)|=sqrt(2)v=sqrt(2)((m)/(M))v_(0)`

`=(sqrt(2))/(4)v_(0)=(v_(0))/(2sqrt(2))`

or `|V_(P)|=(v_(0))/(2sqrt(2))`

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Explanation: