Question

A rod of mass m and length L is kept in a frictionless horizontal floor . A particle of same mass m moving perpendicular to the rod with a speed v strikes at the end of the rod. If coefficient of elasticity for the collision is 1 then calculate angular velocity acquired by the rod. ​Who will able to solve this question I shall mark his/her answer a brainliest answer.​

Answer 1

Answer:

Explanation:

Net torque on the system COM of rod is zero

∴ Applying conservation of angular momentum about COM of rod, we get mv0(L2)=Iω

or mv0L2=ML212ω

or mv0=MLω6 ...(ii)

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

∴12mv20=12Mv2+12Iω2

or mv20=Mv2+ML212ω2 ..(iii)

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

mM=14

v=mv0M and ω=6mv0ML

(b). point P will be rest if xω=v

or x=vω=mv0/M6mv0/ML or x=L/6

∴AP=L2+L6 or AP=23L)

(c). After time t=πL3v0

Angle rotated by rod θ=ωt=6mv0ML.πL3v0

=2π(mM)=2π(14)

∴θ=π2

Therefore, situation is as shown in figure

∴resultant velocity of point P will be

|VP|=2–√v=2–√(mM)v0

=2–√4v0=v022–√

or |VP|=v022–√