A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance travelled by the centre of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic.
Solution
Let initial velocity of the particle = u1
Let final velocity of the particle = v1
Let time taken to move `π/2` angle = t
Given v1 = 0
Let initial velocity of CM of the rod = u2
Given u2 = 0
Let final velocity of the CM of rod = v2
By conservation of linear momentum
mu1 + Mu2 = mv1 + Mv2
`=> v_2 = (m/M)u_1 ...........(1)`
Now by we consider the angular momentum imparted by the particle to the rod.
`m u_1(L/2) = Iω`
For rod about its CM, `I = (ML_2)/12`
mu1(L/2) = ωML2/12
`=> ω = (6m u_1)/(ML) .........(2)`
Now we know
ω = θ/t
θ = π/2
`=> t = θ/ω = (π/2)/((6m u_1)/(ML))`
`=> t = (πML)/(12m u_1)`
Linear distance moved by the CM of the rod will be
s = v2t
By equation 1.
`s = [(πML)/(12m u_1)]× (m/M)u_1`
`= (πL)/12`
Initial KE of the particle is ½m(u1)2
`"KE of the CM of the rod "= 1/2M(v_1)^2 = 1/2M{(m/M)u_1}^2 ......["by equation 1"]`
`= 1/2(u_1)^2m_2/M`
`"KE"_"rot" " of the rod "= 1/2Iω^2`
`"KE"_"rot" = 1/2 ((ML_2)/12)((6m u_1)/(ML))^2`
`= (3/2) m^2(u_1)^2/M`
To be elastic collision KE must be conserved
`1/2m(u_1)^2 = 1/2(u_1)^2m^2/M + (3/2) m^2(u_1)^2/M`
`=> M = 4m`
Hence Proved.
