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 = u_{1}

Let final velocity of the particle = v_{1}

Let time taken to move `π/2` angle = t

Given v_{1}_{ }= 0

Let initial velocity of CM of the rod = u_{2}

Given u_{2}_{ }= 0

Let final velocity of the CM of rod = v_{2}

By conservation of linear momentum

mu_{1} + Mu_{2} = mv_{1} + Mv_{2}

`=> 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`

mu_{1}(L/2) = ωML_{2}/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 = v_{2}t

By equation 1.

`s = [(πML)/(12m u_1)]× (m/M)u_1`

`= (πL)/12`

Initial KE of the particle is ½m(u_{1})^{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.