Question

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.

Answer 1

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
=> v₂ = (m/M)u₁ ---(1)
Now by we consider the angular momentum imparted by the particle to the rod.
mu₁(L/2) = Iω
For rod about its CM, I = ML₂/12
mu₁(L/2) = ωML₂/12
=> ω = 6mu1/ML ---2.
Now we know
ω = θ/t
θ = π/2
=> t = θ/ω = (π/2)/(6mu₁/ML)
=> t = πML/(12mu₁)
Linear distance moved by the CM of the rod will be
s = v₂t
By equation 1.
s = [πML/(12mu₁)]× (m/M)u₁
= πL/12
Initial KE of the particle is ½m(u₁)²
KE of the CM of the rod = ½M(v₁)² = ½M{(m/M)u₁}² [by equation 1.]
= ½(u₁)²m₂/M
KE_rot of the rod = ½Iω²
KE_rot = ½ (ML₂/12)(6mu₁/ML)²
= (3/2) m²(u₁)²/M
To be elastic collision KE must be conserved
½m(u₁)² = ½(u₁)²m^₂/M + (3/2) m²(u₁)²/M
=> M = 4m

Hence Proved.