A uniform smooth rod (mass m and length l ) placed on a smooth horizontal floor is hit by a particle (mass m ) moving on the floor, at a distance l 4 from one end elastically ( e = 1 ). The distance travelled by the centre of the rod after the collision, when it has completed three revolutions, will be

Step-by-step solution

Applying conservation of linear momentum, m v = m v ′ + m V ⇒ v = v ′ + V ........(i) Applying conservation of angular momentum about the point of collision 0 = m l 2 12 ω - m V l 4 ⇒ l ω = 3 V ..........(ii) for a perfectly elastic collision e = 1 v - 0 = V + ω l 4 - v ′ ..........(iii) v = V + 3 V 4 - v - V 2 v = 11 V 4 ⇒ V = 8 v 11 ω = 24 v 11 l Angle rotated by the rod θ = 6 π t = θ ω = 6 π ω t = 6 π × 11 l 24 v = 11