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 t

Question

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

TestHub · Accepted Answer

Applying conservation of linear momentum,

$m v = m v' + m V$

$\Rightarrow v = v' + V$ ........(i)

Applying conservation of angular momentum about the point of collision

$0 = (\frac{m l^{2}}{12}) ω - m V (\frac{l}{4})$

$\Rightarrow l ω = 3 V$ ..........(ii)

for a perfectly elastic collision $e = 1$

$v - 0 = V + \frac{ω l}{4} - v'$ ..........(iii)

$v = V + \frac{3 V}{4} - (v - V)$

$2 v = \frac{11 V}{4}$

$\Rightarrow V = \frac{8 v}{11}$

$ω = \frac{24 v}{11 l}$

Angle rotated by the rod $θ = 6 π$

$t = \frac{θ}{ω} = \frac{6 π}{ω}$

$t = 6 π \times \frac{11 l}{24 v} = \frac{11 l}{4 v}$

Hence, distance travelled by the centre of the rod is

$S = V t$

$S = \frac{8 v}{11} (\frac{11 l}{4 v}) = 2 π l$

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