A uniform rod of mass m and length l has been placed on a smooth table A particle of mass m, travelling perpendicular to the rod, hits it at a distance x=ℓ6 from the centre C of the rod. Collision is elastic Find the angular velocity of the rod and after the collision

Let V1,V2 and be the velocity of particle, velocity of the centre of the rod and angular speed of the rod immediately after collision.In the diagrams the first, second and the third one represent the situations just before collision, just after collision and just after second collision. Momentum conservationmu=mV1+mV2⇒u=V1+V2…………… (i) Apply conservation of angular momentum about a fixed point is space that coincides with the initial position of the centre of the rodmiux¨≡mV1x¨+mℓ212ω[Note: Angular momentum of the rod about said point is ICMω→+R→×MV→2 where the second term is zero in present context ] uℓ6=V1ℓ6+ℓ2ω12 ⇒u=V1+ωℓ26 ....(ii)Collision is elastic hence we can use conservation of kinetic energy or the definition of coefficient of restitution (e). Let’s use the late Relative velocity of point P and the particle after collision Relative velocity of point P and the particle before collision =−1V2+ωx−V1o−u=−1 ⇒V2−V1=u−ωℓ6 .....(iii)Solving (i),(ii) and ((iii)V1=V2=u2 and ω=6uℓ

A uniform rod of mass m and length l has been placed on a smooth table A particle of mass m, travelling perpendicular to the rod, hits it at a distance x=ℓ6 from the centre C of the rod. Collision is elastic Find the angular velocity of the rod and after the collision

Let V1,V2 and be the velocity of particle, velocity of the centre of the rod and angular speed of the rod immediately after collision.In the diagrams the first, second and the third one represent the situations just before collision, just after collision and just after second collision. Momentum conservationmu=mV1+mV2⇒u=V1+V2…………… (i) Apply conservation of angular momentum about a fixed point is space that coincides with the initial position of the centre of the rodmiux¨≡mV1x¨+mℓ212ω[Note: Angular momentum of the rod about said point is ICMω→+R→×MV→2 where the second term is zero in present context ] uℓ6=V1ℓ6+ℓ2ω12 ⇒u=V1+ωℓ26 ....(ii)Collision is elastic hence we can use conservation of kinetic energy or the definition of coefficient of restitution (e). Let’s use the late Relative velocity of point P and the particle after collision Relative velocity of point P and the particle before collision =−1V2+ωx−V1o−u=−1 ⇒V2−V1=u−ωℓ6 .....(iii)Solving (i),(ii) and ((iii)V1=V2=u2 and ω=6uℓ

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A uniform rod of mass m and length $l$ has been placed on a smooth table A particle of mass m, travelling perpendicular to the rod, hits it at a distance $x = \frac{ℓ}{\sqrt{6}}$ from the centre C of the rod. Collision is elastic
Find the angular velocity of the rod and after the collision

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$\frac{\sqrt{6} u}{l}$

$\frac{\sqrt{3} u}{l}$

$\frac{3 u}{2 l}$

$\frac{\sqrt{5} u}{l}$

answer is B.

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Detailed Solution

$Let V_{1}, V_{2} and$ be the velocity of particle, velocity of the centre of the rod and angular speed of the rod immediately after collision.

In the diagrams the first, second and the third one represent the situations just before collision, just after collision and just after second collision. Momentum conservation
$mu = {mV}_{1} + {mV}_{2} \Rightarrow u = V_{1} + V_{2} \dots \dots \dots \dots \dots (i)$
Apply conservation of angular momentum about a fixed point is space that coincides with the initial position of the centre of the rod

$miu \ddot{x} \equiv {mV}_{1} \ddot{x} + \frac{{mℓ}^{2}}{12} ω$

[Note: Angular momentum of the rod about said point is $I_{CM} \vec{ω} + \vec{R} \times M {\vec{V}}_{2}$ where the second term is zero in present context ]

$u \frac{ℓ}{\sqrt{6}} = V_{1} \frac{ℓ}{\sqrt{6}} + \frac{ℓ^{2} ω}{12} \Rightarrow u = V_{1} + \frac{ωℓ}{2 \sqrt{6}} . . . . (ii)$

Collision is elastic hence we can use conservation of kinetic energy or the definition of coefficient of restitution (e). Let’s use the late
$\begin{aligned} Relative velocity of point P and the particle after collision \\ Relative velocity of point P and the particle before collision = - 1 \end{aligned}$

$\frac{V_{2} + ωx - V_{1}}{o - u} = - 1 \Rightarrow V_{2} - V_{1} = u - \frac{ωℓ}{\sqrt{6}} . . . . . (iii)$
Solving (i),(ii) and ((iii)
$V_{1} = V_{2} = \frac{u}{2} and ω = \frac{\sqrt{6} u}{ℓ}$