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Question

# A uniform rod of length l and mass 4m lies on a frictionless horizontal surface on which it is free to move. A ball of mass m moving with speed v as shown in figure, collides with the rod at one of its ends. If the ball comes to rest immediately after the collision, then find the angular velocity ω of the rod just after collision.

A
3v4l
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B
2vl
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C
3v2l
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D
Zero
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Solution

## The correct option is C 3v2lOn the system of (rod + ball), there is no external force. Hence, linear momentum will be conserved. After collision: Pi=Pf ⇒mv=(4m)v′+m(0) ∴v′=v4 ...(1) Also for (rod+ball) system, τext=0. Applying angular momentum conservation about centre (O) of rod : Li=Lf Taking anticlockwise sense of rotation as +ve, +mv(l2)=LTrans+LRot Here LTrans=M(→r×→v0)=0, because vcom i.e v′ passes through centre O, hence (→r×→v0)=0 about point O ⇒mv(l2)=0+ICOMω ...(i) ICOMfor the rod, ICOM=(4m)l212=ml23 Substituting in Eq (i), mv(l2)=ml23×ω ∴ω=3v2l

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