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Question

A uniform rod of length l lies on a frictionless horizontal surface on which it is pivoted about its centre. A ball moving with speed v as shown in figure collides with the rod at one of the ends. After hitting, the ball sticks to the rod. There is a similar uniform rod of length l, pivoted at one of its ends and a similar ball with same speed strikes the other end of the rod and sticks to it. The ratio of final angular momentum in both cases is:


A
12
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B
13
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C
81
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D
14
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Solution

The correct option is A 12
Rod pivoted about its centre:


As we know,
Li=Lf
Initial angular momentum Li=mvR=mvl2
Li=Lf=mvl2 ... (1) (pivoted at A)

Rod pivoted about its end:


As we know,
Li=Lf
Initial angular momentum about end B =mvr=mvl
Li=Lf=mvl (pivoted at B)

Hence, ratio in final angular momentum in the two cases
(Lf)A(Lf)B=(Li)A(Li)B=mvl/2mvl
i.e (Lf)A(Lf)B=12

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