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A particle of mass 2kg moving horizontally with a speed of 2m/s strikes perpendicularly to the uniform rod of mass 10 kg & length 4m lying on a smooth horizontal surface. Particle strikes the rod at a distance of 1m from its centre and comes to the rest after the collision. The coefficient of restitution between the particle & the rod is

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a

3/7

b

7/20

c

1/2

d

7/15

answer is C.

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Detailed Solution

Apply linear momentum conservation & angular momentum conservation.
$\begin{array}{l} {mv}_{0} = Mv \\ {mv}_{0} \times 1 = \frac{{ML}^{2}}{12} \times ω \end{array}$
Solving above equations we get, v=0.4 m/s and $ω$ =0.3 rad/s
$e = \frac{[v + (ω \times 1)] - 0}{v_{0}} = \frac{7}{20}$

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