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On a smooth horizontal plane, a rod PQ o...

On a smooth horizontal plane, a rod PQ of length 6m and mass `2 kg` is placed. The centre of the rod is placed at the origin of the coordinate axes. An impulse of `4N -S` is applied at P perependicular to the length of the rod. Find the distance of a point form centre of the rod which is at rest after the impact (in metres)

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The correct Answer is:
1
`V = (J)/(m)` and `omega = (J(L//2))/(I) = (6JL)/(mL^(2)) = (6J)/(mL)`

At a point which is at rest after the impact, V and `omega` should cancel each other.
So, `r = (V)/(omega) = ((J)/(m)) ((mL)/(6J)) = (L)/(6) = (6)/(6) = 1m`
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