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A mass `m` is moving at speed `v` perpendicular to a rod of length `d` and mass `M= 6m` pivots around a frictionless axle running through its centre and stickes to the end of the rod. The moment of inertia od the rod about its centre is `Md^(2)//12`. Then the angular speed of the system right after the collision us -A. `(2v)/(3d)`B. `(2v)/d`C. `v/d`D. `(3v)/(2d)` |
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Answer» Correct Answer - A By conservation of angular momentum about pivot `L=Iomega` `(mvd)/2=[(Md^(2))/12+m(d/2)^(2)]omega` `=((md^(2))/2+(md^(2))/4)omega` `=3/4md^(2)omegaimplies 2/3v/d=omega` |
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