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A rod of mass M and length L lies on frictionless horizontal table.It is free to move on any way on table.A small body of mass m moving with velocity u collides in elastically with rod. (a)find velocity of center of mass of rod? (b)find angular velocity of rod about center of mass? |
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Answer» : Linear momentum angular momentum and kinetic ENERGY are conserved in the PROCESS. From conservation of linear momentum `Mv'=mv` ltbr. Or `v'=(m)/(M)v` ..(i) Conservation of angular momentum gives, `mvd=sqrt((Ml^(2))/(12))omega` or `omega=((12mvd)/(Ml^(2)))` ...(ii) Collision is elastic, HENCE `e=1` or relative speed of approach `=` relative speed of separation `thereforev=v'+domega` Subsitituting the values, we have `v=(m)/(M)v+(12mvd^(2))/(Ml^(2))` Solving it we get `m=(Ml^(2))/(12d^(2)+l^(2))` plz give brainliest..Explanation: |
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