InterviewSolution
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InterviewSolution
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A uniform bar of length `6a` and mass `8m` lies on a smooth horizontal table. Two point masses `m` and `2m` moving in the same horizontal plane with speeds `2v` and `v` respectively, strike the bar (as shown in figure) and stick to the bar after collision. Calculate (a) velocity of the centre of mass (b) angular velocity about centre of mass and (c) total kinetic energy, just after collision. |
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Answer» (a) As `F_(ext)=0` linear momentum of the system is conserved i.e., `-2mxxv+mxx2v+0=(2m+m+8m)xxV` or `V=0` i.e., velocity of centre of mass is zero. (b) As `tau_(ext)=0` angular momentum of the system is conserved. i.e, `m_(1)v_(1)r_(1)+m_(2)v_(2)r_(2)=(I_(1)+I_(2)+I_(3)) omega` `2mva+m(2v)(2a)=[2m(a)^(2)+8mxx(6a)^(2)//12]omega` ie. `6mva=30ma^(2) omegarArromega=(v/(5a))` (c) From (a) and (b) it is clear that, the system has no translatory motion but only rotatory motion. `E=1/2Iomega^(2)=1/2(30ma^(2))[v/(5a)]^(2)=3/5mv^(2)` |
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