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Rod AB is placed on smooth horizontal plane and two particle each having same mass m as rod, strikes the rod simultaneously and stick as it as shown. Find the loss in kinetic energy due to collision. A. `(mv_(0)^(2))/(3)`B. `(mv_(0)^(2))/(6)`C. `(6mv_(0)^(2))/(7)`D. `(mv_(0)^(2))/(7)` |
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Answer» Loss in K.E. `=(1)/(2)mv_(0)^(2)xx2-(1)/(2)3mv_("cm")^(2)` `2mv_(0)=3m v_(cm)` |
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