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The blocks A and B shown in figure have masses `M_A=5kg` and `M_B=4kg`. The system is released from rest. The speed of B after A has travelled a distance `1m` along the incline is A. `(sqrt(3))/(2) sqrt(g)`B. `(sqrt(3))/(4)sqrt(g)`C. `(sqrt(g))/(2 sqrt(3))`D. `(sqrt(g))/(2)` |
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Answer» Correct Answer - C If `A` moves down the incline by `1` metre, `B`shall move up by `(1)/(2)` metre. If the speed of `B` is `v` then the speed of `A` be `2v`. From conservation of energy : Grain in `K.E` = loss in `P.E`. `(1)/(2) m_(A)(2v)^(2)+(1)/(2) m_(B)v^(2) = m_(A)g xx (3)/(5) -m_(B) g xx (1)/(2)` Solving we get `v = (1)/(2) sqrt((g)/(3))`. |
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