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The potential energy `U`(in `J`) of a particle is given by `(ax + by)`, where `a` and `b` are constants. The mass of the particle is `1 kg` and `x` and `y` are the coordinates of the particle in metre. The particle is at rest at `(4a, 2b)` at time `t = 0`. Find the speed of the particle when it crosses x-axisA. `2sqrt(a^(2) + b^(2))`B. `sqrt(a^(2) + b^(2))`C. `(1)/(2)sqrt(a^(2) + b^(2))`D. `sqrt(((a^(2) + b^(2)))/(2))` |
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Answer» Correct Answer - 1 `vecF = -(d)/(dr) = -a hati - bhatj` `ac c. = (vecF)/(m)=- a hati - bhatj` `(dv)/(dt) = - at hati - bhatj` `vecV = - at hati - bthatj` `vecS=(4a-(a)/(2)t^(2))^(2) hati + (2b-(bt^(2))/(2))hatj` `vecS =(4a-(a)/(2)t^(2))hati +(2b-(bt)/(2))hatj` When it cross x axis y=0 So t= 2 So `V = 2 sqrt((a^(2) + b^(2)))` |
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