A bead of charge q and mass m is free to move on a smooth fixed insulating wire, which is in the shape of parabola y=(2−x2) located in a vertical plane. Uniform gravity g→=10(−j^)m/s2 is present in the space. The bead is pushed slightly on the wire from (0,2m), it reaches on the ground (y=0) with speed V. Electric field E→=α(xi^+yj^)N/C is present in space and qαm=2 ( in SI unit). Find the speed V (in m/s)

Work energy theorem ∫(mg→+qE→).(dxi^+dyj^)=12mv2−0 ∫[mg(−j^)+qα(xi^+yj^)](dxi^+dyj^)=12mv2 −mg∫20dy+qα∫02xdx+qα∫20ydy=12mv2 ⇒6m/s

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A bead of charge q and mass m is free to move on a smooth fixed insulating wire, which is in the shape of parabola $y = (2 - x^{2})$ located in a vertical plane. Uniform gravity $\vec{g} = 10 (- \hat{j}) m / s^{2}$ is present in the space. The bead is pushed slightly on the wire from $(0, 2 m)$ , it reaches on the ground $(y = 0)$ with speed V. Electric field $\vec{E} = α (x \hat{i} + y \hat{j}) N / C$ is present in space and $\frac{q α}{m} = 2$ ( in SI unit). Find the speed V (in m/s)

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answer is 6.

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Detailed Solution

Work energy theorem

$\int (m \vec{g} + q \vec{E}) . (d x \hat{i} + d y \hat{j}) = \frac{1}{2} m v^{2} - 0 \int [m g (- \hat{j}) + q α (x \hat{i} + y \hat{j})] (d x \hat{i} + d y \hat{j}) = \frac{1}{2} m v^{2} - m g \int_{2}^{0} d y + q α \int_{0}^{\sqrt{2}} x d x + q α \int_{2}^{0} y d y = \frac{1}{2} m v^{2} \Rightarrow 6 m / s$