Kinematics

JEE Main level, single correct, medium

258 of 308

Question 258 of 308

A particle is projected with velocity v0{v}_{0} along xx-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma=−αx2.ma=-\alpha {x}^{2}. The distance at which the particle stops:

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Answer: (B) (3mv022α)13{\left(\dfrac{3m{v}_{0}^{2}}{2\alpha }\right)}^{\dfrac{1}{3}}

F=−αx2F=-\alpha {x}^{2}
ma=−αx2ma=-\alpha {x}^{2}
a=−αx2ma=\dfrac{-\alpha {x}^{2}}{m}
vdvdx=−αmx2\dfrac{vdv}{dx}=-\dfrac{\alpha }{m}{x}^{2}
∫v00vdv=∫0x−αmx2dx{\int }_{{v}_{0}}^{0}vdv={\int }_{0}^{x}-\dfrac{\alpha }{m}{x}^{2}dx
(v22)v00=−αm(x33)0x{\left(\dfrac{{v}^{2}}{2}\right)}_{{v}_{0}}^{0}=-\dfrac{\alpha }{m}{\left(\dfrac{{x}^{3}}{3}\right)}_{0}^{x}
−v022=−αmx33\dfrac{-{v}_{0}^{2}}{2}=-\dfrac{\alpha }{m}\dfrac{{x}^{3}}{3}
x=(3mv022α)13x={\left(\dfrac{3m{v}_{0}^{2}}{2\alpha }\right)}^{\dfrac{1}{3}}

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