A particle of mass m and velocity $v_{1}$ in positive $y$ direction is projected on to a belt that is moving with uniform velocity $v_{2}$ in x-direction as shown in figure. Coefficient of friction between particle and belt is $μ$ . Assuming that the particle first touches the belt at the origin of fixed $x - y$ coordinate system and remains on the belt, find the co-ordinates $(x, y)$ of the point where sliding stops.

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$x = v_{2} \frac{\sqrt{v_{1}^{2} + v_{2}^{2}}}{2 μ g}, y = \frac{v_{1} \sqrt{v_{1}^{2} + v_{2}^{2}}}{2 μ g}$

$x = v_{2} \frac{\sqrt{v_{1}^{2} + v_{2}^{2}}}{7 μ g}, y = \frac{v_{1} \sqrt{v_{1}^{2} + v_{2}^{2}}}{2 μ g}$

$x = v_{2} \frac{\sqrt{v_{1}^{2} + v_{2}^{2}}}{5 μ g}, y = \frac{v_{1} \sqrt{v_{1}^{2} + v_{2}^{2}}}{2 μ g}$

$x = v_{2} \frac{\sqrt{v_{1}^{2} + v_{2}^{2}}}{6 μ g}, y = \frac{v_{1} \sqrt{v_{1}^{2} + v_{2}^{2}}}{2 μ g}$

answer is A.

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Detailed Solution

$v_{r} = \sqrt{v_{1} {}^{2}+ {v_{2}}^{2}}$ Retardation $a = μ g$

$∴$ Time when slipping will stop is $t = \frac{v_{r}}{a}$

$t = \frac{\sqrt{v_{1} {}^{2}+ {v_{2}}^{2}}}{μ g}$ ……………(i)

$s_{r} = \frac{{v_{r}}^{2}}{2 a} = \frac{v_{1} {}^{2}+ {v_{2}}^{2}}{2 μ g} x_{r} = - s_{r} \cos θ = - (\frac{v_{1} {}^{2}+ {v_{2}}^{2}}{2 μ g}) (\frac{v_{2}}{\sqrt{v_{1} {}^{2}+ {v_{2}}^{2}}}) = \frac{- v_{2} \sqrt{v_{1} {}^{2}+ {v_{2}}^{2}}}{2 μ g}$

$y_{r} = s_{r} \sin θ = (\frac{v_{1} {}^{2}+ {v_{2}}^{2}}{2 μ g}) (\frac{v_{1}}{\sqrt{v_{1} {}^{2}+ {v_{2}}^{2}}}) = \frac{v_{1} \sqrt{v_{1} {}^{2}+ {v_{2}}^{2}}}{2 μ g}$