Q: A particle of mass 1 kg is fired with velocity 50 m/s at an angle of 60° from horizontal . It is acted by viscous force of 0.2 v during its journey . The horizontal distance distance travelled by it in first 10 sec is

(a) 90 m

(b) 108 m

(c) 125 m

(d) 213 m

Ans: (b)

Sol: Retarding force in X-direction

F_{x} = F cosθ

F_{x} = -0.2 v cosθ = -(v/5)cosθ

a_{x} = -(v/5)cosθ ; (since m = 1 kg)

$\large \frac{dv_x}{dt} = -\frac{v}{5} cos\theta = – \frac{v_x}{5}$

$\large \int_{25}^{v_x} \frac{dv_x}{v_x} = – \frac{1}{5}\int_{0}^{t} dt $ ; (since v_{ox} = 50 cos60° = 25 m/s)

$\large ln\frac{v_x}{25} = – \frac{t}{5}$

$\large \frac{v_x}{25} = e^{- \frac{t}{5}}$

$\large v_x = 25 e^{- \frac{t}{5}}$

$\large \frac{dx}{dt} = 25 e^{- \frac{t}{5}} $

$\large \int_{0}^{x} dx = \int_{0}^{10} 25 e^{- \frac{t}{5}}$

$\large x = 25 [\frac{e^{- \frac{t}{5}}}{-\frac{1}{5}}]_{0}^{10}$

$\large x = -125 [e^{-2} – e^0]$

$\large x = 125 [1 – e^{-2} ]$

$\large x = 125 [1 – \frac{1}{e^2} ]$

x ≈ 108 m