Question

A rat is moving down the slant of a wedge of angle of inclination $θ$ with a velocity $\vec{v}$ as shown in the figure. If the wedge moves towards left with a velocity $\vec{u}$ , find

(a) velocity of the rat relative to ground,
(b) value of $θ$ , if the rate moves vertically downward relative to an observer $G$ fixed with the ground.

Accepted Answer

Given;

Velocity of rat relative to wedge ${\vec{v}}_{R, W} = \vec{v}$

Velocity of wedge ${\vec{v}}_{W} = \vec{u}$

(a) The velocity of rat $R$ relative to the person $G$ standing on ground

${\vec{v}}_{R, W} = {\vec{v}}_{R} - {\vec{v}}_{W} \Rightarrow {\vec{v}}_{R} = {\vec{v}}_{R, W} + {\vec{v}}_{W} = \vec{u} + \vec{v}$
$v_{R} = \sqrt{u^{2} + v^{2} + 2 u v \cos (180 ° - θ)}$

$= \sqrt{u^{2} + v^{2} - 2 u v cosθ}$

(b) If the rat appears to move vertically downward relative to $G$ , then,

$u = v \cos θ$
$\Rightarrow \cos θ = \frac{u}{v} \Rightarrow θ = \cos^{- 1} \frac{u}{v}$

B M Sharma Solutions for Chapter: Kinematics II, Exercise 48: ILLUSTRATION 5.44

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 48: ILLUSTRATION 5.44

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