Question

(a) Show that the velocity acquired by a particle in sliding down an inclined plane is the same as that acquired by a particle falling freely from rest through a distance equal to the height of the inclined plane.

(b) Find the time taken in sliding a particle down the whole length of the incline.

Accepted Answer

$(a)$ Let a particle sliding from $C$ to $A$ along the inclined plane and acquire a final velocity $v_{1}$ , covering a distance $s$ .

If $ heta$ is the angle of inclination, then $\sin θ = \frac{h}{s}$ . Now for the sliding particle, $s = s, u = 0, a = g \sin θ, v = v_{1}$ . [Taking the direction $C$ to $A$ as positive]
Using, $v^{2} = u^{2} + 2 a s$
$\begin{array}{l} \Rightarrow & v_{1}^{2} = 2 (g \sin θ) s = 2 g [\frac{h}{s}] s = 2 g h \end{array}$

$∴ v_{1} = \sqrt{2 g h} . . . (i)$

Again, let a particle fall from rest along a distance $h$ and acquire a velocity $v_{2}$ , then from $v^{2} = u^{2} + 2 a s$ ,
$v_{2}^{2} = 0 + 2 g h \Rightarrow v_{2} = \sqrt{2 g h} . . . (ii)$
From $(i)$ and $(ii)$ , we get the same result. Hence, particle will acquire the same velocity in both cases.

$(b)$ Time taken in sliding a particle down the entire length of the incline: For the motion of particle from $C$ to $A$ , $u = 0, s = s, a$ $= g \sin θ and t = ?$

Using $s = u t + \frac{1}{2} a t^{2}$ , we get $s = \frac{1}{2} (g \sin θ) t^{2}$

$\Rightarrow t^{2} = \frac{2 s}{g \sin θ} = \frac{2 h}{g \sin^{2} θ}$ $[∴ \sin θ = \frac{h}{s}]$

$∴ t = \sqrt{\frac{2 h}{g}} cosec θ$
Thus, the time taken varies directly as the cosecant of the angle of inclination. Now, since cosecant is a decreasing function in the first quadrant with an increase in $θ$ , time $t$ would decrease.

B M Sharma Solutions for Chapter: Kinematics I, Exercise 34: ILLUSTRATION 4.32

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 34: ILLUSTRATION 4.32

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 34: ILLUSTRATION 4.32 with Hints & Solutions

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