A body is released from the top of an inclined plane of inclination theta. It reaches the bottom with velocity V. If keeping the length same the angle of the inclination is doubled what will be the velocity of the body and reaching the ground?

Mar 11, 2018

v_1=sqrt(4*H*g costheta

Explanation:

let the height of incline be initially be $H$ and length of the incline be $l$.and let $\theta$be the initial angle.

The figure show Energy diagram at the different points of the inclined plane.

there for $S \int h \eta = \frac{H}{l}$ $\ldots \ldots \ldots \ldots . . \left(i\right)$
and the $\cos \theta = \frac{\sqrt{{l}^{2} - {H}^{2}}}{l}$ $\ldots \ldots \ldots \ldots . \left(i i\right)$

but ,now after change new angle is (${\theta}_{\circ}$)=$2 \cdot \theta$

Let${H}_{1}$ be the new height of triangle.
$\sin 2 \theta = 2 \sin \theta \cos \theta$=${h}_{1} / l$

[since length of the inclined has not yet changed.]

using (i) and (ii)

we get the new height as ,

${h}_{1} = 2 \cdot H \cdot \frac{\sqrt{{l}^{2} - {H}^{2}}}{l}$

by conserving the Total mechanical energy,

we get,

$m g {h}_{1} = \frac{1}{2} m {v}_{1}^{2}$ [let $_ v 1$ be new speed]

putting ${h}_{1}$ in this ,

${v}_{1} = \sqrt{4 \cdot H \cdot g \cdot \frac{\sqrt{{l}^{2} - {H}^{2}}}{l}}$

or (to reduce variables)

v_1=sqrt(4*H*g costheta

but the initial velocity is

$v = \sqrt{2 g H}$

v_1/v=sqrt(2*costheta

or

v_1=v*sqrt(2*costheta

Hence ,the velocity becomes $\sqrt{2 \cos \theta}$ times the initial.