Question

A small block of mass $m$ is pushed on a smooth track from position $A$ with a velocity $\frac{2}{\sqrt{5}}$ times the minimum velocity required to reach point $D$. When the block reaches point $B$, what is the direction (in terms of angle with horizontal) of acceleration of the block ?

• A. ${\tan }^{-1}\left(\frac{1}{2}\right)$
• B. ${\tan }^{-1}\left(2\right)$
• C. ${\sin }^{-1}\left(\frac{2}{3}\right)$
• D. The block never reaches point $B$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{1}{2}\right)$

$u_{\text{min}}=\sqrt{5gR}$ (to just reach $D$)

Given $u=\frac{2}{\sqrt{5}}\times \sqrt{5gR}$
$=2\sqrt{gR}$
At $B$,
$v_B^2=u^2-2gR(1-\cos {90}^{\circ })$
$=4gR-2gR$
$=2gR$
For $v_{CM},a_t=g\sin \theta$ and $a_C=\frac{v^2}{R}$
At $B,a_t=g\sin {90}^{\circ }=g$
$a_c=\frac{2gR}{R}=2g$
$\tan \alpha =\frac{a_t}{a_c}=\frac{g}{2g}=\frac{1}{2}$