Question

The string shown in the figure is passing over a smooth pulley rigidly attached to the trolley $A$. The speed of the trolley is constant and equal to $v_A$. Speed and magnitude of acceleration of block $B$ at the instant shown in the figure is $v_B$ and $a_B$ respectively, then

• A. $a_B=\frac{6v_A^2}{125}$
• B. $a_B=\frac{16v_A^2}{125}$
• C. $a_B=\frac{16v_A^2}{25}$
• D. $a_B=\frac{24v_A^2}{125}$

Answer 1

The correct option is B $a_B=\frac{16v_A^2}{125}$


$(y-h)+\sqrt{x^2+h^2}=l$
$\frac{dy}{dt}+\frac{x}{\sqrt{x^2+h^2}}\frac{dx}{dt}=0$
$\frac{dy}{dt}=-\frac{x}{\sqrt{x^2+h^2}}\frac{dx}{dt}$
$\frac{dy}{dt}=-\frac{3}{5}{v}_A$
$|v_B|=\frac{3}{5}{v}_A\left(i\right)$
$\frac{d^{2}y}{d{t}^2}=-[\frac{x}{\sqrt{x^2+h^2}}\frac{d^{2}x}{d{t}^2}+\frac{dx}{dt}\left(\frac{d}{dt}\frac{x}{\sqrt{x^2+h^2}}\right)]$

$\frac{d^{2}y}{d{t}^2}=-\left[\frac{dx}{dt}\left(\frac{d}{dt}\frac{x}{\sqrt{x^2+h^2}}\right)\right]$
$(\because \frac{d^{2}x}{d{t}^2}=a_A=0)$

$\frac{d^{2}y}{d{t}^2}=-v_A\left[\frac{(x^2+h^2)\left(dx/dt\right)-\left(x^2\right)\left(dx/dt\right)}{{(x^2+h^2)}^{3/2}}\right]$

$a_B=-\left[v_A^2\frac{h^2}{(x^2+h^2{)}^{\frac{3}{2}}}\right]$

Since we are talking about magnitude of acceleration of $B$,
$\frac{d^{2}y}{d{t}^2}=v_A^2\frac{h^2}{(x^2+h^2{)}^{\frac{3}{2}}}$
$a_B=v_A^2\frac{16}{(5{)}^3}$
$a_B=\frac{16}{125}{v}_A^2\left(ii\right)$