Question

System is shown in the figure. Assume that cylinder remains in contact with the two wedges. The velocity of cylinder is

• A. $\sqrt{19-4\sqrt{3}}\frac{u}{2}m/s$
• B. $\frac{\sqrt{13}u}{2}m/s$
• C. $\sqrt{3}um/s$
• D. $\sqrt{7}um/s$

Answer 1

The correct option is D $\sqrt{7}um/s$

Method - I
Let the velocity of the cylinder be $V$, whose $x$ and $y$ components are $V_x$ and $V_y$ respectively.
As cylinder will remain in contact with wedge $A$ and $B$
$\Rightarrow V_x=2u$



And, from the FBDs
$u\cos {60}^{\circ }=V_y\cos {30}^{\circ }-V_x\cos {60}^{\circ }$
$\Rightarrow u\cos {60}^{\circ }+V_x\cos {60}^{\circ }=V_y\cos {30}^{\circ }$
$\Rightarrow u\cos {60}^{\circ }+2u\cos {60}^{\circ }=V_y\cos {30}^{\circ }$
$\Rightarrow 3u\left(\frac{1}{2}\right)=V_y\left(\frac{\sqrt{3}}{2}\right)$
$\Rightarrow V_y=\sqrt{3}u$
$\therefore$ $V=\sqrt{V_x^2+V_y^2}=\sqrt{7}u$


Method - II
In the relative frame i.e. frame of $A$, the arrangement will be as shown in the below figure

$3u\cos {60}^{\circ }=V_y\cos {30}^{\circ }$
$\Rightarrow V_y=3u\left(\frac{cos{60}^{\circ }}{cos{30}^{\circ }}\right)=\sqrt{3}u$
and $V_x=2u$
$\Rightarrow V=\sqrt{V_x^2+V_y^2}=\sqrt{7}u$