Question

A system is shown in $Fig.6.351$. Assume that the cylinder remains in contact with the two wedges. Then the velocity of cylinder is

• A. $\sqrt{19-4\sqrt{3}}m{s}^{-1}$
• B. $\frac{\sqrt{13}u}{2}m{s}^{-1}$
• C. $\sqrt{7}um{s}^{-1}$
• D. $\sqrt{3}m{s}^{-1}$

Answer 1

The correct option is C $\sqrt{7}um{s}^{-1}$

$\begin{array}{c}AscylinderwillremainincontactwithwedgeA\\ V_x=2u\\ AsitalsoremainincontactwithwedgeB\\ u\sin {30}^0=V_y=\cos {30}^0-V_x\sin {30}^0\\ V_y=V_x\frac{\sin {30}^0}{\cos {30}^0}=\frac{u\sin {30}^0}{\cos {30}^0}\\ V_y=V_x\tan {30}^0+u\tan {30}^0\\ V_y=3u\tan {30}^0=\sqrt{3}u\\ V=\sqrt{{V_x}^2+{V_y}^2}=\sqrt{7}u\end{array}$

Hence,

all option are incorrect answer is $\sqrt{7}u$