question_answer

A system shown in the figure. Assume that cylinder remains in contact with the sedge and block hence the velocity of cylinder is

A) \[\frac{\sqrt{19-4\sqrt{3}}}{2}\text{ m/s}\]

B) \[\frac{\sqrt{13}}{2}\text{ m/s}\]

C) \[\sqrt{3}\text{m/s}\]

D) \[\sqrt{7}\text{ m/s}\]

Correct Answer: D

Solution :

[d] Supposing block is fixed and wedge moves with velocity 3m/s relative to block, and so \[3\sin 30{}^\circ ={{v}_{y}}\cos 30{}^\circ ,\text{     }\therefore {{\text{v}}_{y}}=\sqrt{3}m/s.\] \[\text{and }{{u}_{x}}=\frac{1}{\sin 30{}^\circ }=2m/s\] \[\therefore u=\sqrt{u_{x}^{2}+u_{y}^{2}}=\sqrt{4+3}=\sqrt{7}m/s.\]