Question bank/JEE Main/Physics

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$\sqrt{3}u\ m/s$D$\sqrt{7}u\ m/s$

Show answer and solution▾

[IMAGE] As cylinder will remains in contact with wedge A
Vx = 2u
As it also remain in contact with wedge B
u sin 30° = Vy cos30° - Vxsin30°
Vy = Vx$\frac{Sin30{^\circ}}{Cos30{^\circ}}$ +
$u\frac{Sin30{^\circ}}{Cos30{^\circ}}$
Vy = Vx tan30° + u tan 30°
Vy = 3utan30° = $\sqrt{3}$u
V = $\sqrt{7}$u

Source: Grass-G / P-06 Physics paper 14 October.docx