Question

A block of mass 2 kg slides down the face of a smooth ${45}^{\circ }$ wedge of mass 9kg as shown in figure. The wedge is placed on a frictionless horizontal surface. Determine the acceleration of the wedge.

• A. 2 m/$s^2$.
• B. $\frac{11}{\sqrt{2}}$ m/$s^2$.
• C. 1 m/$s^2$.
• D. none of these

Answer 1

The correct option is C

1 m/$s^2$.

Lets assure wedge goes right with acceleration a

Nsin$\theta =Ma$

$\frac{N}{\sqrt{2}}=9a$

Constraint

Acceleration perpendicular to wedge remains same.

So

mg cos$\theta$ - N =ma sin$\theta$ .....(ii)

solving (i) & (ii) a = 1 m/$s^2$.