Question

A block of mass $2kg$ slides down the face of a smooth ${45}^o$ wedge of mass $9kg$ as shown in the 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 ${}^2$

$a$ is the acceleration of the block w.r.t the wedge
For block:
$N+mA\sin {45}^o=mg\cos {45}^o$

So, $N=(2\times 10\times \frac{1}{\sqrt{2}})-(2\times A\times \frac{1}{\sqrt{2}})$

$N=10\sqrt{2}-A\sqrt{2}$

For wedge : $N\sin {45}^o=MA$

From Equation $\left(i\right)$ and $\left(ii\right)$

$(10\sqrt{2}-A\sqrt{2})\times \frac{1}{\sqrt{2}}=9\times A$

$10-A=9A$

$A=1m/s^2$