Question

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

Answer 1

The correct option is A $1m/s^2$

For block
$N+mA\sin {45}^{\circ }=mg\cos {45}^{\circ }$

$N=10\sqrt{2}-A\sqrt{2}\dots \left(i\right)$
For wedge
$N\sin {45}^{\circ }=MA\dots \left(ii\right)$
Hence, from $\left(i\right)$ & $\left(ii\right)$
$10\sqrt{2}-A\sqrt{2})\times \frac{1}{\sqrt{2}}=9\times A$ $10-A=9A$
$A=1m/s^2$

Final answer: (c)