Question

Find the displacement travelled by the sphere in $1$ second if the acceleration of the cube is $2{\text{m/s}}^2$. System starts from rest and the sphere remains in contact with the cube.

• A. $1\text{m}$
• B. $2\text{m}$
• C. $\sqrt{2}\text{m}$
• D. $\sqrt{3}\text{m}$

Answer 1

The correct option is C $\sqrt{2}\text{m}$

By condition of wedge contraint, the displacment along normal to the common interface is same.


Let ${\delta }_x$ and ${\delta }_y$ be the displacement of sphere in the horizontal and vertical direction respectively.
From the FBD we have,
${\delta }_y\cos \theta ={\delta }_x\cos (90-\theta )={\delta }_x\sin \theta$
$\Rightarrow {\delta }_y={\delta }_x\tan \theta$
$\text{Since}$ $\theta ={45}^{\circ }$
$\Rightarrow {\delta }_y={\delta }_x$
Now, displacement travelled by cube in $1$ second,
$S=ut+\frac{1}{2}a{t}^2$
$\Rightarrow S=0+\frac{1}{2}\left(2\right)(1{)}^2$
$=1\text{m}$
$\therefore$ Net displacement $=\sqrt{{\delta }_x^2+{\delta }_y^2}$
$=\sqrt{(1{)}^2+(1{)}^2}$
$=\sqrt{2}\text{m}$