Question

A block of mass $m$ slides down on an inclined wedge of same mass $m$ as shown in figure. Friction is absent everywhere. Acceleration of center of mass of the block and wedge is

• A. zero
• B. $\frac{g{\sin }^2\theta }{(1+{\sin }^2\theta )}$
• C. $\frac{g{\cos }^2\theta }{(1+{\sin }^2\theta )}$
• D. $\frac{g\sin \theta }{(1+\cos \theta )}$

Answer 1

The correct option is B $\frac{g{\sin }^2\theta }{(1+{\sin }^2\theta )}$

As mass $m$ can slide freely on the wedge. Hence, the block slides downwards with an acceleration $a_r$ relative to the wedge and $(a_r\cos \theta -a)$ with respect to the ground in the horizontal direction.

Since, except gravity as there is no external force, hence to keep the COM of the system to be at rest, the wedge has to move leftward with an acceleration $a.$

Since, net horizontal force on the system is zero. therefore, horizontal component of acceleration of COM of the system is zero.


Since there is no net force in horizontal direction, so

$F_{net,horizontal}=0$

$m(a_r\cos \theta -a)-ma=0$

$2ma=m{a}_r\cos \theta \Rightarrow a_r\cos \theta =2a$

$a=\frac{a_r\cos \theta }{2}........\left(1\right)$


From, FBD of the wedge we get,

$F_x=ma$

$N\sin \theta =ma$

$N=\frac{ma}{\sin \theta }........\left(2\right)$


From FBD of the block we get,

$N+ma\sin \theta =mg\cos \theta ......\left(3\right)$

Putting $\left(2\right)$ in $\left(3\right)$ we get,

$\frac{ma}{\sin \theta }+ma\sin \theta =mg\cos \theta$

$a\left(\frac{1+{\sin }^2\theta }{\sin \theta }\right)=g\cos \theta$

Using eq. $\left(1\right)$ in the above equation we get,

$\frac{a_r\cos \theta }{2}\left(\frac{1+{\sin }^2\theta }{\sin \theta }\right)=g\cos \theta$

$a_r=\frac{2g\sin \theta }{1+{\sin }^2\theta }$

Therefore, acceleration of the block vertically downwards is,

$a_{y,b}=a_r\sin \theta$

$a_{y,b}=\frac{2g{\sin }^2\theta }{1+{\sin }^2\theta }$

Since net force in horizontal direction or $x-$ direction is zero. So acceleration of centre of mass in $x-$ direction will be

$\therefore a_{x,com}=0$

Using relation for acceleration of the centre of mass in $y-$ direction

$a_{y,com}=\frac{m_w{a}_{y,w}+m_b{a}_{y,b}}{m_w+m_b}[\because a_{y,w}=0]$

$a_{y,com}=\frac{m{a}_{y,b}}{2m}=\frac{a_{y,b}}{2}$

Thus, acceleration of COM of the system will be,

$a_{com}=\frac{a_{y,b}}{2}=\frac{g{\sin }^2\theta }{(1+{\sin }^2\theta )}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(B\right)$ is the correct answer.