Question

A block of mass $m$ slides down an inclined wedge of same $m$ shown in figure. Friction is absent everywhere. Magnitude of 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 $\frac{g{\sin }^2\theta }{(1+{\sin }^2\theta )}$

$m{a}_1\to$ psoudo force

$a_2\to acc$ wrt wedge

$N_1=mg+N_{2}cos\theta$ - (1)

$N_{2}sin\theta =m{a}_1$ - (2)

$N_2+m{a}_{1}sin\theta =mgcos\theta$ - (3)

$m{a}_{1}cos\theta +mgsin\theta =m{a}_2$ - (4)

From 2 $N_2=\frac{m{a}_1}{sin\theta }$ - (5)

New from (5) & (3)

$\frac{m{a}_1}{sin\theta }+m{a}_{1}sin\theta =mgcos\theta$

$\therefore a_1(1+si{n}^2\theta )=gsin\theta cos\theta$

$\therefore a_1=\frac{gsin\theta cos\theta }{1+si{n}^2\theta }$ - (6)

From (4) & (6)

$a_1=a_{1}cos\theta +gsin\theta$

$=\frac{gsin\theta co{s}^2\theta +gsin\theta (1+si{n}^2\theta )}{1+si{n}^2\theta }$

$a_2=\frac{2gsin\theta }{1=si{n}^2\theta }\to$ e r t wedge

${\overline{a}}_3={\overline{a}}_2-{\overline{a}}_1\to$ w r t ground

$\therefore a_3=\frac{gsin\theta cos\theta }{1+si{n}^2\theta }\hat{i}-\frac{2gsi{n}^2\theta \hat{j}}{1+si{n}^2\theta }$

$a_3\to$ acc of block wrt ground

Now $a_{com}=\frac{m{\overline{a}}_1+m{\overline{a}}_3}{2m}$

$a_{com}=\frac{1}{2}[{\overline{a}}_1+{\overline{a}}_3]$

$=\frac{1}{2}[-\frac{gsin\theta cos\theta }{1+si{n}^2\theta }\hat{i}+\frac{gsin\theta cos\theta }{1+sin\theta }\hat{i}-\frac{2gsi{n}^2\theta }{1+si{n}^2\theta }\hat{j}]$

$\therefore a_{com}=-\frac{2gsi{n}^2\theta }{1+si{n}^2\theta }\hat{j}$

$\therefore a_{com}=\frac{2gsi{n}^2\theta }{1+si{n}^2\theta }$ in downward direction.