Question

On an inclined plane, two blocks are moving with constant velocities as shown in figure. Neglect the effect of gravity. Find the velocity of their $COM$.

• A. $\frac{4v}{5}\hat{i}+\frac{3v}{5}\hat{j}$
• B. $\frac{3v}{5}\hat{i}+\frac{4v}{5}\hat{j}$
• C. $\frac{4v}{5}\hat{i}-\frac{3v}{5}\hat{j}$
• D. $\frac{3v}{5}\hat{i}-\frac{4v}{5}\hat{j}$

Answer 1

The correct option is A $\frac{4v}{5}\hat{i}+\frac{3v}{5}\hat{j}$

For $2m$ mass block:


$v_{x_1}=-v\cos {37}^{\circ }=\frac{-4v}{5}$ and
$v_{y_1}=-v\sin {37}^{\circ }=\frac{-3v}{5}$

For $4m$ mass block:


$v_{x_2}=2v\cos {37}^{\circ }=2v\times \frac{4}{5}=\frac{8v}{5}$ and
$v_{y_2}=2v\sin {37}^{\circ }=2v\times \frac{3}{5}=\frac{6v}{5}$

As we know,
$v_{COM}=\frac{v_1{m}_1+v_2{m}_2}{m_1+m_2}$

$(v_{COM}{)}_x=$ velocity of $COM$ in $x-$ direction
$=\frac{\left(\frac{-4v}{5}\right)\times 2m+\left(\frac{8v}{5}\right)\times \left(4m\right)}{2m+4m}$
$=\frac{\frac{-8v}{5}+\frac{32v}{5}}{6}$
$=\frac{24v}{30}=\frac{4v}{5}$

$(v_{COM}{)}_y=$ velocity of $COM$ in $y-$ direction
$=\frac{\left(\frac{-3v}{5}\right)\times 2m+\left(\frac{6v}{5}\right)\times \left(4m\right)}{2m+4m}$
$=\frac{\frac{-6v}{5}+\frac{24v}{5}}{6}=\frac{18v}{30}$
$=\frac{3v}{5}$

Hence, $COM$ of the two block system will move with a velocity $v=\frac{4v}{5}\hat{i}+\frac{3v}{5}\hat{j}$

Alternate method:
Along the surface of inclined plane,
$v_{COM}=\frac{\left(4m\right)\left(2v\right)+\left(2m\right)(-v)}{4m+2m}$
$v_{COM}=v$
(taking +ve direction up the incline)

This is along the inclined surface.
But we have to get $v_{COM}$ in $x$ and $y$ direction, so resolve it in $x$ and $y$ direction.


$(v_{COM}{)}_x=v\cos {37}^{\circ }=\frac{4v}{5}$
$(v_{COM}{)}_y=v\sin {37}^{\circ }=\frac{3v}{5}$