Question

Two bodies of mass $m$ are sliding down on the two sides of an equilateral triangular wedge of mass $10m$. Find the velocity of the wedge ($v$) when both the blocks reach the ground. (All surfaces are frictionless)

• A. $v=0$
• B. $v=g\hat{i}$
• C. $v=-g\hat{i}$
• D. $v=g\hat{j}$

Answer 1

The correct option is A $v=0$

When both blocks reach the ground, let their speeds be $v_1$ each and let the velocity of wedge be $v$ towards right.

For mass $m_1=m$, which is sliding towards left, $v_{1x}=-v_1\cos {60}^{\circ }$
For mass $m_2=m$ which is sliding towards right, $v_{2x}=v_1\cos {60}^{\circ }$
For wedge $m_3=10m$ which is sliding in $x$ direction, $v_{3x}=v$

There is no external force in the $x$-direction, so velocity of $(COM{)}_x$ will be zero. (because, initially the system was at rest)
i.e $V_{com}=\frac{m_1{v}_{1x}+m_2{v}_{2x}+m_3{v}_{3x}}{m_1+m_2+m_3}=0$
$\Rightarrow \frac{m(-v_1\cos {60}^o)+m\left(v_1\cos {60}^o\right)+10m\left(v\right)}{m+m+10m}=0$
$\Rightarrow v=0$

Hence, velocity of wedge will be zero.