Question

A block of mass $m$ is placed at rest on the top of a smooth wedge of mass $M$, which in turn is placed at rest on a smooth horizontal surface as shown in figure. Then the distance moved by the wedge as the block reaches the foot of the wedge is

• A. $\left(\frac{M+m}{m}\right)l$
• B. $\left(\frac{M}{m+M}\right)l$
• C. $\left(\frac{m}{M+m}\right)l$
• D. $\left(\frac{M+m}{M}\right)l$

Answer 1

The correct option is C $\left(\frac{m}{M+m}\right)l$

Initially both wedge and block are at rest.
There is no external force in horizontal direction of the wedge block system.
Thus acceleration, $\Rightarrow a_{com}\text{in}x\text{direction is zero}$.
Therefore, $x-$ coordinate of the COM of the system will remain at rest even when block slides down.

Assume that wedge moves $X$ when block reaches the ground.
As $x_{com}$ is not changing.
$m_1{x}_1=m_2{x}_2$

$\Rightarrow MX=m(l-X)$
$X=\frac{ml}{m+M}$