Question

A small 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. Let $h$

be the height of wedge and $\theta$ is the inclination, then the distance moved by the wedge as

the block reaches the foot of the wedge is :

• A. $\frac{Mh\cot \theta }{M+m}$
• B. $\frac{mh\cot \theta }{M+m}$
• C. $\frac{Mh\cos \theta }{M+m}$
• D. $\frac{mh\cos \theta }{M+m}$

Answer 1

The correct option is B $\frac{mh\cot \theta }{M+m}$

As there is zero external force in horizontal direction the COM(Centre of mass) will remain at same position.

The relative distance travelled by the block in right direction is $x=h\cot \theta$

Let the incline move by distance $x^{\prime }$ in left direction

The net distance covered by box is $h\cot \theta -x^{\prime }$

Using $m_1{x}_1=m_2{x}_2$

$m(h\cot \theta -x^{\prime })=M{x}^{\prime }$

$x^{\prime }=\frac{mh\cot \theta }{m+M}$