Question

A wedge of mass $m$, having a smooth semi circular part of radius R is resting on smooth horizontal surface.Now a particle of mass $m/2$ is released from the top point of the semicircular part. Then,

• A. The maximum displacement of the wedge will be R
• B. The maximum displacement of the wedge will be $\frac{2R}{3}$
• C. The wedge will perform simple harmonic motion of amplitude R
• D. The wedge will perform oscillatory motion of amplitude $\frac{R}{3}$

Answer 1

Answer: (B), (D)

The correct options are
B The wedge will perform oscillatory motion of amplitude $\frac{R}{3}$
D The maximum displacement of the wedge will be $\frac{2R}{3}$

As there is no net external force acting on the system in horizontal direction , momentum is conserved in this direction and equal to zero .
Hence , displacement of center of mass is zero .

Displacement of center of mass is given by
${\left(S_x\right)}_{cm}=\frac{m_1{S_x}_1+m_2{S_x}_2}{m_1+m_2}$
$0=\frac{m_{1}x+m_2(x-2R)}{m_1+m_2}$
Hence , $x=\frac{2R}{3}$
As the mean position is taken as zero displacement and the maximum displacement is $\frac{2R}{3}$ , the wedge is in an oscillatory motion of amplitude $\frac{R}{3}$ about its mean position.