Question

A block of mass 'm' is initially lying on a wedge of mass 'M' with an angle of inclination $\theta$, as shown in the figure. Calculate the displacement of the wedge when the block reaches bottom of the wedge.

• A. $\frac{mh}{M+m}$
• B. $\frac{mhcot\theta }{M+m}$
• C. $\frac{mhcos\theta }{M+m}$
• D. $\frac{mhsin\theta }{M+m}$

Answer 1

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



$l=hcos\theta$
Let displacement of wedge be x towards right:
$\Delta x_{CM}=\frac{m(l+x)+Mx}{M+m}$
$0=ml+(M+m)x$
$x=\frac{-ml}{M+m}=-\frac{mhcot\theta }{M+m}$