Question

A wedge $Q$ of mass $M$ is placed on a horizontal frictionless surface $A B$ and a block $P$ of mass $m$ is released on its frictionless slope. As $P$ slides by a length $L$ on this slope of inclination $\theta, Q$ would slide by a distance of

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

Answer 1

Answer: (D)

Let $Q$ slides by a distance $x$ towards left, then net horizontal displacement of $P$ w.r.t. ground
$=L \cos \theta-x$
Now apply $m_{1} x_{1}=m_{2} x_{2} $
$\Rightarrow M x=m(L \cos \theta-x)$
$\Rightarrow x=\frac{m L \cos \theta}{M+m}$