Question

A block of mass $m$ is pushed towards a movable wedge of mass $\eta m$ and height $h$, with a velocity $u$. All surfaces are smooth. The minimum value of $u$ for which the block will reach the top of the wedge is

• A. $\sqrt{2gh}$
• B. $\sqrt{2gh(1+\frac{1}{\eta })}$
• C. $\sqrt{2gh(1-\frac{1}{\eta })}$
• D. $\eta \sqrt{2gh}$

Answer 1

The correct option is B $\sqrt{2gh(1+\frac{1}{\eta })}$

The centre of mass of the 'block plus wedge' must move with speed $\frac{mu}{m+\eta m}=\frac{u}{1+\eta }=v_{CM}$.
$\therefore \frac{1}{2}m{u}^2-mgh=\frac{1}{2}(m+\eta m)v_{CM}^2$
$u=\sqrt{2gh(1+\frac{1}{\eta })}$
The answer is (C)