Question

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

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

Answer 1

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

When block just reaches the top, it is at rest wrt wedge, so that $u$ is minimum.
Let common velocity be $v^{\prime }$

From conservation of linear momentum,
$mu=(m+nm)v^{{}^{\prime }}$
$v^{{}^{\prime }}=\frac{u}{1+n}$
From energy conservation
$\frac{1}{2}m{u}^2=mgh+\frac{1}{2}(m+nm)v^{{{}^{\prime }}^2}$
$u=\sqrt{2gh(1+\frac{1}{n})}$