Question

A block of mass ‘m’ is pushed towards a movable wedge of mass ‘2m’ 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. $2\sqrt{2}g$
• B. $\sqrt{3gh}$
• C. $\sqrt{6gh}$
• D. $\sqrt{\frac{3}{2}}gh$

Answer 1

The correct option is B

$\sqrt{3gh}$

When block just reaches the top of the wedge, velocity of block w.r.t. wedge at the top of the wedge will be zero. Let ‘n’ be the horizontal velocity of both at this instant.

Applying energy conservation,

$\frac{1}{2}m{u}^2=\frac{1}{2}\left(3m\right)v^2+mgh$ .....(i)

Applying momentum conservation, we get

(2m + m)v = mu ….(ii)

$\Rightarrow v=\frac{u}{3}$

Substitute the value of ‘v’ in equation (i) to get,

$\frac{1}{2}m{u}^2=\frac{1}{2}\left(3m\right)\frac{u^2}{9}+mgh$

$\Rightarrow [\frac{u^2}{2}-\frac{3u^2}{18}]=gh$

$\Rightarrow h=\frac{6u^2}{18g}$

or $u=\sqrt{3gh}$