Question

A block of mass $m$ is moved towards a movable wedge of mass $M=Km$ and height $h$ with velocity $u$ (All the surface are smooth). If the block just reaches the top of the wedge, the value of $u$ is:

• A. $\sqrt{2gh}$
• B. $\sqrt{\frac{2ghK}{1+K}}$
• C. $\sqrt{\frac{2gh(1+K)}{K}}$
• D. $\sqrt{2gh[1-\frac{1}{K}]}$

Answer 1

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

Let finally the system moves horizontally with velocity $v$.

At the maximum height, the velocity of the block w.r.t the wedge is zero but w.r.t ground is equal to $v$ horizontally.

Now applying conservation of linear momentum $p_i=P_f$

$mu+0=mv+Mv$

$mu+0=mv+\left(Km\right)v⟹v=\frac{u}{K+1}$ ............(1)

Also using work-energy theorem: $W_{allforces}=\Delta K.E$ (w.r.t ground)

$-mgh=\frac{1}{2}M{v}^2+\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

$-mgh=\frac{1}{2}\left(Km\right)\times \frac{u^2}{(K+1{)}^2}+\frac{1}{2}m\frac{u^2}{(K+1{)}^2}-\frac{1}{2}m{u}^2$ (using 1)

$⟹u=\sqrt{\frac{2gh(K+1)}{K}}$