Q. A block of mass $m$ is pushed with a velocity $u$ towards a movable wedge of mass $nm$ and height $h$, figure. All the surfaces are smooth. The minimum value of $u$ for which the block will reach the top of wedge is
A) $\sqrt{2gh\left( 1 - \frac{1}{n} \right)}$
B) $\sqrt{2gh\left( 1 + \frac{1}{n} \right)}$
C) $\sqrt{3gh}$
D) $\sqrt{2gh}$
Explanation
(b) If $v$ is common velocity of the block and movable wedge, then applying the principle o conservation of linear momentum we get,
$mu + 0 = (m + nm)v$
$v = \frac{mu}{m(1 + n)} = \frac{u}{1 + n}$
This infact, can be taken as velocity of centre of mass of the block and wedge
$ie,\ v_{CM} = v = \frac{u}{1 + n}$
Applying the principle of conservation of energy
$\frac{1}{2}mu^{2} = mgh + \frac{1}{2}m(1 + n)\frac{u^{2}}{(1 + n)^{2}}$
or $u^{2} = 2gh + \frac{u^{2}}{1 + n}$
$u = \sqrt{2gh\left( 1 + \frac{1}{n} \right)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.