Question

Two identical balls $A$ and $B$ of equal masses, are lying on a smooth surface as shown in figure.


Ball $A$ hits ball $B$ [which is initially at rest] and sticks to it. What should the minimum velocity of ball $A$, so that both balls reach the highest point of inclined place? Ignore the impact at the corner of the inclined surface.

• A. $v=\sqrt{2gh}\text{m/s}$
• B. $v=\sqrt{4gh}\text{m/s}$
• C. $v=\sqrt{8gh}\text{m/s}$
• D. $v=\sqrt{5gh}\text{m/s}$

Answer 1

The correct option is C $v=\sqrt{8gh}\text{m/s}$

Velocity of the ball $A=v\text{m/s}$ along $+vex-$ axis direction.
Let mass of each ball be $m\text{kg}$
Let $v^{\prime }$ be the velocity of the system of balls $(A+B)$ after sticking.

Applying momentum conservation in $x-$ direction:
$mv=(m+m)v^{\prime }$
$\therefore v^{\prime }=\frac{v}{2}...\left(i\right)$

$\because$ Work done by normal force $W_N=0$ along the inclined plane and gravity is a conservative force, so applying energy conservation on ($A+B$), from the bottom to the highest point on the inclined plane:


$\text{Loss in KE of system}=\text{Gain in PE of system}$
$\frac{1}{2}(m+m)v^{\prime 2}=(m+m)gh$
$\frac{1}{2}\left(2m\right)v^{\prime 2}=2mgh$
$\Rightarrow v^{\prime }=\sqrt{2gh}$
From Eq. $\left(i\right)$, putting value of $v^{\prime }$,
$\left(\frac{v}{2}\right)=\sqrt{2gh}$
$\therefore v=2\sqrt{2gh}=\sqrt{8gh}\text{m/s}$
Option (c) is correct.