Question

Two identical balls each of radii $r$ are kept on a horizontal plane with their centers $d$ distance apart. A third identical ball collides elastically with both the balls symmetrically as shown in the figure. If the third ball comes to rest after the collision, $d$ should be

• A. $3r$
• B. $2\sqrt{2}r$
• C. $(\sqrt{2}+1)r$
• D. $(\sqrt{2}+2)r$

Answer 1

The correct option is B

$2\sqrt{2}r$

Let $v$ be velocity of the third ball before collision
$v_1,v_2$ be velocities of first and second balls after collision
From principle of conservation of momentum (COM) along x-axis:
$mv=m{v}_{1}cos\theta +m{v}_{2}cos\theta$
COM along y-axis:
$0=m{v}_{1}sin\theta -m{v}_{2}sin\theta \Rightarrow v_1=v_2=v_o\Rightarrow v=2v_{o}cos\theta$
As collision is elastic along common normal,
Co-efficient of restitution $e=1$
$\Rightarrow Velocityofseperation=Velocityofapproach$
$\Rightarrow v_0-0=(vcos\theta -0)\Rightarrow v_o=vcos\theta \therefore v_o=2v_{o}co{s}^2\theta \Rightarrow \theta ={45}^{\circ }$

$\Rightarrow d=Hypotenuse=2\sqrt{2}r$