Question

In the given figure, all the collisions are elastic. Find the velocity of $n^{th}$ ball after collision.

• A. $2^{n-1}u$
• B. $3^{n-1}u$
• C. ${\left(\frac{2}{3}\right)}^{n-1}u$
• D. ${\left(\frac{3}{2}\right)}^{n-1}u$

Answer 1

The correct option is C ${\left(\frac{2}{3}\right)}^{n-1}u$

Collision between $1^{st}$ and $2^{nd}$ balls :
Before collision :
The velocity of mass $m=u$
After collision :
Let the velocity of mass $m=v_1$,
velocity of mass $2m=v_2$

From linear momentum conservation :
$mu=m{v}_1+2m{v}_2$
$\Rightarrow v_1+2v_2=u........\left(i\right)$

Newton's law of restitution :
$e=1=\frac{-(v_2-v_1)}{(0-u)}$
$\Rightarrow v_2-v_1=u....\left(ii\right)$
From eq (i) & (ii)
$2u=3v_2$
$\Rightarrow v_2=\frac{2}{3}u$
Collision between $2^{nd}$ and $3^{rd}$ balls:


Linear momentum conservation
$2m\times \frac{2}{3}u=2m{v}_1+4m{v}_2$
$\frac{2}{3}u=v_1+2v_2....\left(iii\right)$
For elastic collision,
$e=1=\frac{-(v_2-v_1)}{0-\frac{2u}{3}}$
$\Rightarrow \frac{2u}{3}=v_2-v_1....\left(iv\right)$
From eq (iii) & (iv),
$3v_2=\frac{4}{3}u\Rightarrow v_2=\frac{4}{9}u={\left(\frac{2}{3}\right)}^{2}u$
Similarly, after collision between $4m$ & $8m$ balls
$v_2={\left(\frac{2}{3}\right)}^{3}u$
After $n^{th}$ collision, velocity of ball of mass $2^{n-1}m$ is :
$v={\left(\frac{2}{3}\right)}^{n-1}u$
i.e, velocity of the $n^{th}$ ball will be ${\left(\frac{2}{3}\right)}^{n-1}u$