Question

$N$ identical balls are placed on a smooth horizontal surface. Another ball of same mass collides elastically with velocity $u$ with first ball of $N$ balls. A process of collision is thus started in which first ball collides with second ball and the second ball with the third ball and so on. The coefficient of restitution for each collision is $e$. Find speed of $N^{th}$ ball:

• A. $\frac{(e+1{)}^{N-1}}{2^{N-1}}u$
• B. $(1+e{)}^{N}u$
• C. $u^N(1+e{)}^N$

Answer 1

The correct option is A $\frac{(e+1{)}^{N-1}}{2^{N-1}}u$

Step 1: Find the velocity after first collision.
After the first collision, velocity of first ball becomes $u$.

Using conservation of momentum
$mu=m{v}_1+m{v}_2$

$u=v_1+v_2\dots \left(i\right)$

$e=\frac{-(v_1-v_2)}{u}$

$e=\frac{-(u-v_2-v_2)}{u}$

$eu=2v_2-u$

$v_2=\frac{(e+1)u}{2}$

Step 2: Find the velocity after second collision.
Conservation of momentum

$m{v}_2=m{v}_2^{\prime }+m{v}_3$

$v_2=v_2^{\prime }+v_3$

$v_2^{\prime }=v_2-v_3$

$e=\frac{-(v_2^{\prime }-v_3)}{v_2}$

$e{v}_2=-(v_2-v_3-v_3)$

$e{v}_2+v_2=2v_3$

$v_3=\frac{v_2(e+1)}{2}$

$v_3=\frac{(e+1)}{2}\frac{(e+1)}{2}u$

$v_3=\frac{(e+1{)}^2}{2^2}u$

Step 3: Find the velocity after $N^{th}$ collision.

$v_3=\frac{(e+1{)}^2}{2^2}u$

Similarly,

$v_4=\frac{(e+1{)}^3}{2^3}u$

$v_n=\frac{(e+1{)}^{N-1}}{2^{N-1}}u$