In the above question how they have written the ratio in first step , and what is the use of it .

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Initial KE of particle 1 is $\frac12m_1u_1^2$. Final KE is $\frac12m_1v_1^2$. Fractional KE *retained* after collision is $(v_1/u_1)^2$.

Loss of momentum by particle 1 equals gain of momentum by particle 2 :

$m_1(u_1-v_1)=m_2(v_2-u_2)$

$u_1-v_1=\mu(v_2-u_2)$ ... (i)

where $\mu=m_2/m_1$.

Law of Restitution with $e=1$ for elastic collisions :

$v_2-v_1= -e(u_2-u_1) = u_1-u_2$

$u_1+v_1=v_2+u_2$ ... (ii)

Combining (i) and (ii) :

$2v_1=(1-\mu)v_2+(1+\mu)u_2$

$2u_1=(1+\mu)v_2+(1-\mu)u_2$.

Assuming that the target is initially at rest $(u_2=0)$ then

$v_1/u_1=(1-\mu)/(1+\mu)$.

In this case $\mu=m_2/m_1=2$ so

$(v_1/u_1)^2=(-1/3)^2=\frac19$.

After $n$ collisions the KE of particle 1 is $(\frac19)^n$ of its initial KE :

$(\frac19)^n=\frac{1}{6561}$

$9^n=6561=9^4$.

The number of collisions required is 4.

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