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.
