Collision in Spring Mass System!

The coefficient of restitution is defined as the ratio of the speed after to the speed before a collision. So, in this case, the answer is half of the speed before the collision (which we can find from conservation of energy).

$0.5k(a^2-(0.5a)^2)=0.5mV_i^2\Rightarrow V_i=a\sqrt{\frac{3k}{4m}}$

$V_f=0.5V_i=a\sqrt{\frac{3k}{16m}}\Rightarrow n=3$

We neglect the effect of the spring during the collision by letting the duration of the collision go to zero.

I'm astonished to see this problem at level 4. This is not even level 3 man!!

$\text {P.E. converted to } K.E.=\dfrac 1 2 *k\{a^2-(\dfrac a 2 )^2\}= \dfrac 1 2 *k\{\dfrac 3 4 *a^2\} \\ K.E.~gained~=\dfrac 1 2 *m*V^2.\\\implies~V=a\sqrt{\dfrac k m *\dfrac 3 4 } .\\But ~velocity~ AFTER~ collision =\dfrac 1 2 *V=\dfrac 1 2 * a\sqrt{\dfrac k m *\dfrac 3 4 }\\=a\sqrt{\dfrac 1 4 *\dfrac k m *\dfrac 3 4 }\\ =a\sqrt{ \dfrac { \color{#3D99F6}{3} *k} {16m} }=a\sqrt{ \dfrac { \color{#3D99F6}{n} *k} {16m} }. ~~~~~ ~~~~~ n= \Large \color{#D61F06}{3}$