A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall to a horizontal position and released. The ball hits the wall, the coefficient of restitution, being $(2/\sqrt{5})$. What is the minimum number of collisions after which the amplitude of oscillation becomes less than $60°?$

When simple   pendulum   released from position  A strikes   the wall with   velocity v then by conservation   of mechanical  energy. 
$mgL+0=\frac{1}{2}m{v}^2+0$, i.e.  $v=\sqrt{2gL}$
Now  as  coefficient   of restitution  is e so, speed of   pendulum  after first  collision  will be
 $v_1=ev=e\sqrt{2gL}$
Now   after completing   oscillation   in accordance with conservation  of mechanical   energy  it will strike the wall  with same velocity  and so its velocity  after second  collision  will be

$v_2=e{v}_1=e\left(e\sqrt{2gL}\right)=e^2\sqrt{2gL}$

Let after n collisions, the angular amplitude is 60^∘ when the bob again moves towards the wall from C, the velocity V_B′ before collision is $\mathrm{mg}\frac{1}{2}=\frac{1}{2}{\mathrm{mV}}_{\mathrm{B}}^{\text{'}2}⟹{\mathrm{V}}_{\mathrm{B}}^{\mathrm{\prime }}=\sqrt{\mathrm{gl}}$

${\left(\frac{2}{\sqrt{5}}\right)}^{\mathrm{n}}\times \sqrt{2\mathrm{gl}}=\sqrt{\mathrm{gl}}⟹{\left(\frac{2}{\sqrt{5}}\right)}^{\mathrm{n}}=\frac{1}{\sqrt{2}}$

Taking log on both sides we get

$\mathrm{nlog}\left(\frac{2}{\sqrt{5}}\right)=\log \frac{1}{\sqrt{2}}⟹\mathrm{n}=4$