Block A of mass 2 kg moving at 10 m s ^ -1 on a smooth horizontal...

Since the collision is perfectly elastic and the masses of Block A and Block B are equal, they exchange their velocities.

Velocity of Block B just after the collision, $v = 10$ m s$^{-1}$.

As Block B slides on the rough surface, it experiences a retardation $a = \mu g$.

$a = 0.5 \times 10 = 5$ m s$^{-2}$

Using the equation of motion $v_f^2 = v^2 - 2as$ for the stopping distance $s$ (where final velocity $v_f = 0$):

$0 = (10)^2 - 2 \times 5 \times s$

$100 = 10 s$

$s = 10$ m.

Answer: $10$ m