The blocks shown in the figure possess equal masses. The contact...

Given that the blocks have equal masses ($m_A = m_B = m$) and the collision is perfectly elastic, the velocities of the blocks are exchanged after the collision.

Initial velocity of block A, $u_A = 10 \text{ m/s}$
Initial velocity of block B, $u_B = 0$

After the collision, the velocity of block B becomes $v_B = u_A = 10 \text{ m/s}$.

Block B moves on a rough surface with a coefficient of friction $\mu = 0.10$.
The retardation of block B is given by $a = \mu g$.
Substituting the given values, $a = 0.10 \times 10 = 1 \text{ m/s}^2$.

Using the equation of motion $v^2 = u^2 - 2as$ for block B coming to rest ($v = 0$):
$0 = (10)^2 - 2(1)s$
$2s = 100$
$s = 50 \text{ m}$

The distance travelled by block B is $50 \text{ m}$.