A block of mass m_1 moving with a velocity v on a smooth horizontal...

By the principle of conservation of linear momentum, the initial momentum of the system is equal to the final momentum.

Initial momentum $= m_1 v + m_2(0) = m_1 v$

Since the collision is completely inelastic, the two blocks stick together and move with a common velocity.

Final momentum $= (m_1 + m_2) \left(\dfrac{v}{5}\right)$

Equating initial and final momentum:
$m_1 v = (m_1 + m_2) \left(\dfrac{v}{5}\right)$

Cancelling $v$ from both sides and cross-multiplying:
$5m_1 = m_1 + m_2$

$4m_1 = m_2$

$\dfrac{m_1}{m_2} = \dfrac{1}{4}$

Thus, the ratio $m_1 : m_2$ is $1:4$.

Answer: $1:4$