There is a 2.0-kg disk travelling at 3.0 m/s on frictionless ice. It strikes a 1.0-kg stick of length 4.0 m that is lying flat on the same frictionless ice. Assume elastic collision and that the disk does not deviate from its original line of motion, find the translational speed of the disk, $v_{df}$. Let $m_d$ be the mass of the disk and $m_s$ be the mass of the stick. Let $v_{d}$ be the velocity of the disk and $v_{s}$ be the velocity of the stick.

The solution used the conservation of linear momentum, conservation of angular momentum and conservation of mechanical energy:

$$m_dv_{di}= m_dv_{df}+m_sv_s $$

$$\frac{l}{2} m_dv_{di} = \frac{l}{2} m_dv_{df} + I_{CM} \omega $$

$$\frac{1}{2} m_dv_{di}^2 = \frac{1}{2} m_dv_{df}^2 +\frac{1}{2} m_sv_{s}^2+ I_{CM} \omega^2 $$

Which, after solving, got a solution of:

$v_{df} = \frac{7}{3} m/s^2 $

My question is: Why does the relation $u_1 -u_2 = v_2 - v_1$ not work in this case? Isn't it an elastic collision? When I used this:

By conservation of linear momentum,

$$m_dv_{di}= m_dv_{df}+m_sv_s $$

$$(2.0)(3.0) = (2.0)(v_{df}) +(1.0)(v_s)$$ hence $$v_s = 6-2 v_{df}$$

from $u_1 -u_2 = v_2 - v_1$,

$$3.0 - (0) = (v_s)-(v_{df})$$ hence $$v_s = 3+v_{df}$$

Combining and doing some basic manipulation, $$v_{df}=1.0 m/s^2$$

Which contradicts with the solution.

Please let me know where I may have made some wrong assumptions. Thanks!