Step 1: Normal Direction (x-axis)

The collision is perfectly inelastic ($e=0$), so the final relative velocity along the line of impact is zero. Since the masses are identical ($m$), they will share the linear momentum equally.

Initial Momentum $P_x = mu$. Final Momentum $P_x’ = 2m v_x’$.

$$ v_{1x}’ = v_{2x}’ = \frac{u}{2} $$

Step 2: Tangential Direction (y-axis)

Friction acts to stop the relative slipping. Let $J_y$ be the impulse due to friction exerted on Disc 2 in the $+y$ direction (and on Disc 1 in the $-y$ direction).

Velocities at Contact Point $C$:

• Disc 1 (Right): Center stationary initially. Spin $\omega$. $v_{c1} = \omega r$ (up, $+y$).
• Disc 2 (Left): Moving right. No spin. $v_{c2} = 0$ (in $y$).
• Relative velocity $v_{rel} = \omega r$ (up). Friction opposes this, pushing Disc 1 down ($-y$) and Disc 2 up ($+y$).

Impulse Equations:

For Disc 1 (Force $-J_y$, Torque $-J_y r$):

1. $m v_{1y}’ = -J_y \implies v_{1y}’ = -\frac{J_y}{m}$
2. $I (\omega_1′ – \omega) = -J_y r \implies \frac{1}{2}mr^2(\omega_1′ – \omega) = -J_y r \implies \omega_1′ = \omega – \frac{2J_y}{mr}$

For Disc 2 (Force $+J_y$, Torque $-J_y r$ due to force at right edge):

1. $m v_{2y}’ = J_y \implies v_{2y}’ = \frac{J_y}{m}$
2. $I \omega_2′ = -J_y r \implies \omega_2′ = -\frac{2J_y}{mr}$

Step 3: Solving for “No Slip”

Slipping ceases, so tangential velocities at contact match:

$$ v_{c1y}’ = v_{c2y}’ $$

$$ v_{1y}’ + \omega_1′ r = v_{2y}’ – \omega_2′ r $$

Substituting the impulse expressions:

$$ -\frac{J_y}{m} + (\omega – \frac{2J_y}{mr})r = \frac{J_y}{m} – (-\frac{2J_y}{mr})r $$

$$ -\frac{J_y}{m} + \omega r – \frac{2J_y}{m} = \frac{J_y}{m} + \frac{2J_y}{m} $$

$$ \omega r – \frac{3J_y}{m} = \frac{3J_y}{m} \implies \omega r = \frac{6J_y}{m} $$

Therefore, the impulse is $J_y = \frac{m \omega r}{6}$.

Step 4: Final Velocities

Substitute $J_y$ back into the equations:

• $\omega_1′ = \omega – \frac{2}{mr}(\frac{m\omega r}{6}) = \omega – \frac{\omega}{3} = \frac{2\omega}{3}$
• $\omega_2′ = -\frac{2}{mr}(\frac{m\omega r}{6}) = -\frac{\omega}{3}$
• $v_{1y}’ = -\frac{1}{m}(\frac{m\omega r}{6}) = -\frac{\omega r}{6}$
• $v_{2y}’ = \frac{\omega r}{6}$

The speed of the center is $|v| = \sqrt{v_x’^2 + v_y’^2}$:

$$ |v_1| = |v_2| = \sqrt{(\frac{u}{2})^2 + (\frac{\omega r}{6})^2} = \frac{1}{2} \sqrt{u^2 + \frac{\omega^2 r^2}{9}} = \frac{1}{2} \sqrt{u^2 + \left(\frac{\omega r}{3}\right)^2} $$