We have two identical cylinders. Cylinder 1 is moving with velocity $u$ and rolling without slipping ($\omega = u/r$). Cylinder 2 is at rest. Since the collision is elastic and the masses are identical ($m_1 = m_2$), the velocities of the centers of mass are exchanged instantly. However, because there is no friction between the cylinders, the angular velocity cannot change impulsively during the collision.

State just after collision ($t=0$):

• Cylinder 1: Velocity $v_1 = 0$, Angular Velocity $\omega_1 = u/r$ (keeps spinning).
• Cylinder 2: Velocity $v_2 = u$, Angular Velocity $\omega_2 = 0$ (slides without spin).

Cylinder 1: It has zero linear velocity but spins clockwise. The point of contact slips backward with speed $u$. Kinetic friction $f_1 = \mu mg$ acts forward.

• Linear Acceleration: $a_1 = \frac{f_1}{m} = \mu g$ (forward)
• Angular Deceleration: $\tau = f_1 r \Rightarrow I\alpha = \mu mgr \Rightarrow \frac{1}{2}mr^2 \alpha = \mu mgr \Rightarrow \alpha = \frac{2\mu g}{r}$ (opposing spin)

Cylinder 2: It moves forward with speed $u$ but has no spin. The point of contact slips forward. Kinetic friction $f_2 = \mu mg$ acts backward.

• Linear Deceleration: $a_2 = -\mu g$ (backward)
• Angular Acceleration: $\tau = f_2 r \Rightarrow \alpha = \frac{2\mu g}{r}$ (increasing spin)

Friction stops acting when pure rolling begins ($v = r\omega$).

For Cylinder 1:

$v_1(t) = 0 + a_1 t = \mu g t$
$\omega_1(t) = \frac{u}{r} – \alpha t = \frac{u}{r} – \frac{2\mu g}{r} t$
Condition $v_1 = r\omega_1$:
$\mu g t = r(\frac{u}{r} – \frac{2\mu g}{r} t) \Rightarrow \mu g t = u – 2\mu g t \Rightarrow 3\mu g t = u$
Time $t_1 = \frac{u}{3\mu g}$

For Cylinder 2:

$v_2(t) = u – \mu g t$
$\omega_2(t) = 0 + \frac{2\mu g}{r} t$
Condition $v_2 = r\omega_2$:
$u – \mu g t = 2\mu g t \Rightarrow 3\mu g t = u$
Time $t_2 = \frac{u}{3\mu g}$

Both cylinders achieve pure rolling at the same instant $t = \frac{u}{3\mu g}$. At this instant, friction vanishes for both, so their accelerations ($a_1$ and $a_2$) become zero. Thus, the relative acceleration vanishes.

We need the distance between them at $t = \frac{u}{3\mu g}$.

Displacement of Cylinder 2:

$$S_2 = u t – \frac{1}{2} (\mu g) t^2$$

Displacement of Cylinder 1:

$$S_1 = 0 t + \frac{1}{2} (\mu g) t^2$$

Separation $\Delta S$:

$$\Delta S = S_2 – S_1 = u t – \mu g t^2$$

Substitute $t = \frac{u}{3\mu g}$:

$$\Delta S = u \left( \frac{u}{3\mu g} \right) – \mu g \left( \frac{u}{3\mu g} \right)^2$$ $$\Delta S = \frac{u^2}{3\mu g} – \mu g \left( \frac{u^2}{9\mu^2 g^2} \right)$$ $$\Delta S = \frac{u^2}{3\mu g} – \frac{u^2}{9\mu g} = \frac{3u^2 – u^2}{9\mu g} = \frac{2u^2}{9\mu g}$$