1. Interaction with First Cube

System: Bullet ($m=12\text{ g} = 0.012\text{ kg}$) and Cube 1 ($M=240\text{ g} = 0.24\text{ kg}$).
Event: Bullet passes through Cube 1. $u = 180\text{ m/s}$. Emerging speed = $0.5u = 90\text{ m/s}$.

Using Conservation of Momentum for the first collision:

$$m u = M V_1 + m (0.5u)$$ $$V_1 = \frac{0.5 m u}{M} = \frac{0.5 \times 0.012 \times 180}{0.240}$$ $$V_1 = \frac{1.08}{0.24} = 4.5 \text{ m/s}$$

Cube 1 starts moving at $4.5\text{ m/s}$ immediately after the bullet emerges.

2. Motion between Cubes

The bullet travels the distance $d = 90\text{ cm} = 0.9\text{ m}$ to reach the second cube. Neglecting interaction time means the bullet instantly emerges and starts traveling the gap.

Time for bullet to reach 2nd Cube ($t_1$):

$$t_1 = \frac{d}{v_{bullet\_emerge}} = \frac{0.9}{90} = 0.01 \text{ s}$$

During this time $t_1$, Cube 1 travels a distance $x_1$:

$$x_1 = V_1 \times t_1 = 4.5 \times 0.01 = 0.045 \text{ m}$$

At $t=t_1$, the separation between cubes is reduced to $d’ = d – x_1 = 0.9 – 0.045 = 0.855 \text{ m}$.

3. Interaction with Second Cube

System: Bullet ($m$) and Cube 2 ($M$).
Event: Bullet hits Cube 2 and gets embedded.

Conservation of Momentum:

$$m (0.5u) = (M + m) V_2$$ $$V_2 = \frac{0.012 \times 90}{0.240 + 0.012} = \frac{1.08}{0.252}$$ $$V_2 = \frac{1080}{252} = \frac{30}{7} \approx 4.286 \text{ m/s}$$

4. Collision of Cubes

Now, both cubes are moving to the right on a frictionless floor.
Cube 1 is at $x = 0.045\text{ m}$ with velocity $V_1 = 4.5\text{ m/s}$.
Cube 2 is at $x = 0.9\text{ m}$ with velocity $V_2 = \frac{30}{7}\text{ m/s}$.

Relative Velocity:

$$V_{rel} = V_1 – V_2 = 4.5 – \frac{30}{7} = \frac{9}{2} – \frac{30}{7} = \frac{63 – 60}{14} = \frac{3}{14} \text{ m/s}$$

Time to collide ($t_2$):

$$t_2 = \frac{\text{Separation}}{V_{rel}} = \frac{0.855}{3/14} = 0.855 \times \frac{14}{3}$$ $$t_2 = 0.285 \times 14 = 3.99 \text{ s}$$

5. Total Time

The question asks “How long after the bullet emerges from the first cube…”. This is the total time $T = t_1 + t_2$.

$$T = 0.01 + 3.99 = 4.00 \text{ s}$$