Given: $M=m$, $L=0.4$ m, $u=10$ m/s, $k=0.5$ kg/s. Drag force $F=-kv$ acts on the box.

1. Motion Cycle

Because $M=m$ and collisions are elastic, the block and box simply exchange velocities upon impact.

• Step 1 (Block moves): Block travels distance $L$. Box is stationary (no drag). Velocity remains $v$.
• Collision: Velocity $v$ transfers to Box. Block stops.
• Step 2 (Box moves): Box travels distance $L$. Drag reduces velocity.

2. Velocity Loss per Cycle

When the box moves distance $L$ under drag $F = -kv$: $$ M \frac{dv}{dt} = -kv \implies M v \frac{dv}{dx} = -kv \implies dv = -\frac{k}{M} dx $$ $$ \Delta v = -\frac{k}{M} L $$ Substituting values: $\Delta v = – \frac{0.5}{0.49} (0.4) \approx -0.408 \text{ m/s}$.

3. Total Collisions

Number of full box-movements possible: $$ n = \frac{u}{|\Delta v|} = \frac{10}{0.408} \approx 24.5 $$ The system completes 24 full cycles (Block Move + Box Move).
Collisions so far: $24 \times 2 = 48$.
After 24 cycles, residual velocity exists. It transfers to the block (Collision 48 complete). Block moves remaining distance (Collision 49). Box moves partial distance and stops.
Total Collisions = 49.

4. Total Distance of Box

Using Impulse-Momentum theorem for the total retarding force on the box: $$ \int F dt = \Delta P_{system} $$ $$ \int k v dt = M u_{initial} $$ Since $\int v dt = X$ (total distance traveled by box): $$ k X = M u $$ $$ X = \frac{M u}{k} = \frac{0.49 \times 10}{0.5} = 9.8 \text{ m} $$