1. Initial setup

A block of mass $m$ moves with speed $v$ and hits a stationary block of mass $m$. All collisions are perfectly inelastic, so after each collision the colliding bodies stick together.

We need the total kinetic energy lost by the time the last block of mass $8m$ starts moving.

The masses in line are: $m,\ 2m,\ 4m,\ 8m$ with an incoming moving mass $m$.

So the sequence of sticking is: $m \to (m+m) \to (2m+2m) \to (4m+4m) \to (8m+8m)$

2. Initial kinetic energy

Only the incoming block of mass $m$ is moving initially.

$K_i = \frac{1}{2}mv^2$

3. First collision: $m$ hits stationary $m$

Using conservation of momentum: $mv = (2m)v_1$ $v_1 = \frac{v}{2}$

Kinetic energy after first collision: $K_1 = \frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 = \frac{mv^2}{4}$

4. Second collision: mass $2m$ moving at $v/2$ hits stationary $2m$

Momentum conservation: $(2m)\cdot \frac{v}{2} = (4m)v_2$ $mv = 4mv_2$ $v_2 = \frac{v}{4}$

Kinetic energy after second collision: $K_2 = \frac{1}{2}(4m)\left(\frac{v}{4}\right)^2 = \frac{mv^2}{8}$

5. Third collision: mass $4m$ moving at $v/4$ hits stationary $4m$

Momentum conservation: $(4m)\cdot \frac{v}{4} = (8m)v_3$ $mv = 8mv_3$ $v_3 = \frac{v}{8}$

Kinetic energy after third collision: $K_3 = \frac{1}{2}(8m)\left(\frac{v}{8}\right)^2 = \frac{mv^2}{16}$

6. Fourth collision: mass $8m$ moving at $v/8$ hits stationary $8m$

Momentum conservation: $(8m)\cdot \frac{v}{8} = (16m)v_4$ $mv = 16mv_4$ $v_4 = \frac{v}{16}$

This is the moment when the last block of mass $8m$ has started moving.

Kinetic energy after this collision: $K_f = \frac{1}{2}(16m)\left(\frac{v}{16}\right)^2 = \frac{mv^2}{32}$

7. Total energy loss

Initial energy: $K_i = \frac{1}{2}mv^2$

Final energy: $K_f = \frac{mv^2}{32}$

Energy lost: $\Delta K = K_i - K_f = \frac{1}{2}mv^2 - \frac{mv^2}{32}$

$\Delta K = \frac{16mv^2 - mv^2}{32} = \frac{15mv^2}{32}$

Percentage loss: