A bullet crosses a plank and loses \( \left( \frac{1}{10} \right)^{\text{th}} \) part of its initial velocity. Calculate the minimum number of required planks such that the bullet cannot cross the last plank.

1. Why use the $v^2$ formula?

When a constant force opposes motion over a fixed distance $d$ (plank thickness), energy ideas or the kinematic equation

$v^2 = u^2 - 2ad$

work perfectly, because $a$ (retardation) is constant inside the plank.

2. Extracting the retardation value from the first plank

Let the bullet start with speed $u$. After one plank it emerges with $v_1 = 0.9u$ (lost $\tfrac1{10}$ of its speed).

Plug into the formula for one plank of thickness $d$:

$v_1^2 = u^2 - 2ad$ $\bigl(0.9u\bigr)^2 = u^2 - 2ad$ 0.81u^2 = u^2 - 2ad\;\Longrightarrow\;2ad = 0.19u^2\tag{1}

3. Repeat plank by plank (same $a$, same $d$)

Each new plank removes the same kinetic‐energy chunk $2ad = 0.19u^2$ from the square of the speed.

Let $v_n$ be the speed after the $n$-th plank. Then

$v_{n}^2 = v_{n-1}^2 - 0.19u^2$

Compute successively:

1. $v_1^2 = 0.81u^2$
2. $v_2^2 = 0.81u^2 - 0.19u^2 = 0.62u^2$
3. $v_3^2 = 0.62u^2 - 0.19u^2 = 0.43u^2$
4. $v_4^2 = 0.43u^2 - 0.19u^2 = 0.24u^2$
5. $v_5^2 = 0.24u^2 - 0.19u^2 = 0.05u^2$
6. $v_6^2 = 0.05u^2 - 0.19u^2 = -0.14u^2 < 0$

A negative value means the bullet cannot even finish the 6-th plank: it stops inside it.

4. Conclusion

Therefore, minimum 6 planks are needed so that the bullet fails to come out of the last one.

What is happening?

Imagine you run through a long row of cardboard walls. Every time you pass one wall, you slow down because it pushes against you. The bullet is like you, and the wooden planks are the cardboard walls.

Key idea in baby‐steps

1. Rule of the game: After crossing one plank, the bullet keeps only 9 out of every 10 centimetres of speed it had before that plank.
2. Same push in every plank: Each plank gives the same push (same force and same thickness), so the slowing effect is the same each time.
3. Count planks until it stops: We just keep checking, plank by plank, how the speed drops. The moment the math says the speed becomes zero inside a plank, that plank is the last one it will enter but never come out of.

Tiny table

Let the bullet enter the first plank with speed $u$.

Distance inside one plank = $d$, retardation = $a$ (assumed constant in every plank).

For the first plank:

$v_1^2 = u^2 - 2ad\qquad(\text{with }v_1 = 0.9u)$

0.81u^2 = u^2 - 2ad\;\Longrightarrow\;2ad = 0.19u^2\tag{A}

Thus each plank reduces the term $v^2$ by $0.19u^2$.

Now proceed plank by plank.

1. After first plank: $v_1^2 = 0.81u^2$

2. After second plank: $v_2^2 = v_1^2 - 0.19u^2 = 0.81u^2 - 0.19u^2 = 0.62u^2$

3. After third plank: $v_3^2 = 0.62u^2 - 0.19u^2 = 0.43u^2$

4. After fourth plank: $v_4^2 = 0.43u^2 - 0.19u^2 = 0.24u^2$

5. After fifth plank: $v_5^2 = 0.24u^2 - 0.19u^2 = 0.05u^2$

6. Attempting sixth plank: $v_6^2 = 0.05u^2 - 0.19u^2 = -0.14u^2 < 0$

Negative $v^2$ is impossible; it means the bullet comes to rest within the 6-th plank.

Hence, the minimum number of identical planks required = 6.

[\boxed{6}]