A bullet of mass m passing through a wooden plank of…

A bullet of mass $m$ passing through a wooden plank of thickness $D = 5\ \mathrm{cm}$ slows from $v_0 = 350\ \mathrm{m/s}$ to $v_1 = 250\ \mathrm{m/s}$. The resistance force is proportional to the square of the speed, $F = -kv^2$. Find (1) the bullet's speed as a function of the distance travelled in the plank, (2) the time it takes to pass through.

Explanation of Method

A bullet slowed by a plank with drag ∝ v².

Step 1: Speed against distance

\[ m v\frac{dv}{dx} = -kv^2, \]

\[ v = v_0\,e^{-kx/m}. \]

From $v(D) = v_1$: $\dfrac{k}{m} = \dfrac{1}{D}\ln\dfrac{v_0}{v_1} = \dfrac{\ln 1.4}{0.05\ \mathrm{m}} \approx 6.73\ \mathrm{m^{-1}}$.

\[v(x) = v_0\left(\frac{v_1}{v_0}\right)^{x/D} \approx 350\,e^{-6.73\,x}\]

(in m/s, with $x$ in metres)

Step 2: Time in the plank

\[ t = \int_0^D \frac{dx}{v} = \frac{m}{k v_0}\left(e^{kD/m} - 1\right), \]

\[ t = \frac{m}{k}\left(\frac{1}{v_1} - \frac{1}{v_0}\right). \]

\[ t = \frac{D\,(v_0 - v_1)}{v_0 v_1 \ln(v_0/v_1)} = \frac{0.05 \cdot 100}{87\,500 \cdot 0.3365}. \]

\[t \approx 1.70 \cdot 10^{-4}\ \mathrm{s} = 0.170\ \mathrm{ms}\]

Check

Average speed $ D/t = 0.05/1.70 \cdot 10^{-4} \approx 294\ \mathrm{m/s} $, between 250 and 350 m/s $\checkmark$

$v\ (\mathrm{m/s})$ $x\ (\mathrm{cm})$ $0$ $1$ $2$ $3$ $4$ $5$ $250$ $350$ $v = 350\,e^{-6.73x}$ $\text{plank, } D = 5\ \mathrm{cm}$ $t \approx 0.17\ \mathrm{ms}$

Speed inside the plank (with $x$ in metres in the formula). With a drag force proportional to $v^2$, the speed falls by the same factor over every equal distance.

Result

\[\boxed{\,v(x) = v_0\left(\frac{v_1}{v_0}\right)^{x/D} \approx 350\,e^{-6.73\,x}\,}\]

\[\boxed{\,t \approx 1.70 \cdot 10^{-4}\ \mathrm{s} = 0.170\ \mathrm{ms}\,}\]

Remark

About the original solution. Part (1) is correct. In (2) it used $e^{kD/m} = v_1/v_0$ instead of $v_0/v_1$ when simplifying, which gives $t = \tfrac{D}{v_0}\,\tfrac{1 - v_1/v_0}{\ln(v_0/v_1)} \approx 0.121\ \mathrm{ms}$. That would mean an average speed of 412 m/s, faster than the bullet's entry speed. The correct time is about 0.170 ms.