Irodov 1.101: Trading Depth for Time in a Plank – Amit & Deepali Sinha Foundation

Problem Irodov 1.101

A bullet enters a plank of thickness $h$ with speed $v_0$ and emerges from the far side with a smaller speed $v$ . While inside the plank, the bullet is slowed by a resistance force proportional to the square of its instantaneous speed. Find how long the bullet spends inside the plank.

The bullet crossing the plank of thickness h: it enters with v_0 and leaves with v<v_0. At a general point x inside, the resistance F=bv^2 opposes the instantaneous velocity v(x). — The bullet crossing the plank of thickness $h$ : it enters with $\vect v_0$ and leaves with $\vect v<v_0$ . At a general point $x$ inside, the resistance $F=bv^2$ opposes the instantaneous velocity $\vect v(x)$ .

Given

$h$
$v_0$
$v$
$F=bv^2$ opposing the motion, $b$ unknown

Find

The time

t

the bullet spends inside the plank

Solution

Step 1Newton’s second law with a quadratic resistance

Let $b$ be the (unknown) proportionality constant of the resistance, so the plank exerts $F=bv^2$ on the bullet, directed opposite to its velocity. Along the line of motion, with $v$ the scalar speed,

m\frac{dv}{dt} = -bv^2 \tag{1}

Step 2Eliminating

b

using the crossing depth

Rewrite the acceleration with the chain rule, $dv/dt=v\,dv/dx$ , and substitute into the equation of motion (1):

mv\frac{dv}{dx} = -bv^2 \quad\Longrightarrow\quad \frac{dv}{v} = -\frac{b}{m}\,dx.

Integrating from the entry point $x=0$ , $v=v_0$ , to the exit point $x=h$ , $v=v$ ,

\int_{v_0}^{v}\frac{dv'}{v'} = -\frac{b}{m}\int_0^h dx' \quad\Longrightarrow\quad \ln\frac{v}{v_0} = -\frac{b}{m}h.

\dfrac{b}{m} = \dfrac{1}{h}\ln\dfrac{v_0}{v}

(2)

Step 3The time from the same equation of motion

Separate the equation of motion (1) in time instead of space and integrate from $t=0$ , $v=v_0$ , to the exit instant $t$ , $v=v$ :

\int_{v_0}^{v}\frac{dv'}{v'^2} = -\frac{b}{m}\int_0^{t} dt' \quad\Longrightarrow\quad \left(\frac{1}{v_0}-\frac{1}{v}\right) = -\frac{b}{m}t.

Solving for $t$ and writing the left side over a common denominator,

t = \frac{m}{b}\left(\frac{1}{v}-\frac{1}{v_0}\right) = \frac{m}{b}\cdot\frac{v_0-v}{v_0 v}.

Substituting $m/b$ from (2),

t = \frac{h}{\ln(v_0/v)}\cdot\frac{v_0-v}{v_0 v}.

t = \dfrac{h(v_0-v)}{v_0 v\,\ln(v_0/v)}

(3)

Check

Dimensionally, $h(v_0-v)/(v_0v)$ has units $\mathrm m\cdot(\mathrm{m/s})/(\mathrm{m/s})^2=\mathrm{m\cdot s/m}=\mathrm s$ , a time, and $\ln(v_0/v)$ is dimensionless, so the boxed result checks. As $v\to v_0$ (a very thin, or very low resistance, plank barely slowing the bullet), write $v=v_0(1-\varepsilon)$ with $\varepsilon\to0^+$ : then $v_0-v\to v_0\varepsilon$ and $\ln(v_0/v)\to\varepsilon$ , so $t\to h\varepsilon/(v_0\varepsilon)=h/v_0$ , the time for a bullet crossing at the unchanged speed $v_0$ , exactly as it must.

Remark

Writing the acceleration as $v\,dv/dx$ instead of $dv/dt$ turns the equation of motion (1) into one relating $v$ directly to position, with time absent altogether, so it can be integrated using only the crossing depth $h$ and the two speeds. That is what lets $h$ , $v_0$ and $v$ alone fix the ratio $b/m$ before the time integral is even touched, so the unknown resistance constant never has to be found on its own, only in the combination the time equation actually needs.