Question

Having gone through a plank of thickness \[h\] , the bullet changed its velocity from ${v_0}$ to $v$ . The time of motion of the bullet in the plank is $t = \dfrac{{xh\left( {{v_0} - v} \right)}}{{{v_0}v\ln \left( {\dfrac{{{v_0}}}{v}} \right)}}$ , assuming the resistance force to be proportional to the square of the velocity. Find $x$ .

Answer 1

HintFrom Newton’s second law of motion Force $F = ma = m\dfrac{{dv}}{{dt}}$ where $m$ is the mass of body, $a$ is the acceleration which can also be written as $\dfrac{{dv}}{{dt}}$ where $v$ is the velocity of body and $t$ is the time. $\dfrac{{dv}}{{dt}}$ can be further written as $v\dfrac{{dv}}{{ds}}$ .
The resistance force can be written as $F = - k{v^2}$ where $k$ is a constant.

 Complete step-by-step solution:As given in the question the resistance force to be proportional to the square of the velocity, so it can be written as $F = - k{v^2}$ where $k$ is a constant. Here –ve sign represents that the force is against the motion as resistance is applied against the direction of motion.
Now as we know from Newton’s second law of motion Force $F = ma = m\dfrac{{dv}}{{dt}}$ where $m$ is the mass of body, $a$ is the acceleration which can also be written as $\dfrac{{dv}}{{dt}}$ where $v$ is the velocity of body and $t$ is the time.
Now we equate both type of forces,
$m\dfrac{{dv}}{{dt}} = - k{v^2}$
On arranging we have,
$m\dfrac{{dv}}{{{v^2}}} = - kdt$
Now, as given in the question that the bullet changed its velocity from ${v_0}$ to $v$, so integrating within limits we have
$\int\limits_{{v_0}}^v {\dfrac{{dv}}{{{v^2}}} = - \dfrac{k}{m}} \int\limits_0^t {dt} $
On solving the integration and simplifying we have
$t = \dfrac{m}{k}\dfrac{{\left( {{v_0} - v} \right)}}{{{v_0}v}}$ ……(1)
Now we have to find the value of $k$ . For this we know that $\dfrac{{dv}}{{dt}}$ can be further written as $v\dfrac{{dv}}{{ds}}$ .
So, again equating the forces we have
$mv\dfrac{{dv}}{{ds}} = - k{v^2}$
On arranging we have
$\dfrac{{dv}}{v} = - \dfrac{k}{m}ds$
Now as given in the question that the bullet has gone through the plank whose thickness is \[h\].
So, on integrating within limits we have
$\int\limits_{{v_0}}^v {\dfrac{{dv}}{v}} = - \dfrac{k}{m}\int\limits_0^h {ds} $
On solving the integration and simplifying we have
$\ln \dfrac{v}{{{v_0}}} = - \dfrac{k}{m}h$
On further simplification we have $k = \dfrac{m}{h}\ln \dfrac{{{v_0}}}{v}$
Now substituting this value of k in equation (1) we have
$t = \dfrac{{h\left( {{v_0} - v} \right)}}{{{v_0}v\ln \dfrac{{{v_0}}}{v}}}$
Now comparing this expression of t with one given in the question i.e. $t = \dfrac{{xh\left( {{v_0} - v} \right)}}{{{v_0}v\ln \left( {\dfrac{{{v_0}}}{v}} \right)}}$ we have the value of $x$ .
Hence, $x = 1$.

Note: In the expression of resistance force $F = - k{v^2}$ , the - ve sign represents that the force is against the motion as resistance is applied against the direction of motion.
The force $F = ma$ can always be written as $F = m\dfrac{{dv}}{{dt}}$ and $F = mv\dfrac{{dv}}{{ds}}$ .