A bullet enters a wooden block with velocity | vedclass

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(B) Given, resistance force $F \propto \sqrt{v}$, so $ma = -k\sqrt{v}$, which implies $a = -c\sqrt{v}$ where $c$ is a constant.
We know $a = \frac{dv

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) Given, resistance force $F \propto \sqrt{v}$, so $ma = -k\sqrt{v}$, which implies $a = -c\sqrt{v}$ where $c$ is a constant.
We know $a = \frac{dv}{dt} = -c\sqrt{v}$.
Rearranging, $\frac{dv}{\sqrt{v}} = -c \, dt$.
Integrating from $t=0$ $(v=u)$ to $t=T$ $(v=0)$:
$\int_{u}^{0} v^{-1/2} \, dv = \int_{0}^{T} -c \, dt \Rightarrow [2\sqrt{v}]_{u}^{0} = -cT \Rightarrow -2\sqrt{u} = -cT \Rightarrow c = \frac{2\sqrt{u}}{T}$.
Now, $v \frac{dv}{ds} = -c\sqrt{v} \Rightarrow \sqrt{v} \, dv = -c \, ds$.
Integrating from $s=0$ $(v=u)$ to $s=S$ $(v=0)$:
$\int_{u}^{0} v^{1/2} \, dv = \int_{0}^{S} -c \, ds \Rightarrow [\frac{2}{3} v^{3/2}]_{u}^{0} = -cS \Rightarrow -\frac{2}{3} u^{3/2} = -cS \Rightarrow S = \frac{2u^{3/2}}{3c}$.
Substituting $c = \frac{2\sqrt{u}}{T}$:
$S = \frac{2u^{3/2}}{3(2\sqrt{u}/T)} = \frac{uT}{3}$.
Given $u = 120 \,m/s$ and $T = 1.5 \,s$:
$S = \frac{120 	imes 1.5}{3} = \frac{180}{3} = 60 \,m$.
Since $60 \,m$ is not in the options, the provided options are incorrect.

$A$ bullet enters a wooden block with velocity $120 \,m/s$. The bullet travels $1.5 \,s$ in the block before its velocity reduces to zero due to a resistance force which is proportional to the square root of the velocity. The distance travelled by the bullet in the wooden block is (in $\,m$)

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