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A bullet enters in a piece of wood with velocity $v_0$ and the resistive force acting on the bullet in the wood is proportional to $v^{\frac{1}{3}}$. If the total distance travelled by the bullet is proportional to $\left(v_0\right)^\beta$, then the value of $\beta$ is
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The correct answer is:
$\frac{5}{3}$
Resistive force $=-k v^{\frac{1}{3}}$
So, retardation of bullet is
$$
\begin{gathered}
a=\frac{d v}{d t}=-\frac{k}{m} v^{\frac{1}{3}} \Rightarrow \frac{d v}{d x} \cdot \frac{d x}{d t}=-\frac{k}{m} v^{\frac{1}{3}} \\
\Rightarrow \frac{d v}{d x} \cdot v=-\frac{k}{m} \cdot v^{\frac{1}{3}} \Rightarrow v^{\frac{2}{3}} d v=-\frac{k}{m} d x
\end{gathered}
$$
Integrating with limits, we get $(s=$ total displacement).
$$
\int_{v_0}^0 v^{\frac{2}{3}} d v=-\frac{k}{m} \int_0^s d x
$$
$$
\begin{aligned}
& \quad\left[\frac{3}{5} v^{\frac{5}{3}}\right]_{v_0}^0=-\frac{k}{m}[x]_0^s \Rightarrow-\frac{3}{5} v_0^{\frac{5}{3}}=-\frac{k \cdot s}{m} \\
& \text { or } \quad s \propto v^{\frac{5}{3}}
\end{aligned}
$$
So, retardation of bullet is
$$
\begin{gathered}
a=\frac{d v}{d t}=-\frac{k}{m} v^{\frac{1}{3}} \Rightarrow \frac{d v}{d x} \cdot \frac{d x}{d t}=-\frac{k}{m} v^{\frac{1}{3}} \\
\Rightarrow \frac{d v}{d x} \cdot v=-\frac{k}{m} \cdot v^{\frac{1}{3}} \Rightarrow v^{\frac{2}{3}} d v=-\frac{k}{m} d x
\end{gathered}
$$
Integrating with limits, we get $(s=$ total displacement).
$$
\int_{v_0}^0 v^{\frac{2}{3}} d v=-\frac{k}{m} \int_0^s d x
$$
$$
\begin{aligned}
& \quad\left[\frac{3}{5} v^{\frac{5}{3}}\right]_{v_0}^0=-\frac{k}{m}[x]_0^s \Rightarrow-\frac{3}{5} v_0^{\frac{5}{3}}=-\frac{k \cdot s}{m} \\
& \text { or } \quad s \propto v^{\frac{5}{3}}
\end{aligned}
$$
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