Centre of Mass 4 Question 8
9. A cylindrical solid of mass $10^{-2} kg$ and cross-sectional area $10^{-4} m^{2}$ is moving parallel to its axis (the $x$-axis) with a uniform speed of $10^{3} m / s$ in the positive direction. At $t=0$, its front face passes the plane $x=0$. The region to the right of this plane is filled with stationary dust particles of uniform density $10^{-3} kg / m^{3}$. When a dust particle collides with the face of the cylinder, it sticks to its surface. Assuming that the dimensions of the cylinder remain practically unchanged and that the dust sticks only to the front face of the cylinder find the $x$-coordinate of the front of the cylinder at $t=150 s$.
$(1998,5$ M)
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Answer:
Correct Answer: 9. $10^{5} m$
Solution:
- Given, $m _0=10^{-2} kg, A=10^{-4} m^{2}, v _0=10^{3} m / s$
and $\rho _{\text {dust }}=\rho=10^{-3} kg / m^{3}$.
$$ \begin{aligned} m & =m _0+\text { mass of dust collected so far } \\ & =m _0+A x \rho _{dust} \\ m & =m _0+A x \rho \end{aligned} $$
The linear momentum at $t=0$ is
$$ \begin{gathered} p _0=m _0 v _0 \\ \text { and momentum at } t=t \text { is } \\ p _t=m v=\left(m _0+A x \rho\right) v \end{gathered} $$
From law of conservation of momentum
$$ \begin{array}{lc} & p _0=p _t \\ \therefore & m _0 v _0=\left(m _0+A x \rho\right) v \\ \therefore & m _0 v _0=\left(m _0+A x \rho\right) \frac{d x}{d t} \\ \text { or } & \left(m _0+A \rho x\right) d x=m _0 v _0 d t \\ \text { or } & \int _0^{x}\left(m _0+A \rho x\right) d x=\int _0^{150} m _0 v _0 d t \\ \Rightarrow & m _0 x+A \rho \frac{x^{2}}{2}=\left(m _0 v _0 t\right) _0^{150} \\ \text { Hence, } & m _0 x+A \rho \frac{x^{2}}{2}=150 m _0 v _0 \end{array} $$
Solving this quadratic equation and substituting the values of $m _0 A$, $\rho$ and $v _0$, we get positive value of $x$ as $10^{5} m$. Therefore, $x=10^{5} m$.
