A clcan body o[ mass 100 g starts with a velocity of 2 m / s on a smooth horizontal planc, acc

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A clcan body o[ mass 100 g starts with a velocity of 2 m /...

A clcan body o[ mass $100 \mathrm{~g}$ starts with a velocity of $2 \mathrm{~m} / \mathrm{s}$ on a smooth horizontal planc, accumulating dust at the rate of $5 \mathrm{~g} / \mathrm{s}$. Find the velocity at the end of 20 scconds and the distanec travclled during that period. Solution Herc $\frac{d}{d t}(m v)-0 . \quad$ So, $\quad m \frac{d v}{d t}+v \frac{d m}{d t}-0 .$ After $t$ seconds, mass of the body $=100+5 t$ So, $(100 \mid 5 \mathrm{t}) \frac{d v}{d t}+5 v-0 \quad$ or $\quad \frac{d v}{v}-\frac{5 d t}{100+5 t}$ Integrating $[\ln v]_{200}^{+}-[\ln (10015 t)]_{0}^{30}$ $\therefore \quad \ln \frac{v}{200}=-\ln 200+\ln 100=\ln \frac{100}{200} \therefore \quad \frac{v}{200}=\frac{1}{2} \quad \Rightarrow \quad v=100 \mathrm{~cm} / \mathrm{sec} .=\operatorname{lm} / \mathrm{sec}$ Nlso, $\ln v=-\ln (100+5 t)+\ln C$ When $t=0, \quad v=200 \mathrm{~cm} / \mathrm{sec}$ $\therefore \ln 200=-\ln 100+\ln C \therefore \quad \ln C=\ln 200+\ln 100=\ln (200 \times 100)$ $\therefore \quad \ln v=-\ln (100+5 t)+\ln (200 \times 100)$ $\therefore \quad v-\frac{200 \times 100}{10015 t}-\frac{d s}{d t} \quad \therefore \quad d s-\frac{200 \times 100 d t}{10015 t} \mathrm{~cm}-\frac{200 d t}{10015 t} \mathrm{~m}$ $\therefore \quad \int d s-s-\int_{0}^{20} \frac{200 d t}{10015 t}-40[\ln (100 \mid 5 t)]_{0}^{20}-40\left[\ln \frac{200}{100}\right]-40 \ln 2$ $=40 \times 0.3010 \times 2.303=27.7 \mathrm{~m}$

A clcan body o[ mass $100 \mathrm{~g}$ starts with a velocity of $2 \mathrm{~m} / \mathrm{s}$ on a smooth horizontal planc, accumulating dust at the rate of $5 \mathrm{~g} / \mathrm{s}$. Find the velocity at the end of 20 scconds and the distanec travclled during that period.
Solution Herc $\frac{d}{d t}(m v)-0 . \quad$ So, $\quad m \frac{d v}{d t}+v \frac{d m}{d t}-0 .$
After $t$ seconds, mass of the body $=100+5 t$ So, $(100 \mid 5 \mathrm{t}) \frac{d v}{d t}+5 v-0 \quad$ or $\quad \frac{d v}{v}-\frac{5 d t}{100+5 t}$
Integrating $[\ln v]_{200}^{+}-[\ln (10015 t)]_{0}^{30}$
$\therefore \quad \ln \frac{v}{200}=-\ln 200+\ln 100=\ln \frac{100}{200} \therefore \quad \frac{v}{200}=\frac{1}{2} \quad \Rightarrow \quad v=100 \mathrm{~cm} / \mathrm{sec} .=\operatorname{lm} / \mathrm{sec}$
Nlso, $\ln v=-\ln (100+5 t)+\ln C$
When $t=0, \quad v=200 \mathrm{~cm} / \mathrm{sec}$
$\therefore \ln 200=-\ln 100+\ln C \therefore \quad \ln C=\ln 200+\ln 100=\ln (200 \times 100)$
$\therefore \quad \ln v=-\ln (100+5 t)+\ln (200 \times 100)$
$\therefore \quad v-\frac{200 \times 100}{10015 t}-\frac{d s}{d t} \quad \therefore \quad d s-\frac{200 \times 100 d t}{10015 t} \mathrm{~cm}-\frac{200 d t}{10015 t} \mathrm{~m}$
$\therefore \quad \int d s-s-\int_{0}^{20} \frac{200 d t}{10015 t}-40[\ln (100 \mid 5 t)]_{0}^{20}-40\left[\ln \frac{200}{100}\right]-40 \ln 2$
$=40 \times 0.3010 \times 2.303=27.7 \mathrm{~m}$

Key Concepts

Variable Mass Systems

Variable mass systems are those in which the mass of the object changes over time due to processes such as accumulation or expulsion of material. In these systems, the standard form of Newton's second law must be modified to account for the mass change, often leading to additional terms in the dynamic equations. This concept is crucial for understanding phenomena like rocket propulsion, dust accumulation, or any problem where the mass is time-dependent.

Conservation of Momentum in Variable Mass Systems

In problems involving variable mass systems, the conservation of momentum must be applied carefully. The change in momentum is not solely due to the velocity change but also includes the contribution from the changing mass. This leads to an equation featuring both the time derivative of velocity and the derivative of mass, typically expressed as m(dv/dt) + v(dm/dt) when no external forces act on the system.

Differential Equations in Dynamics

When analyzing systems with changing mass or non-constant forces, the dynamics are often described by differential equations. These equations model the relationship between the rate of change of physical quantities and their current values. In the context of variable mass systems, the differential equation incorporates both mass and velocity as functions of time, requiring techniques like separation of variables for solution.

Separation of Variables

Separation of variables is a mathematical technique used to solve first-order differential equations by isolating the variables on different sides of the equation. This method is particularly useful in dynamics problems where the equation governing the motion can be rearranged into a form that enables direct integration. It simplifies the process of finding an explicit expression for the variable of interest, such as velocity as a function of time.

Application of Initial Conditions

The use of initial conditions is fundamental in solving differential equations that emerge in physical problems. Initial conditions provide specific values of variables at a starting time, allowing the determination of constants of integration after the general solution is found. This step ensures that the resulting model accurately represents the particular scenario under consideration, such as the initial mass and velocity in a variable mass problem.

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