5.0 Variable Mass

We have so far dealt with system having constant mass.

Now, let us consider the system in which the mass of the system is variable i.e. either the mass of the system increases or decreases.

A very common example is a cart filled with sand and the sand is either added to the cart or coming out of the cart.

Let us understand the concept of variable mass with few examples.

Due to the small hole in the cart, sand comes out of the cart with a velocity $\left( {{{\vec v}_r}} \right)$ relative to the cart.

Initial linear momentum of the system, $${\overrightarrow p _i} = mv\widehat i\quad ...(i)$$

At any time $t$, Let $dm$ mass of the sand comes out of the cart. So, the mass and velocity of the cart becomes $(m-dm)$ & $(v+dv)$ respectively.

${\overrightarrow v _C}$: Velocity of the cart

${\overrightarrow v _S}$: Velocity of the sand

${\vec v_{SC}}$: Velocity of the sand relative to the cart

So, $$\begin{equation} \begin{aligned} {\overrightarrow v _{SC}} = {\overrightarrow v _S} - {\overrightarrow v _C} \\ {\overrightarrow v _r} = {\overrightarrow v _S} - {\overrightarrow v _C} \\ {\overrightarrow v _S} = {\overrightarrow v _r} + {\overrightarrow v _C} \\\end{aligned} \end{equation} $$ or $${\overrightarrow v _S} = {\overrightarrow v _r} + (v + dv)\widehat i\quad ...(ii)$$

Linear momentum of the system at any time $t$, $$\begin{equation} \begin{aligned} {\overrightarrow p _f} = {\overrightarrow p _{cart}} + {\overrightarrow p _{sand}} \\ {\overrightarrow p _f} = (m - dm)(v + dv) + dm\left( {{{\overrightarrow v }_S}} \right) \\\end{aligned} \end{equation} $$ or $$\begin{equation} \begin{aligned} {\overrightarrow p _f} = (m - dm)(v + dv)\widehat i + dm\left[ {{{\overrightarrow v }_r} + (v + dv)\widehat i} \right] \\ {\overrightarrow p _f} = (mv + mdv - vdm - dm.\;dv + vdm + dm.\;dv)\widehat i + {\overrightarrow v _r}dm \\ {\overrightarrow p _f} = (mv + mdv)\widehat i + {\overrightarrow v _r}dm\quad ...(iii) \\\end{aligned} \end{equation} $$

From equation $(i)$ & $(iii)$ we get, $$\begin{equation} \begin{aligned} {\overrightarrow p _i} = {\overrightarrow p _f} \\ mv\widehat i = (mv + mdv)\widehat i + {\overrightarrow v _r}\;dm \\ mdv\widehat i = - {\overrightarrow v _r}\;dm \\\end{aligned} \end{equation} $$

Dividing both side by $dt$, we get, $$\begin{equation} \begin{aligned} m\frac{{dv}}{{dt}}\widehat i = - {\overrightarrow v _r}\frac{{dm}}{{dt}} \\ {\overrightarrow F _T} = - {\overrightarrow v _r}\frac{{dm}}{{dt}} \\\end{aligned} \end{equation} $$ or $${\overrightarrow F _T} = {\overrightarrow v _r}\left( { - \frac{{dm}}{{dt}}} \right)$$ where,

${\overrightarrow F _T}$: Thrust force

$\left( {\frac{{dm}}{{dt}}} \right)$: Rate at which the mass changes

The negative sign in $\left( { - \frac{{dm}}{{dt}}} \right)$ implies that the mass of the system is decreasing.

If ${\overrightarrow v _r}$ is in the backward direction, then the thrust force ${\overrightarrow F _T}$ acts in the forward direction and vice versa.

The above statement will be better understood with the help of the diagram as shown below.

Consider a cart partially filled with sand. The total mass of the care is $m$ and the cart is moving with velocity $v$. A vehicle contains sand is continuously adding sand to the cart. Sand is added to the cart with a velocity ${\overrightarrow v _r}$ relative to the cart.

Initial linear momentum of the system, $$\begin{equation} \begin{aligned} {\overrightarrow p _i} = {\overrightarrow p _{cart}} + {\overrightarrow p _{sand}} \\ {\overrightarrow p _i} = mv\widehat i + dm\left( {{{\overrightarrow v }_r}} \right)\quad ...(i) \\\end{aligned} \end{equation} $$

At any time $t$, $dm$ mass of the sand is added to the cart. So, the mass and velocity of the cart becomes $(m+dm)$ & $(v+dv)$ respectively.

${\overrightarrow v _C}$: Velocity of the cart

${\overrightarrow v _S}$: Velocity of the sand

${\vec v_{SC}}$: Velocity of the sand relative to the cart

So, $$\begin{equation} \begin{aligned} {\overrightarrow v _{SC}} = {\overrightarrow v _S} - {\overrightarrow v _C} \\ {\overrightarrow v _r} = {\overrightarrow v _S} - {\overrightarrow v _C} \\ {\overrightarrow v _S} = {\overrightarrow v _r} + {\overrightarrow v _C} \\\end{aligned} \end{equation} $$ or $${\overrightarrow v _S} = {\overrightarrow v _r} + (v + dv)\widehat i\quad ...(ii)$$

Linear momentum of the system at any time $t$, $$\begin{equation} \begin{aligned} {\overrightarrow p _f} = {\overrightarrow p _{cart}} \\ {\overrightarrow p _f} = (m + dm)(v + dv)\widehat i\quad ...(iii) \\\end{aligned} \end{equation} $$

From equation $(i)$, $(ii)$ & $(iii)$ we get, $$\begin{equation} \begin{aligned} {\overrightarrow p _i} = {\overrightarrow p _f} \\ mv\widehat i + dm\left( {{{\overrightarrow v }_r}} \right) = (m + dm)(v + dv)\widehat i \\\end{aligned} \end{equation} $$ or $$\begin{equation} \begin{aligned} mv\widehat i + dm\left[ {{{\overrightarrow v }_r} + \left( {v + dv} \right)\widehat i} \right] = (m + dm)(v + dv)\widehat i \\ mdv\widehat i = {\overrightarrow v _r}\;dm \\\end{aligned} \end{equation} $$

Dividing both side by $dt$ we get, $$\begin{equation} \begin{aligned} m\frac{{dv}}{{dt}}\widehat i = {\overrightarrow v _r}\frac{{dm}}{{dt}} \\ {\overrightarrow F _T} = {\overrightarrow v _r}\frac{{dm}}{{dt}} \\\end{aligned} \end{equation} $$ where,

${\overrightarrow F _T}$: Thrust force

$\left( {\frac{{dm}}{{dt}}} \right)$: Rate at which the mass changes

If ${\overrightarrow v _r}$ is in the forward direction, then the thrust force ${\overrightarrow F _T}$ also acts in the forward direction and vice versa.

The above statement will be better understood with the help of the diagram as shown below.

When the mass is removed from the system, $${\overrightarrow F _T} = {\overrightarrow v _r}\left( { - \frac{{dm}}{{dt}}} \right)$$

-ve sign indicates that the mass is removed from the system.

When the mass is added to the system, $${{\vec F}_T} = {{\vec v}_r}\left( {\frac{{dm}}{{dt}}} \right)$$

Find the thrust force experienced by the driver. Also find the direction in which the thrust force acts.

Solution: Velocity of the oil pouring out relative to the tanker $\left( {{{\overrightarrow v }_r}} \right) = - 5\widehat i\;m/s$

Rate of change of mass $\left( { - \frac{{dm}}{{dt}}} \right) = - 3kg/s$

Note: $-ve$ sign indicates that the mass of the system is decreasing.

As we know that the thrust ${\overrightarrow F _T}$ is, $$\begin{equation} \begin{aligned} {{\vec F}_T} = {{\vec v}_r}\left( { - \frac{{dm}}{{dt}}} \right) \\ {\overrightarrow F _T} = \left( { - 5\widehat i} \right)\left( { - 3} \right) \\ {\overrightarrow F _T} = 15\widehat i\;N \\\end{aligned} \end{equation} $$