Name: Variable Mass System
Uploaded: 2019-07-19
Duration: 12 min 25 s
Description: In this video, we will look at how the laws of motion and their mathematical formulation apply to a variable mass system like a rocket. Don't miss it.

Question

Two blocks initially at rest having masses $m_{1} and m_{2}$ are connected by spring of spring constant $k = \frac{2}{3} N / m$ (as shown in the figure). The block of mass $m_{1}$ is pulled by a constant force $F_{1} = 4 N$ and the other block is pulled by a constant force $F_{2} = 2 N$ Find the maximum elongation of the spring (the spring is initially relaxed) Assuming $m_{2} = 2 m_{1}$

Accepted Answer

Given,
Mass of the block is related as, $m_{2} = 2 m_{1}$
Spring constant is, $k = \frac{2}{3} N m^{- 1}$
Force on block $1$ is, $F_{1} = 4 N$
Force on block $F_{2} = 2 N$ is, $F_{2} = 2 N$

Acceleration of system is given by, $a = \frac{F_{1} - F_{2}}{m_{1} + m_{2}}$
$\Rightarrow a = \frac{4 - 2}{3 m_{1}} = \frac{2}{3 m_{1}}$

We will do the problem in centre of mass frame, in this frame as both block stops simultaneously then at that instant spring is maximum elongated.
Let us assume, $x_{1} & x_{2}$ are the maximum elongation of block $1 & 2$ respectively. The case is shown in below figure,

In centre of mass frame
Force acting on block $1$ is,
$F_{1}' = F_{1} - m_{1} a$
$\Rightarrow F_{1}' = F_{1} - m_{1} \times \frac{2}{3 m_{1}}$
$\Rightarrow F_{1}' = 4 - \frac{2}{3} = \frac{10}{3} N$

Force on block $2$ is,
$F_{2}' = F_{2} + m_{2} a$
$\Rightarrow F_{2}' = 2 + 2 m_{1} \times \frac{2}{3 m_{1}}$
$\Rightarrow F_{2}' = 2 + \frac{4}{3} = \frac{10}{3} N$

Now from work energy theorem from COM frame, we have
$\Rightarrow F_{1}' x_{1} + F_{2}' x_{2} = \frac{1}{2} k {(x_{1} + x_{2})}^{2}$
$\Rightarrow \frac{10}{3} \times (x_{1} + x_{2}) = \frac{1}{2} \times \frac{2}{3} {(x_{1} + x_{2})}^{2}$
$\Rightarrow (x_{1} + x_{2}) = 10 m$ , so maximum elongation is given by, $x_{m a x} = x_{1} + x_{2} = 10 m$

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