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 block of masses $m_{1} and m_{2}$ are connected with the help of a spring of spring constant $k$ initially the spring in its natural length as shown. A sharp impulse is given to mass $m_{2}$ so that it acquires a velocity $v_{0}$ towards right. If the system is kept on smooth floor then find

(a) the velocity of the centre of mass,
(b) the maximum elongation that the spring will suffer?

Accepted Answer

Given:

The masses are $m_{1}$ and $m_{2}$ .

Spring constant $= k$ .

The initial velocity of $m_{1}$ is zero.

We know that the velocity of the centre of mass is given by, $v_{cm} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

So, $v_{cm} = \frac{m_{1} \times 0 + m_{2} \times v_{0}}{m_{1} + m_{2}}$ $= \frac{m_{2} v_{0}}{m_{1} + m_{2}}$

Let the maximum elongation be $z$ .

So, when the system will be elongated maximum, the bodies will move with the velocity of centre of mass, $v_{cm}$ .

So, by conservation of energy,

$\frac{1}{2} m_{2} v_{0}^{2} = \frac{1}{2} (m_{1} + m_{2}) v_{cm}^{2} + \frac{1}{2} k z^{2}$

$\Rightarrow \frac{1}{2} m_{2} v_{0}^{2} = \frac{1}{2} (m_{1} + m_{2}) {(\frac{m_{2} v_{0}}{m_{1} + m_{2}})}^{2} + \frac{1}{2} k z^{2}$

$\Rightarrow m_{2} v_{0}^{2} = \frac{{(m_{2} v_{0})}^{2}}{m_{1} + m_{2}} + k z^{2}$

$\Rightarrow (m_{1} + m_{2}) k z^{2} = (m_{1} + m_{2}) m_{2} v_{0}^{2} - (m_{2} v_{0})^{2}$

$\Rightarrow (m_{1} + m_{2}) k z^{2} = m_{1} m_{2} v_{0}^{2} + (m_{2} v_{0})^{2} - (m_{2} v_{0})^{2}$

$\Rightarrow k z^{2} (m_{1} + m_{2}) = m_{1} m_{2} v_{0}^{2}$

So, $z = \sqrt{\frac{m_{1} m_{2} v_{0}^{2}}{(m_{1} + m_{2}) k}} = v_{0} \sqrt{\frac{m_{1} m_{2}}{(m_{1} + m_{2}) k}}$

Hence, the maximum elongation that the spring suffers, $v_{0} \sqrt{\frac{m_{1} m_{2}}{(m_{1} + m_{2}) k}}$

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