Question

Mass ${m_1}$ hits and sticks with ${m_2}$ while sliding horizontally with velocity $v$ along the common line of centres of three equal masses $\left( {{m_1} = {m_2} = {m_3} = m} \right)$. Initially masses ${m_2}$ and ${m_3}$ are stationary and the spring is unstretched. Minimum kinetic energy of ${m_2}$ is $\dfrac{{ym{v^2}}}{{36}}$. Find $y$.

Answer 1

Hint
In this problem, there are two types of energy. First the kinetic energy of the first mass transferred to the second mass. So, the kinetic energy of the two masses gives the potential energy to the spring, so by equating the kinetic energy of the two masses with the potential energy of the spring, the solution can be determined.
The kinetic energy is given by,
$\Rightarrow KE = \dfrac{1}{2}m{v^2}$
Where, $KE$ is the kinetic energy, $m$ is the mass of the block and $v$ is the velocity of the block.
The potential energy of the spring is,
$\Rightarrow PE = \dfrac{1}{2}k{x^2}$
Where, $PE$ is the potential energy of the spring, $k$ is the spring constant and $x$ is the compression or expansion of the spring.

Complete step by step answer
Given that, The mass ${m_1}$ hits the mass ${m_2}$, so that the velocity of the mass ${m_1}$ is reduced by half.
Three masses are equal $\left( {{m_1} = {m_2} = {m_3} = m} \right)$.
When the mass ${m_1}$ hits the mass ${m_2}$ and the mass ${m_2}$ compresses the spring, then the kinetic energy of the two masses is equal to the potential energy of the spring. Then,
$\Rightarrow \dfrac{1}{2}\left( {2m} \right){\left( {\dfrac{v}{2}} \right)^2} = \dfrac{1}{2}k{x^2}$
Both the masses are equal so $2m$ and the velocity is reduced by half when it hits the second mass so $\dfrac{v}{2}$.
From the above equation,
$\Rightarrow \dfrac{1}{2}\left( {2m} \right)\dfrac{{{v^2}}}{4} \times \dfrac{2}{k} = {x^2}$
By cancelling the terms, then
$\Rightarrow \dfrac{{m{v^2}}}{{2k}} = {x^2}$
The above equation is written as,
$\Rightarrow x = \sqrt {\dfrac{{m{v^2}}}{{2k}}} $
Since, the compression of the spring is $\dfrac{2}{3}$ times of maximum of ${m_3}$ from the centre of mass, then the kinetic energy of the ${m_3}$ gets from the potential energy of the spring, then
Kinetic energy of ${m_3}$ is $ \Rightarrow \dfrac{1}{2}k{\left( {\dfrac{2}{3}x} \right)^2}$
Substituting the value of $x$ in the above equation, then
$\Rightarrow \dfrac{1}{2}k{\left( {\dfrac{2}{3} \times \left( {\sqrt {\dfrac{{m{v^2}}}{{2k}}} } \right)} \right)^2}$
By squaring the terms, then
$\Rightarrow \dfrac{1}{2} \times k \times \dfrac{4}{9} \times \dfrac{{m{v^2}}}{{2k}}$
By cancelling the terms, then
$\Rightarrow \dfrac{{m{v^2}}}{9}$
Here multiplying and dividing by $4$, then
$\Rightarrow \dfrac{{m{v^2}}}{9} \times \dfrac{4}{4}$
On multiplying the above equation, then
$\Rightarrow \dfrac{{4m{v^2}}}{{36}}$
By comparing the term given in the question $\dfrac{{ym{v^2}}}{{36}}$, then the value of $y$ is, $y = 4$.

Note
The energy can neither be created nor destroyed. The kinetic energy of the first mass gives the kinetic energy to the second mass, and the second mass gives the potential energy to the spring. That potential energy of the spring gives again kinetic energy to the third mass. Thus, the energy gets transferred but not created.