Question

Two blocks of mass $m_1$ and $m_2$ $(m_1<m_2)$ are connected with an ideal spring on a smooth horizontal surface as shown in figure. At $t=0$ $m_1$ is at rest and $m_2$ is given a velocity $v$ towards right. At this moment, spring is in its natural length. Then choose the correct alternative:


(Assume all surfaces are smooth)

• A. Block of mass $m_2$ will be finally at rest after some time
• B. Block of mass $m_2$ will never come to rest
• C. Both the blocks will be finally at rest.
• D. None of these

Answer 1

The correct option is B Block of mass $m_2$ will never come to rest

Case (i)

Let velocity of $m_2$ is zero, let velocity of $m_1$ at this moment be $v^{{}^{\prime }}$
By momentum conservation
$m_1{v}^{{}^{\prime }}=m_{2}V$
$v^{{}^{\prime }}=\frac{m_{2}V}{m_1}$
Now, Final kinetic energy of $m_1$
$=\frac{1}{2}{m}^1{v}^{{{}^{\prime }}^2}=\frac{1}{2}{m}_1{\left(\frac{m_2}{m_1}\right)}^2{V}^2=\frac{1}{2}\left(\frac{m_2}{m_1}\right)m_2{V}^2=\left(\frac{m_2}{m_1}\right)\frac{1}{2}{m}_2{V}^2$
Final kinetic energy of $m_1>$ initial mechanical energy of system

Hence, case (i) is not possible.
$b$ is correct.

Case(ii)

Both blocks are at rest.

If both blocks are at rest final momentum of system is $0$, while initial momentum is $m_{2}v$.

As there is no external force, conservation of linear momentum has to be obeyed.
Case (ii) is not possible.
$c$ is incorrect