Question

In a two block system shown in figure match the following-

Answer 1

$m_1=1$; $m_2=2$;$v_1=5$;$v_2=10$;
Let $\mu$ be the coeeficient of frction between the blocks.
Total initial momentum=$\overrightarrow{p_i}=m_1{v}_1\hat{i}-m_2{v}_2\hat{i}$;
$\therefore \overrightarrow{p_i}=1\ast 10-2\ast 5=0$;
$v_{cm}=\frac{\overrightarrow{p_i}}{m_1+m_2}=0$;
$\Rightarrow p_{cm}$=0;
Frictional force on both the blocks will be equal and opposite, hence, on the two blocks together considered as a system, they will be internal. But, velocity of each block will decrease until both the blocks cross each other. Then, they velocities will be constant unless the velocities become zero before the blocks lose contact with each other. The acceleration of each of the block will be negative and the velocity will be zero after time
$v_1-\mu \ast g{t}_1$;
$\Rightarrow t_1=\frac{v_1}{\mu \ast g}$;
$v_2-\frac{\mu \ast m_{1}g}{m_2}{t}_2$;
$\Rightarrow t_2=\frac{v_2\ast \frac{m_2}{m_1}}{\mu \ast g}$;
Putting the values for $m_1$, $m_2$, $v_1$, $v_2$ we observe that
$t_1=t_2=\frac{1}{\mu }$;
Thus after time $t=\frac{1}{\mu }$ both the blocks will stop. The assumption here is the blocks do not separate from each other before reaching zero velocity.
Options:
A->3,4 since 0 is a constant also.
B->3,4 since 0 is a constant also.
C->2
D->2