The elastic collision between two bodies, A and B can be considered using the model in which A and B are free to move along a common line without friction. When separation between the surfaces is greater than d = 1m, the interacting force is zero. However, when their distance less than d, a constant repulsive force F=6 N is present. The mass of body A is $m_A=1kg$ and it is initially at rest. The mass of body B is $m_B=3kg$ and it is approaching towards A with a speed $v_0=2m{s}^{-1}$ Then select the correct statement(s).

At minimum separation, the common velocity attained by the bodies equals the velocity of centre of mass. So,

$v_{cm}=v_0=\frac{3\times 2}{1+3}=1.5m{s}^{-1}$

At time t, for body A, we have

$v_A=u_A+a_{A}t=0+\left(\frac{F}{m_A}\right)t$

$\Rightarrow v_A=6t\_\_\_\_\_\_\_\_\_\_\_\left(1\right)$

At time t, for body B, we have

$v_B=u_B+a_{B}t=2-\left(\frac{6}{3}\right)t$

$\Rightarrow v_B=2-2t\_\_\_\_\_\_\_\_\_\_\left(2\right)$

Since $v_A=v_B$

So we get $6t=2-2t\Rightarrow t=0.25s$

 Change in kinetic energy is given by

  $\Delta K=\frac{1}{2}(1+3){\left(1.5\right)}^2-\frac{1}{2}\left(3\right){\left(2\right)}^2=-1.5J$

During this time let displacement of A be $\Delta s$, then from work-energy theorem, work done equals change in kinetic energy, so

$W=\Delta K=F\Delta s\cos \left({180}^0\right)$

$\Rightarrow -1.5=6\Delta s(-1)\Rightarrow \Delta s=\frac{1.5}{6}=0.25m$

 Minimum separation is 1-0.25= 0.75 m

  Hence (b) and (d) are correct.