Two particle A and B, of mass 3m and 2m respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed pulley of negligible mass. After the system is released and A falls through a distance L, it hits a horizontal inelastic table so that its speed is immediately reduced to zero. Assume that B never hits the table or the pulley. If the time for which A is resting on the table after the first collision and before it is jerked off is $t_0$ . and  the difference between the total kinetic energy of the system immediately before A first hits the table and total kinetic energy immediately after A starts moving upwards for the first time is $△K, then$ 

Acceleration before A hits the table is

$\mathrm{a}=\left(\frac{3\mathrm{m}-2\mathrm{m}}{3\mathrm{m}+2\mathrm{m}}\right)\mathrm{g}=\frac{\mathrm{g}}{5}$
Speed of A just before hitting the table is
${\mathrm{u}}_0=\sqrt{2\mathrm{aL}}=\sqrt{\frac{2}{5}\mathrm{gL}}$
This is also the speed of B at this moment. Now A comes to rest and string becomes slack. Now B moves up with  retardation g. B stops and then starts falling down. When B acquires speed $u_0$ and comes to
 

position shown in fig. (iii), the string is about to regain tension. As soon as the string becomes taut, the speed of both  A and B will becomes same, say V The impulse, I applied by the string causes sudden change in momentum of A and B.
For A –I = 3mV                              ……..(i)
For B –I = 2mV – 2m$u_0$                 ……..(ii)
From (i) and (ii)  $\mathrm{V}=\frac{2}{5}{\mathrm{u}}_0$
$\left(a\right)$The required time = time of flight of a particle projected vertically with speed $u_0$

$=\frac{2{\mathrm{u}}_0}{\mathrm{g}}=\frac{2}{\mathrm{g}}\sqrt{\frac{2}{5}\mathrm{gL}}=\sqrt{\frac{8}{5}\frac{\mathrm{L}}{\mathrm{g}}}$
(b) Loss in kinetic energy is  

$\mathrm{\Delta k}=\frac{1}{2}\left(5\mathrm{m}\right){\mathrm{u}}_0^2-\frac{1}{2}\left(5\mathrm{m}\right){\mathrm{V}}^2$

$$\begin{gathered} =\frac{5\mathrm{m}}{2}{\mathrm{u}}_0^2\left[1-\frac{4}{5}\right]\left[\therefore \mathrm{V}=\frac{2}{5}{\mathrm{u}}_0\right] \\ =\frac{1}{2}{\mathrm{mu}}_0^2=\frac{\mathrm{mgL}}{5}\left[\therefore {\mathrm{u}}_0=\sqrt{\frac{2}{5}\mathrm{gL}}\right] \end{gathered}$$