Two blocks A and B of masses $1 \mathrm{kg}$ and $2 \mathrm{kg}$ respectively are connected by a string, passing over a light frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that s-p  string remains just taut. At moment t=0, a force F=20t newton starts acting on the pulley along vertically upward direction as shown in figure. Calculate velocity of A when B loses contact with the floor.(Take, $g=10 \mathrm{m}/{\mathrm{s}}^2$).

 Let T be the tension in the string. Then,

$$\begin{gathered} 2T=20t \\ T=10t\text{ newton } \end{gathered}$$

$\text{ Let the block }A\text{ loses its contact with the floor at time }t=t_1\text{. This happens when the tension }$

$$\begin{gathered} \text{ in string becomes equal to the weight of }A\text{. Thus, } \\ T=mg \\ 10t_1=1\times 10 \\ {t}_1=1 \mathrm{s} \end{gathered}$$

$$\begin{gathered} \text{ or } \\ \text{ Similarly, for block }B\text{, we have } \\ \text{ or }10t_2=2\times 10 \\ \text{ i.e. the block }B\text{ loses contact at }2 \mathrm{s}.\text{ For block }A,\text{ at time }t\text{ such that }t\ge t_1\text{ let }a\text{ be its } \\ \text{ acceleration in upward direction. Then, } \\ 10t-1\times 10=1\times a=(dv/dt) \\ \text{ or } \\ \text{ Integrating this expression, we get } \end{gathered}$$

or ${\int }_0^{v}dv=10{\int }_1^t(t-1)dt$

Substituting $t=t_2=2 \mathrm{s}$

$v=5t^2-10t+5$

v=20-20+5

 = 5 $\mathrm{m}/\mathrm{s}$