At $$t=0$$, the lower end of the bar A is just above the upper end of bar B (mass$$ of bar $$A=3$$ kg, mass of bar $$B=113kg). Find time (in$$ $$sec$$$$) when upper end of block A just crosses the lower end of $$.8$$. (Assume$$ the system was released at $$t=0).$$g=10m/s2$$

Application of Newtons laws

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At t=0, the lower end of the bar A is just above the upper end of bar B (mass of bar A=3 kg, mass of bar $B = \frac{11}{3}$ kg). Find time (in sec) when upper end of block A just crosses the lower end of .8. (Assume the system was released at t=0). $[g = 10 m / s^{2}]$

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Solution

$Acceleration of A and B = (\frac{\frac{11}{3} - 3}{\frac{11}{3} + 3}) g = 1 m / s^{2}$
$Relative acceleration of A and B = 2 m / s^{2}$
$Now by using s = u t + \frac{1}{2} a t^{2}; 4 = 0 + \frac{1}{2} (2) (t)^{2} \Rightarrow t = 2 s$

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