Answer
a). $1.624s$
b). $1.193m$
Work Step by Step
a). This is the same Atwood machine again.
Hence,
$a=\frac{m_{2}-m_{1}}{m_{1}+m_{2}}g=\frac{0.255-0.215}{0.255+0.215}9.8=0.834m/s^{2}$
Now, $S=ut+\frac{1}{2}at^{2}$
$1.1=0+0.5\times0.834t^{2}$
$t=1.624s$
b). For $m_{1}$, $v=0+at=0.834\times1.624=1.35m/s$
Even after ascending 1.1m, it will travel some more distance till it comes to rest.
$v^{2}=u^{2}+2aS$
$1.35^{2}=0+2\times9.8S$
$S=0.093m$
So, total ascend = $1.1+0.093=1.193m$
