Note:
#T = "string tension"#, we don't need to calculate this.
First find #mg#
#1.2g# and #0.8g#
Don't evaluate #g#
Using #F=ma#
Write down the resultant force in the direction of motion for each particle.
#1.2g-T=1.2a# #[1]#
#T-0.8g=0.8a# #[2]#
Solve simultaneously:
Adding #[1] and [2]#
#1.2g-0.8g=1.2a+0.8a#
#0.4g=2a#
#a=(0.4g)/(2)=0.2gms^(-2)#
Now we know the rate of acceleration, we can find the distance travelled in 1 second.
Using:
#S=ut+1/2at^2#
Initial velocity is zero.
#u=0#
#S=(0)1+1/2(0.2g)(1)^2#
#S=0.1g#
#g=9.81#
#S=(0.1)(9.81)=0.981m#
From this we can see that, when the #1.2kg# mass reaches the floor, the #0.8kg# mass will stop accelerating, but will continue to travel upwards until gravity stops it.
We now need to find the time it takes the #1.2kg# mass to travel the distance #0.64m#, and consequently this will be the time the #0.8kg# mass is accelerating upwards. The remaining time will be the duration the #0.8kg# mass is decelerating and we need to find the distance travelled during this period.
Duration of time #1.2kg# mass takes to travel #0.64m#
Using:
#s=ut+1/2at^2#
#0.64=(0)t+1/2(0.2g)t^2=>t=sqrt((1.28)/(0.2g)#
#g=9.81#
#:.#
#t=sqrt((1.28)/(0.2(9.81)))=0.8077100438s=0.808s# 3 d.p.
Next the velocity of the #0.8kg# mass after this time:
Using:
#v=u+at#
#v=0+(0.2g)(0.808)=>v=0.1616g=1.585296 \ \ms^(-1)#
Remaining time in which it decelerates due to gravity.
#t=1-0.808=0.192s#
Distance travelled during this period.
Using:
#v^2=u^2+2as#
#(0)^2=(1.585)^2+2(-g)s#
#s=(0^2-(1.585)^2)/(2*(-g))=0.128044m#
To get total distance travelled by the #0.8kg# mass, we add:
#0.64m+0.128044m=0.768044m#
This is as close as I can get to #0.776m#. This could be due to rounding errors.
