MHB Mechanics- connected particles

[Shah 72]

Two particles A and B are attached to the ends of a light inextensible string, which passes over a smooth pulley. Particle A has mass 8 kg and particle B has mass 5kg. Both the particles are held 1.2m above the ground. The system is released from rest and the particles move vertically.
a) when particle A hits the ground, it does not bounce. Find the max height reached by particle B
b) when particle A hits the ground, the string is cut. Find the total time from being released from rest until B hits the ground.
I calculated (a) and got the max height traveled by B when A hits the ground would be 1.2 +1.2 +0.27= 2.68m
Iam not getting the right ans for (b) which is 1.99 s
t1= 1.02s
When the string is cut, Tension=0N , max height reached by B =2.68m, a=-g=-10m/s^2, u=0m/s on the way down.

 

[topsquark]

Where did you get the 0.27 m in part a)?

We've asked you to show your work any number of times. Please tell us how you calculated a) and why you did what you did.

And always always always sketch a diagram.

-Dan

 

[Shah 72]

Where did you get the 0.27 m in part a)?

We've asked you to show your work any number of times. Please tell us how you calculated a) and why you did what you did.

And always always always sketch a diagram.

-Dan

Ok I will post my working.
a)So by drawing free body diagram of both A and B, I calculated a=2.308m/s^2
Then I calculated the velocity with which A hits the ground by taking u= 0 m/s , a= 2.308m/s^2, s=1.2m, v= 2.35 m/s
B also moves up with the same velocity of 2.35 m/ s and also moves 1.2m upwards.
I need to calculate the X m it travels upwards and then comes down.
So I took u= 2.35m /s, v= 0 m/s, a=-g=-10m/s^2, s=?
I got s=0.27m
I added this 1.2+1.2+0.27 and got max height = 2.68m