Edexcel M1

I did this question but I could only get the answer in the book, when I divide the answer I worked out by two. Can someone check it?
My working:
For A: R(upwards) T-1.2g=1.2a
For B: R(downwards) 0.9g+0.7g-T= 0.7a+0.9a
Adding both equations:
0.9g+0.7g-T+ T-1.2g = 1.2a+0.7a+0.9a
0.8g=2.8a
a=0.8g/2.8
a=2.8ms^-1

But book's answer is 1.4ms^-1.

0.9g+0.7g-1.2g = ???? It's not 0.8g

Can you help me in this question?The c) part

When Q hits the floor, P will carry on upwards now moving freely under gravity as the string will be slack. It will reach a certain height, come to instantaneous rest, and then fall back until we get to the point where the string becomes taut again.
Note: You know the speed of P at this point.


Now use conservation of momentum.

m₁u₁+m₂u₂=m₁v₁+m₂v₂

m₁=0.1
m₂=o.2

What is the initial speed of P? Can I apply the value as 0m/s? ... and u₂ as 1.81.
Should I find the value v₁ using suvat equation using 9.8 as the value of a?

To be clear, we are applying conservation of momentum at the time the string becomes taut again.

Since you have m1=0.1, then you're using the subscript "1" to refer to P.

u1 will be the velocity of P just before the string becomes taut. I don't know where you've got 1.81 from, but if that was the speed that Q hit the ground at, then that's correct. When Q hit the ground, P carried on up at that same speed, and when it came down to the same point at which the string became slack, then its speed will be the same, thus 1.81.

u2 will be zero.

After the string becomes taut they will both move at the same speed, hence v1=v2.