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Main problem is with part (b). I got it right up to the accelerations with -11.6 and 4.4 respectively.

Then I dont really see the reason why we have to break it up to two parts as given in the solution. We should be able just to use the constant acceleration formulae and just to find the time where the displacement is the SAME (time just before collison). Even if Q stops moving up, it would change direction then going downwards...(That's the difference between mine answer and the "correct" answer?)

What am I thinking why?

Model answer given below.

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CasperYC
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    You assume that $Q$ experiences the same $11.6\ \mathrm{ms}^{-2}$ acceleration during the entire time until the collision. But $Q$ only has that acceleration on the way up. As soon as it starts sliding down again, its acceleration is only $4.4\ \mathrm{ms}^{-2}$, just like $P$. The key fact is that $Q$ stops moving up before the particles meet. – David K Apr 01 '22 at 01:12
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    TRICKY!!!!!! Now I get it, I missed the fact that the friction, uR, changed the direction once Q starts to slide down, hence the change in the acceleration.......@DavidK – CasperYC Apr 01 '22 at 01:24

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