A1/A2 Mechanics - Pulleys MadasMaths Question

In the Question posted, i got part a and b correct but for part c, according to the solution, they worked out a new acceleration using a forces diagram of -13.3ms^-2 and i dont understand why they would need to work out a new acceleration. I though that when A reaches the floor, the string will go slack and hence then B would be moving under gravity so the acceleration would be -9.8 and that we would then input this into SUVAT equation and work out the additional distance (x) particle B moves up.
So then the final answer would be 1.54 + 1.54 + (x)
But i guess im wrong so i would really appreciate it if someone could help me out greatly by explaining it to me.

(edited 11 months ago)

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There is a constant air resistance of 7N, so that will causes a greater deceleration in the upwards phase of the particle motion. Youre correct that the string is slack so the motion is determined by gravity and air resistance on the particle.

(edited 11 months ago)

Reply 3

Blue_Iris

Original post by mqb2766

So do we only consider the acceleration to be -9.8 if there were no resistive forces acting on the particle or any other force except from the weight of the particle acting downwards?

Reply 4

mqb2766

Original post by Blue_Iris

So do we only consider the acceleration to be -9.8 if there were no resistive forces acting on the particle or any other force except from the weight of the particle acting downwards?

Gravity exerts a force of mg on the particle. Newton 2 says the acceleration (ma) is equal to the net or resultant force on the particle. So you sum all the forces acting on the particle and that gives the resultant force or acceleration (multplied by m).