AQA AS Physics Unit 2 (PHYA2) June 9th 2016 Resit paper

yeah i put the initial gradient as the same, but said the the acceleration would continue increasing for a longer period, and my sketch on the graph had the same initial gradient and stayed steeper than the other gradient for longer, until it reached a constant velocity, a bit higher up than the lorry without the leak.

It said you can draw ON the graph, not draw a graph haha. So I drew a dotted line between when the gradient is increasing and when the gradient is 0. Then labelled resistive force=driving force and resistive force < driving force. Hope I'm right

Yeah i just drew another curve on the graph

What did yours look like?

The embankment one:

R (Reaction force at embankment) = 0
Resolve from the rock.
Vertical force at the left much be greater than the right for the boy to stand without falling

Mass of boy = 46kg Distance to boy = 0.8
Mass of plank = 40kg Distance to C.O.M = 1.2

46g * 0.8 = 40g * 1.2 + C
313.6 = 361 + C
C is negative so the boy can stand.

That's how I did it.

What did yours look like?

What did you guys get for the max vertical height in the projectile motion question?

i Think alot of people in this thread are forgetting that for the 6 mark question, The initial Acceleration at t= 0 will be the same as the one that is not leaking because the mass is the same as no water has leaked out already. I also mentioned that Force = Coefficient of friction x reaction force and that reaction force = weight so therefore the Force is dependant on the mass of the truck so the maximum frictional force will be less because of the smaller mass, this will lead to a greater velocity reached.

If you're doing it from the bank you need the support force of the rock which would be 86g I think?

Did anyone get 2.9ms^-1 for the rocket question which means that the engine will manage to decelerate to the safe speed?

This is how I drew it:

I resolved from the rock so you can ignore all the forces through it,.

yes