You want the distance from O to 5m above the ground, which will be 5m less than the distance you just calculated, so 17.3-5=12.3m. Let this equal s, they gave you u in the question, you know a=g as it's free fall, and you want v. So using suvat equation v2=u2 + 2as, v2=4.22+2*9.81*12.3. V should come out as 16.1ms-1 if I've done that calculation correctly.
Gosh I didn't read the bit about Q being faster than p by 1ms^-1. Thanks for your help
How would you answer January 2010 Q2? The first one I was thinking cosine rule. Not sure about the second bit?
Well I had to look at the mark scheme for that. You have to have two components of the 19N. One in the horizontal and one in the vertical. The question is so badly worded.
So for the first part
R^2 = sqrt [(12 + 19cos60)^2 + (19sin60)^2 ]
and you get R as 27.1 N
Part b
You have a triangle. The adjacent is (12 + 19cos60) and the opposite is (19sin60.)
Sorry, I was concentrating on my economics essay. I think someone already answered this, but I'm doing mechanics for a bit now if you need help with anything else, let me know
Can someone explain how you work out change in momentum
For 3ii. Jan 2010.
Thanks
Change in momentum = momentum before - momentum after = 9*0.096 - -3.5*0.096 = 1.2kgms-1. Remember to account for direction, as the particle changes its direction of motion, it changes to a negative momentum.
Change in momentum = momentum before - momentum after = 9*0.096 - -3.5*0.096 = 1.2kgms-1. Remember to account for direction, as the particle changes its direction of motion, it changes to a negative momentum.
I don't understand question 4a on the same paper
I just got 0.4gsin60 and 0.4gcos60. Where does the other 0.3gsin60 and 0.3gcos60 come from? I thought it was only considered particle p.