Is the tension in the string the same even though it is only half the length now?
Same as what? It's different to the first part.
The length of the portion of the string does not effect the tension. It's only the masses and their position (and the rate of revolution) that effects the tension. And there are now two masses.
The length of the portion of the string does not effect the tension. It's only the masses and their position (and the rate of revolution) that effects the tension. And there are now two masses.
Hmm I think I've got it cheers.
Do you mind helping with 6ii.
I think I have the time for p, but I don't understand how to get the time for q.
S = 250 u = 86sin62.3 v = ? a = -9.8 t = ? S= ut -4.9t^2 250 = 86sin62.2t - 4.9t^2 t = 10.8 s
IF the vertical speed initially was 86ms^-1 (which it isn't - it's less), then in 10 seconds it's vertical velocity would have become negative i.e 86-10g. But it isn't, it's 30 upwards, so t=10.8 can't be right.
Time from P to the ground s = -250 u = 30 v =? a = -9.8 t = ? -250 = 30t - 9.8t^2 t = 6.8s
If you're going to work out the time from P to the ground, then this is the actual time of flight of Q.
IF the vertical speed initially was 86ms^-1 (which it isn't - it's less), then in 10 seconds it's vertical velocity would have become negative i.e 86-10g. But it isn't, it's 30 upwards, so t=10.8 can't be right.
If you're going to work out the time from P to the ground, then this is the actual time of flight of Q.
Hmm Im not too sure. I've attached the mark scheme if you want to have a look.