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M2 help

Can someone explain 4ii to me I just don't get it.

Like there is no point giving me hints, I don't get what is going on.
Original post by Super199
Can someone explain 4ii to me I just don't get it.

Like there is no point giving me hints, I don't get what is going on.


Since you've done the first part, you know the general set up.

For the second part there is an additional mass added to the string at its midpoint.

Then tension in the AP part of the string will be what's necessary for the two masses to move in a circle.
Reply 2
Original post by ghostwalker
Since you've done the first part, you know the general set up.

For the second part there is an additional mass added to the string at its midpoint.

Then tension in the AP part of the string will be what's necessary for the two masses to move in a circle.


Is the tension in the string the same even though it is only half the length now?
Original post by Super199
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.
(edited 7 years ago)
Reply 4
Original post by ghostwalker
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.


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.

http://www.ocr.org.uk/Images/57767-question-paper-unit-4729-01-mechanics-2.pdf
Original post by Super199
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.

http://www.ocr.org.uk/Images/57767-question-paper-unit-4729-01-mechanics-2.pdf


Q was launched when P was above A.

So, how long did it take for P to get to the point above A? Knock that off the total flight time fro P, and you have the flight time for Q.
Reply 6
Original post by ghostwalker
Q was launched when P was above A.

So, how long did it take for P to get to the point above A? Knock that off the total flight time fro P, and you have the flight time for Q.


Time for P to get to A.

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

Time from P to the ground
s = -250
u = 30
v =?
a = -9.8
t = ?
-250 = 30t - 9.8t^2
t = 6.8s

Total time = 17.6
17.6 - 16.8 = 10.8s
But where does the time for p - 15.5s come from?
Original post by Super199
Time for P to get to A.

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.
(edited 7 years ago)
Reply 8
Original post by ghostwalker
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.

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