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M1 ON 13 MAY ,WHO HAS THE BEST PREPATION,Post your doubts here?

AlishaK

AlishaK

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sagar65265 said:
According to what the marking scheme says, the graph is more like an upward slope, then an inverted mountain:

If t = 0, the initial velocity = 5, so the graph must start at t = 0 and v = 5.

After that, from t till T1, the question states that the velocity is increasing. This can be shown on the graph as a curve going upwards to a positive maximum velocity.

From T1 till T2, the velocity decreases, so the graph should then show a decreasing velocity (however, the velocity must NOT be negative because the velocity only decreases, it doesn't change in direction) till the positive minimum, which they've asked you to calculate - it's the velocity at T2, 5 ms-1.

Lastly, since it says that the velocity increases after T2, the graph should show an increasing velocity curve.

This is what i've been able to gather from the mark scheme, which is not so strict on the values at which it goes to a maximum and minimum, etc.

Hope this Helped!
Good Luck for all your exams!
Click to expand...
Thanks a ton but is it possible for you to show me the graph. If you can...Anyways, Thanks.
 
Ahmedraza73

Ahmedraza73

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Z.S said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_41.pdf

Help with number 3 please
Click to expand...
first dealing with the right hand forces:
ignore the weight of the ring.and consider the forces of string T N
RESOLVING FORCES :horizontal=T cos thetha
for Vertical=T sin thetha
adding both will give the resultant: T cos thetha +Tsin thetha=15.5..........(1)
in the same way deal with the left hand side forces:
for horizontal=-T cos thetha
For Vertical: T sin thetha
adding both will give you : -T cos thetha + T sin thetha=8.5N.........(2)
SOLVE Simultaneously:
you will get T sin thetha=12
put the 12 in the first equation then yu will get : T cos thehta=3.5
Now fing the angle:put equal to both: Sin thetha/Cos thetha= 12/3.5
since Sin thetha /Cos thetha= Tan thetha
then thetha= tan inverse (12/3.5)= 73.7
 
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AlishaK

AlishaK

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Ahmedraza73 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_42.pdf
help me with Question 6 (i)
and question :5 (ii)
Comon people its urgent anyone :(
Click to expand...
For question 6 i) Use cosine rule to find the angles A and B.Once u know them....resolve the forces as usual...PS: the angles u found (A and B) are also the angles that the strings make with the horizontal x axis. So, now resolving makes sense eh? both the strings have the same tension as they have the same length nd they make the same angle with the horizontal. So finally it will be 2Tsin36.9 = 6.....T=5N
And 5 ii) i showed someone else bfore i'll link u there.

Best of luck!
 
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sagar65265

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Joker101 said:
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20%289709%29/9709_s12_qp_43.pdf
Question 7 part III please
Click to expand...

For this part, it is necessary to find out how long A is in the air until the time when it reaches a height of 1.3 meters above the ground for the second time.
When B has fallen 0.65 meters in 0.5 seconds, A is already at a height of 1.3 meters above the ground level. Since the string is inextensible, A also has a velocity of 2.6 ms^-1 after these 0.5 seconds of motion.

A continues moving up, but there is no tension to keep pulling it since there is no slack in the string, and thus the acceleration is - 10 ms^-2.

u = 2.6 ms^-1
v = 0 ms^-1
a = - 10 ms^-2

so 0 = 2.6 - 10t
Therefore, the time taken for A to move to it's highest point is 2.6/10 = 0.26 seconds.
A moves from a height of 1.3 meters above ground level to it's maximum height in 0.26 seconds, so to get from this maximum height to 1.3 meters above ground level again, it takes another 0.26 seconds (because air resistance does not play a role in the motion considered in this question, the time taken to go fro 1.3 meters above ground level to it's maximum height for A is the same as the time taken for the reverse journey).
So once the string is slack, it takes 0.26 + 0.26 = 0.52 seconds for A to go to a height of 1.3 m above ground level for the second time.
Note, however, that the question asks for the total time that A has been in motion, so we have to include the first 0.5 seconds when B and A are both moving.

Therefore the total time T = 0.52 +0.5 = 1.02 seconds.

For the graph, when B is released, A has a positive (upwards) velocity with a constant positive acceleration until 0.5 seconds, when the string becomes slack. Therefore, for this region of the graph from t = 0 seconds to t = 0.5 seconds, the gradient is positive and the line is above the x - axis.

However, after 0.5 seconds, the velocity decreases, since the acceleration is downwards and at 0.5 + 0.26 = 0.76 seconds, the velocity is zero at the maximum height.
Therefore, for this region of the graph from t = 0.5 seconds to t = 0.76 seconds, the graph has a negative gradient (deceleration) but the line is still above the x - axis because the ball is moving upwards, not downwards. This line reaches the x - axis at 0.76 seconds because the velocity decreases to zero at this point.

After 0.76 seconds, the ball is still accelerating downwards but this time, it has a negative velocity, so the gradient is still the same as for the previous region but the line will go below the x - axis and will end at T = 1.02 seconds.

Hope this helped!
Good Luck for all your exams!
 
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