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hey its very simple.........>
)i mean to find range we inject the limits and then chk if a vertex lies in the domain....so in dis case it wud be f(0) and f(pi) for limits and for vertex f(-b/2a) where did f(90) com frm?
s10 qp11 Q5 ii
but ur ans is wrong
Thank you! Have a nice day, everyone. Wish you best health for the upcoming exam! AoA wr wbHello every one!
I have a really hard question on Mechanics 1, which I am dying for some help. Here it is:
3 Particles P and Q are attached to opposite ends of a light inextensible string which passes over a fixed smooth pulley. The system is released from rest with the string taut, with its straight parts vertical, and with both particles at a height of 2 m above horizontal ground. P moves vertically downwards and does not rebound when it hits the ground. At the instant that P hits the ground, Q is at the point X, from where it continues to move vertically upwards without reaching the pulley. Given that P has mass 0.9 kg and that the tension in the string is 7.2 N while P is moving, find the total distance travelled by Q from the instant it first reaches X until it returns to X. [6]
I hope I could learn something from you all, because Mechanics 1 is killing me
the ans which I qouted is write yaar see the mark schemebut ur ans is wrong
lol
i think i was looking at another variant ms sorry
Thank you so much for your time. I am really appreciate it. You save me!AoA wr wb
make a sketch of the problem first and give P the left side so the solution is identical
Considering P: Take downwards +
Resolving : 9 - 7.2 = 0.9a
a=2
Now considering Q: Take upwards +
s 2 ( because it travels same distance as P)
u 0 (strts frm rest)
v v (this we hav to find atm)
a 2 (see top)
t no need
v^2 = u^2 +2as
put values
v=√8
now make a sketch in ur mind the question says when P hits the ground Q is at X and continues to move up means it doesnt stop..use common sense when p is at ground and q still moving it means the string has become slack so gravity will act on it
at X:
s (we hav to find)
u √8 (the v found earlier becomes u)
v 0 (at max height v=0)
a -10
t noneed
v^2=u^2 + 2as
u get s = 0.4
use logic.. 0.4 when Q flies up when string is slack so it will be s = 0.4 x 2 = 0.8m for it to become taut again
that took a lot of tym pray for me now
Please can anybody help me with this?
Whenever there are odd number of possibilities you always do n+1 hence it will be 13+1=14. Then take 75% of it which is equal to 10.5. As the answer is not a whole number, you have to take the 10th and 11th number and then divide it by two:
Thanksq3ii-: This should be alright if you know the formula. So you want the volume between two points. If you sketched the curve properly, you should get your limits: 0 and 2.
Now use the formulae π∫(y^2)dx. You know y= (x-2)^2. So substitute that into your formula:
1. π∫((x-2)^2)^2 dx = π∫((x-2)^4)dx.
2. Integrate that, and you should get: π [((x-2)^5)/5] (<---- I assume you know how to do that integration
3. Now substitute in your limits in the place of x (remember, the bigger one first): π {((2-2)^5)/5) - ((0-2)^5)/5)}
4. That comes to ---> π[ (0) - (-32/5) ] = 32π/5
That's 3ii solved! Hope that helped. Remember to look at the mark schemes, those are usually helpful. (That's all I have time for today. Maybe I'll get back to you later)
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