Mechanics M1: Post your doubt here

[hussamh10]

i had troubles with accuracy in this exam..math and in the morning on a FRIDAY like come on..for sure my brain wont be concentrating -_-

Me too bro just incase rechecked and had them correct i missed my Juma also

 

[iKhaled]

Me too bro just incase rechecked and had them correct i missed my Juma also

it's ok man god would understand u had a serious reason for missing it...

 

[hussamh10]

it's ok man god would understand u had a serious reason for missing it...

Yeah ....

 

[balay]

dude the friction was 1000N u need to find DF( driving force) so DF-mgsin(theta)-Fr=ma

...................i forgot the q......was resistance mentioned or did we have to calculate it?

 

[iKhaled]

guys i just wanna know who is gonna do M2 mechanics after finishing our M1 today

 

[Amy Bloom]

First of all analyze the question this is hell of a question trust me
we hve to build two equation then solve the simultaneously (always consider this option if you cannot figure out the question)
so
the K.E at B =Mgh
mg(H-2.2)=1/2 x m x Vb

And then consider Whole journey from X till ground

mgH=1/2 x m x(2.55Vb)^2 Vg/Vb=2.55 so (speed at the ground)Vg=2.55^2

elimintate Vb by
1/2 x m xVb^2 =mg(h-2.2)
making Vb^2 the subject we get
Vb^2=20(H-2.2)
putting it in the second equation
1/2 x m x 2.55^(20(H-2.2))=10mH
By solving we get H=2.6 (i hope this is the answer by the way which paper is this?)

Yeah indeed its hell! thanks buddy.

 

[Just visiting]

If I scored a B in pure and an A in mechanics, will I get an A overall. And what us I lost 1 or 2 marks below threshold?

 

[asum O_o]

Guys please post the topic wise past paper for mechanics M1. I have no idea where to search for it. Please reply as soon as possible. I am in real need of it

 

[sweetiepie]

https://www.xtremepapers.com/community/threads/help-needed-for-mechanics-m1.23845/ help here

 

[Prince77]

AsA,
i NEED M1 formulas, Urgent!
Anyone please help me!

and how to solve 4 (ii) : http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_42.pdf

 

[Silent Hunter]

Nice and informative thread. Should be made sticky for the coming exams preparations

 

[Thampi4]

Hi
Can someone explain q5 for m/j 2012 p42 pls


Thanks

 

[sagar65265]

AsA,
i NEED M1 formulas, Urgent!
Anyone please help me!

and how to solve 4 (ii) : http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_42.pdf

Just one little question; have you completed the first part of the question fully? (4 (i))
Anyways, about question 4 (ii):

Part 4 (i) yields two answers:
Normal Component = Tcos65 + 40
Horizontal Component = Tcos25

These two answers can be obtained by using Newton's Second Law and setting the acceleration to zero (you can do this as the system is in equilibrium).

For the second part, they are asking the value of the Tension in the string.
The formula for the Horizontal force exerted on the ring by the rod, which is the frictional force acting on it, is given by:

Frictional Force = (coefficient of friction) * Normal Force exerted by surface

You've already given expressions for the Frictional Force (Horizontal force in part (i)) and the Normal Force (Vertical force in part (i)), and they have given you the value of the coefficient of friction, so all you have to do is substitute the values:

Tcos25 = 0.4 * (Tcos65 + 40)

Tcos25 = 0.4Tcos65 + 16

T (cos25 - 0.4cos65) = 16

T = 16/(cos25 - 0.4cos65) = 21.701 N

= 21.7 Newtons
Hope this helped!

Good Luck for all your exams!

 

[sagar65265]

Hi
Can someone explain q5 for m/j 2012 p42 pls


Thanks

For 5 (i), you can consider each one of the two blocks separately, A separately and B separately.
Since the system is given to be in equilibrium, the acceleration of every part (A and B) of the system will be zero.


Thus, Newton's Second Law becomes
Net Force = m x a = m x 0 = 0

We can assign a sign to each direction, for example, upwards is positive and downwards is negative.

So if the Net Force is Zero for both blocks, the forces on them cancel each other out.
For Block A, there are two forces acting, the blocks weight and the tension in the string S1 pulling it upwards:
Weight = 30 N
Tension(S1) = ?
Since the weight acts downward, it is a "negative" force relative to the tension, which acts upwards and is thus a "positive" force. So,

Tension(S1) - 30 N = 0 So Tension(S1) = 30 N

Now let's focus on block B. There are 3 forces acting on it, Tension from S1 (Downwards), Weight (Downwards) and Tension from S2 (Upwards).
So Tension(S1) and Weight are "negative" while Tension(S2) is "positive". Therefore,

Tension(S2) - Tension(S1) - Weight = 0
Tension(S2) = Weight + Tension(S1) (Here Weight is the weight of the 2 kg block B)
Tension(S2) = 20 + 30 = 50 N

So now that we know the tensions, question (i) is over and done with.

For part (ii), we need to repeat the same process again, except that we use different values and the blocks are accelerating, so we can't set acceleration to zero.
Block A: 3 forces, Weight (Downwards), Air Resistance (Upwards) and Tension (Upwards).

Therefore,

1.6 N + Tension - 30 = 3a
So Tension = 3a + 28.4

For Block B: 3 forces, Weight (Downwards), Air Resistance (Upwards) and Tension (Downwards).

Therefore,

4 - Tension - 20 = 2a
So Tension = - 16 - 2a

So since Tension is the same at both ends of the thread,

- 16 - 2a = 3a + 28.4
5a = - 44.4
So a = - 8.88 ms^-2

The reason this acceleration is negative is because it's a downwards acceleration and earlier, the downward direction was defined to be negative relative t0 the upwards direction.

Tension is therefore = 3a + 28.4 = 3(-8.88) + 28.4 = 1.76 N

Therefore the Tension = 1.76 N.

Hope this helped!

Good Luck for your exams!

 

[minie23]

sagar65265

would you mind explaining me no. 7 ? :s

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_42.pdf

 

[applepie1996]

when resolving forces ....how do we know when to consider the ring and when to leave out the force in the ring ??

 

[JalalKaiser]

Part 4 (i) yields two answers:
Normal Component = Tcos65 + 40
Horizontal Component = Tcos25

These two answers can be obtained by using Newton's Second Law and setting the acceleration to zero (you can do this as the system is in equilibrium).

Good Luck for all your exams!

JazakAllah Khair! One teensy-tiny question though, how'd the value of 40 come about? The weight's 4kg. I'm a bit of a dunce, haha.

 

[JalalKaiser]

Oh and, could you please explain the marking scheme method for the second part? I'd like to know both ways tbh.

 

[GorgeousEyes]

Guys what's the hardest paper u have ever solved ?

 

[JalalKaiser]

Guys what's the hardest paper u have ever solved ?

Probably the 2012 Paper 42.