Mechanics M1: Post your doubt here

[haha101]

Do June 2011 upto Nov 2012 !

i did only them

 

[blueberryyums]

I did only 2 years

:O Are we ready or are we going to bring the GT down?

 

[haha101]

:O Are we ready or are we going to bring the GT down?

Well i did whole papers of 2 years . other than that did some topicals . Just hope that nothing hard comes

 

[Hubsy16]

hey can anyone clarify that 'a-v-s' integration/differentiation rule in the kinematics questions. Like for instance, to get velocity you integrate it with the value of acceleration I think.....I'm really confused.....Could use some help.....Also does anyone have any idea about how many significant figs you're usually supposed to give an answer in?

 

[Abu mota]

i dont understand how to solve questions in which there is a force acting on two bodies.
when do u use both bodies as a whole body and when do u use them separately?

 

[Abu mota]

like these kinds of questions http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_43.pdf Q.7
also http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_42.pdf Q.5

 

[Msbh22]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s08_qp_4.pdf

QUESTION NO.6 iii
HELP NEEDED

 

[Saad Talaat]

hey can anyone clarify that 'a-v-s' integration/differentiation rule in the kinematics questions. Like for instance, to get velocity you integrate it with the value of acceleration I think.....I'm really confused.....Could use some help.....Also does anyone have any idea about how many significant figs you're usually supposed to give an answer in?

differentiation is from s to v and v to a
integration is from a to v and v to s
best is 3 sig. fig.

 

[leosco1995]

i dont understand how to solve questions in which there is a force acting on two bodies.
when do u use both bodies as a whole body and when do u use them separately?

You can use the whole system to get the acceleration in one step.

You can use them separately if you need to find the tension.. you just need to solve the simultaneous equations you get.

 

[Joker101]

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 ,,

 

[haha101]

guys while integrating acceleration for velocity and velocity to acceleration , at wht instances do we have to include the integrtion constant "C" ?? Please

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_4.pdf

QUESTION NO.6 iii
HELP NEEDED

The total work done on the ball while it's going up and coming done can be related to the change in it's kinetic energy according to the work - kinetic energy theorem:

The change in K.E. while going upwards:

0.5 * 0.6 * (0^2 - 5.2^2) = - 8.112 J
Therefore, the total work done on the ball must equal - 8.112J.
Total work = Work done against motion by gravitational force + Work done by air resistance against motion
Total Work = - 0.6 * 10 * 1.3 + Work done against motion by air resistance.

(1.3 is the change in height of the ball above the ground when it's going upwards.)

- 8.112 = - 7.8 + Work done against motion by air resistance
So the work done by air resistance while the ball is going up is - 8.112 + 7.8 = -0.312 J of work.

While going down, the ball has an initial velocity of 0 ms^-1 and a final velocity of 12 ms^-1. So the kinetic energy has increased:

Change in K.E. while falling downwards = 0.5 * 0.6 * (12^2 - 0^2) = 43.2 J

So the total work done must be equal to 43.2 joules.

Work done by Gravitational force aiding motion + Work done by air resistance against motion = 43.2 J
0.6 * 10 * 7.5 + Work done by air resistance against motion = 43.2 J
Work done by air resistance against motion = 43.2 J - 45 J = - 1.8 J
The total work done against motion on the balls journey = 1.8 J + 0.312 J = 2.112 J = 2.11J

The reason the answer is positive is that air resistance is always against motion, so there is no need to ascribe a negative value to the work it does.

Hope this helped!
Good Luck for all your exams!

 

[minie23]

guys while integrating acceleration for velocity and velocity to acceleration , at wht instances do we have to include the integrtion constant "C" ?? Please

the constant C is referred as the initial velocity 'u' !
If in the questions it says, the body starts at rest, just replace c = o
Or if the questions says the body has initial velocity 2m/s , just replace c = 2

Hope I helped

 

[snowbrood]

guys while integrating acceleration for velocity and velocity to acceleration , at wht instances do we have to include the integrtion constant "C" ?? Please

if the body start from a point that is some distance from origin point O

 

[haha101]

the constant C is referred as the initial velocity 'u' !
If in the questions it says, the body starts at rest, just replace c = o
Or if the questions says the body has initial velocity 2m/s , just replace c = 2

Hope I helped

if the body start from a point that is some distance from origin point O

Thanks guys
minie23 what if we have to integrate velocity to distance then ?

 

[Manobilly]

Thanks guys
minie23 what if we have to integrate velocity to distance then ?

If there is a distance at rest include it or sometimes u are given the d and at a particular time ,subsitute it to find C.

 

[leosco1995]

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 ,,

From part (ii) you found that the maximum acceleration is 2 so if you consider the system as a whole,

ΣF = ma
P - 3150 = (200+250) * 2
P - 3150 = 900
P = 4050 N

Otherwise you can just consider box B. Draw a diagram showing the resistive forces. There exists friction between box A and box B, and friction between box B and the floor.

The co-efficient between the boxes is 0.2 and reaction force is from box A which is 2000 N so F = 0.2(2000) = 400

ΣF = ma
P - (friction between B and floor) - (friction between box A and B) = (250 * 2)
P - 3150 - 400 = (250) * 2
P = 4050 N

 

[Saahil Parikh]

Hello everybody.
I'm having a bit of a problem in mechanics, Resolving Forces to be precise. If someone can post notes on Resolving Forces it'll be really helpful. Thanks

 

[Alyjohn]

Please post the topic wise past paper for Mechanics M1. I am in real need of it. Please reply

 

[leosco1995]

Please post the topic wise past paper for Mechanics M1. I am in real need of it. Please reply

I plan to make a mobile app with solutions to all of these questions... year and topic-wise but it might take a while. Would there be any interest in this for P1 and P3 too? I can't do Statistics though since I suck at that.