Post your AS-Level Mathematics (P1 and M1) doubts here.

[axetreme.O]

Bro post it.

What is the probability that a RCB supporter will be forced to support CSK in this IPL?
I think it's pretty high.

 

[BeBeskii108]

How was M1 42 for you guys ? I found the paper bit challenging.

 

[Mstudies]

M1.
Formulae for different energy situations. I still don't get when to use net work done = increase in KE...
Like for all the different possibility :S
Can you please write these up?

The unit used to measure work done is Joules (J) or newton meters (which is also joules).
The unit used to measure kinetic energy or gravitational potential energy is also joules.

I'll explain in detail

Work and Kinetic Energy

• When a net force does work on a rigid body, it causes the body's speed to change.
• The work done by the net force is the same as the sum-total of the work done by the action of every force acting the body. If you add up the work done by each of the forces acting on a body you will get the same value as the work done by the net force.
• The work done on a body that caused the body to be set in motion with some speed v can be expressed as function of the body's final speed v and mass m, independent of type of force that acted on the body. We call this function the body's Kinetic Energy.

K.E = 0.5mv²

• The energy a body possesses by virtue of its motion -- its Kinetic Energy -- is not dependent upon how the object reached its state of motion, only upon its current state of motion

Work-Energy Theorem:

• The energy associated with the work done by the net force does not disappear after the net force is removed (or becomes zero), it is transformed into the Kinetic Energy of the body. We call this the Work-Energy Theorem.

W.d = Change in K.E
W.D = ( K.E at the top - K.E at the bottom)

If the body's speed increases, then the work done on the body is positive and we say its Kinetic Energy has increased. Whereas if the body's speed decreases, then it kinetic energy decreases and the change in kinetic energy Difference in KE is negative. In this case the body does positive work on the system slowing it down or alternately the work done on the body is negative.

• If the object is not rigid and any of the forces acting on it deforms the object, then the Work-Energy Theorem will no longer be valid. Some of the energy transferred to the object has gone into deforming the object and is no longer available to increase or decrease the object's Kinetic Energy.

The following will be a bit complex
Derivation of Work-Energy Theorem using a Variable Net Force
Frame of Referenceirection of the Net Force.

To reduce the complexity of the derivation, we will assume that the direction of the Net Force is constant while the work is being done. The Work-Energy Theorem is still valid if the net force changes direction as well as magnitude while the work is being done, provided the body is rigid.

The key step is to convert the calculus definition for acceleration into an expression that is a derivative of x.
Plug this and the Second Law into the definition for Work, and integrate.

Hope u understood. Diagrams from Physics edu.

 

[Oishee Asif]

The unit used to measure work done is Joules (J) or newton meters (which is also joules).
The unit used to measure kinetic energy or gravitational potential energy is also joules.

I'll explain in detail

Work and Kinetic Energy

• When a net force does work on a rigid body, it causes the body's speed to change.
• The work done by the net force is the same as the sum-total of the work done by the action of every force acting the body. If you add up the work done by each of the forces acting on a body you will get the same value as the work done by the net force.
• The work done on a body that caused the body to be set in motion with some speed v can be expressed as function of the body's final speed v and mass m, independent of type of force that acted on the body. We call this function the body's Kinetic Energy.

K.E = 0.5mv²

• The energy a body possesses by virtue of its motion -- its Kinetic Energy -- is not dependent upon how the object reached its state of motion, only upon its current state of motion

Work-Energy Theorem:

• The energy associated with the work done by the net force does not disappear after the net force is removed (or becomes zero), it is transformed into the Kinetic Energy of the body. We call this the Work-Energy Theorem.

W.d = Change in K.E
W.D = ( K.E at the top - K.E at the bottom)

If the body's speed increases, then the work done on the body is positive and we say its Kinetic Energy has increased. Whereas if the body's speed decreases, then it kinetic energy decreases and the change in kinetic energy Difference in KE is negative. In this case the body does positive work on the system slowing it down or alternately the work done on the body is negative.

• If the object is not rigid and any of the forces acting on it deforms the object, then the Work-Energy Theorem will no longer be valid. Some of the energy transferred to the object has gone into deforming the object and is no longer available to increase or decrease the object's Kinetic Energy.

The following will be a bit complex
Derivation of Work-Energy Theorem using a Variable Net Force
Frame of Referenceirection of the Net Force.

To reduce the complexity of the derivation, we will assume that the direction of the Net Force is constant while the work is being done. The Work-Energy Theorem is still valid if the net force changes direction as well as magnitude while the work is being done, provided the body is rigid.

The key step is to convert the calculus definition for acceleration into an expression that is a derivative of x.
Plug this and the Second Law into the definition for Work, and integrate.

Hope u understood. Diagrams from Physics edu.

Unfortunately, the M1 exam already passed.
But JazaakAllah Khayr.

 

[tania]

Sorry yar thori der hogyi .. xD Here u go.
And JUST PRAY SOMETHING LIKE THIS DOESN'T COME ! Huhu

Thanks fro replying although i just saw it

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s11_qp_43.pdf

Q3 pls

 

[NinjaInPyjamas]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s11_qp_43.pdf

Q3 pls

i) Find time taken for Q and P to reach the bottom.

s=ut+0.5at^2
3.2=0+ (0.5 x 10 x t^2)
t= 0.8s

Acceleration of P:
a=gsinθ
=10sin30
= 5

Find u now.
6.4 = 0.8u + {0.5 x 5 x (0.8)^2}
u = 6 m/s

ii) Speed of P at bottom

v= u + at
= 6 + (5 x 0.8)
= 10 m/s

 

[Haya Ahmed]

http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_s13_qp_42.pdf
Q6 (iii)

 

[TheJDOG]

http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_s13_qp_42.pdf
Q6 (iii)

Hope its clear, sorry had to squeeze it in last lines.

 

[Haya Ahmed]

http://onlineexamhelp.com/wp-content/uploads/2014/02/9709_w13_qp_43.pdf
q5 (ii)

 

[Haya Ahmed]

Hope its clear, sorry had to squeeze it in last lines.

Thanks a lot! Sorry for the late reply :3
But anyhow can you explain the 2nd part? why shading the small triangle from 20 to 26 only is wrong and why should we shade the whole graph? thanks in advance ^_^

 

[Haya Ahmed]

http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_s13_qp_42.pdf

Q6(ii) can someone explain why should we shade the whole graph not only the part from 20 to 26?

here is the ms : http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_s13_ms_42.pdf

 

[brewerj7]

has anyone got any notes for work and power equations in mechanics? please help

 

[itsaxdrey]

A particle P moves in a straight line. It starts from rest at a point O and moves towards a point A on
the line. During the first 8 seconds P’s speed increases to 8 m s−1 with constant acceleration. During
the next 12 seconds P’s speed decreases to 2 m s−1 with constant deceleration. P then moves with
constant acceleration for 6 seconds, reachingA with speed 6.5 m s−1
.
(i) Sketch the velocity-time graph forP’s motion. [2]
The displacement of P from O, at time tseconds after P leaves O, iss metres.
(ii) Shade the region of the velocity-time graph representing sfor a value oft where 20 ≤ t ≤ 26.
[1]
(iii) Show that, for 20 ≤ t ≤ 26,
s = 0.375t
2 −13t +202.

can i get the solutions