ALL CIE PHYSICS DOUBTS HERE !!!

[WARRIOR]

Could you tell which paper these questions are from? it would make it easier.

its questions from my school book :/

[missbeautiful789]

For mass m₁:

Resultant force = mass x acceleration

The only force on m₁ is the tension on the string. Its weight is being "neutralised" by the reaction force of the table, so forget about its weight.
so
T=3xa

For mass M₂:

Its hanging, so the string is pulling it upward and its weight is acting downward. So its RF= T-mg
assuming g=10

T-20=2xa
T=2xa+20

T is the same so we can say:

3a=2a+20
a=20

B) final velocity=initial velocity + acceleration x time

v=u+at

from rest, so u=o

V=at
v=20x1.2
V=24

c) S=ut+0.5xaxt²

s=0.5x20x1.2²
s=14.4

d) s=0.5x20x0.4²
s=1.6

______________________________________________

2500=1400xa
a=1.79
_________________________________________


resolved force of weight down plane= mg sin@

v²=u²+2aS

625=2xax50

a=6.25

RF=ma        so

mg sin@=mx6.25
mx10xsin@=mx6.25
sin@=0.625
@=38.7 degrees

Looks like homework..don't forget the units

[WARRIOR]

For mass m₁:

Resultant force = mass x acceleration

The only force on m₁ is the tension on the string. Its weight is being "neutralised" by the reaction force of the table, so forget about its weight.
so
T=3xa

For mass M₂:

Its hanging, so the string is pulling it upward and its weight is acting downward. So its RF= T-mg
assuming g=10

T-20=2xa
T=2xa+20

T is the same so we can say:

3a=2a+20
a=20

B) final velocity=initial velocity + acceleration x time

v=u+at

from rest, so u=o

V=at
v=20x1.2
V=24

c) S=ut+0.5xaxt²

s=0.5x20x1.2²
s=14.4

d) s=0.5x20x0.4²
s=1.6

______________________________________________

2500=1400xa
a=1.79
_________________________________________


resolved force of weight down plane= mg sin@

v²=u²+2aS

625=2xax50

a=6.25

RF=ma        so

mg sin@=mx6.25
mx10xsin@=mx6.25
sin@=0.625
@=38.7 degrees

Looks like homework..don't forget the units

yes it was infact hw!

thanks ! + rep

that last one was pretty hard for me! even with your answers inftont of me

thanks again

[WARRIOR]

yo guys !

i have a tension test and a vector and projectiles test

if anyone has any exams or notes or question and can be kind enough to post them..CHEERS!

[Deadly_king]

For mass m₁:

Resultant force = mass x acceleration

The only force on m₁ is the tension on the string. Its weight is being "neutralised" by the reaction force of the table, so forget about its weight.
so
T=3xa

For mass M₂:

Its hanging, so the string is pulling it upward and its weight is acting downward. So its RF= T-mg
assuming g=10

T-20=2xa
T=2xa+20

T is the same so we can say:

3a=2a+20
a=20

B) final velocity=initial velocity + acceleration x time

v=u+at

from rest, so u=o

V=at
v=20x1.2
V=24

c) S=ut+0.5xaxt²

s=0.5x20x1.2²
s=14.4

d) s=0.5x20x0.4²
s=1.6

She made just a small mistake there.

Actually the tension is upwards towards the pulley, but the acceleration is downwards due to the weight of M₂.
Hence the equation should have been 20 - T = 2a

Now you can solve it with the first equation T = 3a and you'll be getting the acceleration to be 4ms^-2

b) All the other answers were affected since the acceleration was not correct. Otherwise the method was perfect

v = u + at ----> v = 0 + 4(1.2) = 4.8ms^-1

c) v² = u² + 2aS ----> 4.8²= 0 + 2(4)S
Hence the distance S is found to be 2.88m

d) Displacement is given by S = ut + 0.5at²
S = 0 + 0.5(4)(0.4)² ---> S = 0.32m

[missbeautiful789]

An acceleration of 20m/s!! down  , that should've raised a few flags. Its falling at twice g !!!  

Thanks for that DK
+rep

[Deadly_king]

An acceleration of 20m/s!! down  , that should've raised a few flags. Its at falling twice g 

Thanks for that DK
+rep

Hehe.............i guess you were not very attentive. Otherwise you would have noticed your mistake right away

No problem dear............just doing my job

[WARRIOR]

hello guys !
!
I need some notes on resolving vectors along wth questions and answers !

i have a test tomorrow !

sorry for the short notice~

[Dania]

Can someone please help me on this question on stationary waves. It's no. 4 (c)
Refer to the attached pdf.

[Deadly_king]

Can someone please help me on this question on stationary waves. It's no. 4 (c)
Refer to the attached pdf.

Sound with lowest frequency is produced => An antinode is found at the end of the tube and it is the only one.

Hence there should be a node at the piston. Distance between a node and the consecutive antinode is given to be lamda/4

Lambda/4 = 45 cm -----> lambda is given to be 180cm = 1.8m

V = f x lambda

Therefore f = v/lambda = 330/1.8 = 180 Hz

Frequency is given to the nearest tenth since v = 330 and lambda=1.8 (nearest 10)

Hope it helps

[Dania]

Thanks

Also, there's something I always struggle to grasp. There are questions where they give a diagram with a wave on a graph & tell you that there's another wave, giving you the phase angle of the latter one. I don't know if you get me. But what I'm trying to say is that how would I make the necessary calculations to enable me to draw the second wave? Like the time difference, etc. Would you happen to have a link that could help me?

Ref. October/November 2002, Paper 2, Q. (b) (i).
Thanks in advance.

[ruby92]

M/j 2010 physics p41
q2 part c(iii)

[ashish]

M/j 2010 physics p41
q2 part c(iii)

I will give you a tip try it

Internal energy = PEs + KEs

since ideal gas does not have intermolecular forces of attraction PE =o

Internal energy= KEs
                     = number of gas molecules * KE of one atom

[TJ-56]

Can sumone help me with diagrams in nov 2003 qp2, the first one is q3 (c) (i) just wanna make sure if its right and explain if you can pls,
the second one is q5 (c) why isn't it a straight line from origin, why  does it have a curve as it says on the mark scheme?
thanks so much for any help in advance

[Deadly_king]

Thanks

Also, there's something I always struggle to grasp. There are questions where they give a diagram with a wave on a graph & tell you that there's another wave, giving you the phase angle of the latter one. I don't know if you get me. But what I'm trying to say is that how would I make the necessary calculations to enable me to draw the second wave? Like the time difference, etc. Would you happen to have a link that could help me?

Ref. October/November 2002, Paper 2, Q. (b) (i).
Thanks in advance.

Yeah, I understand what you mean. Once it was rather complicated for me too.

You should know this formula => Phase angle = 2(pie)/lambda

This formula applies when you have a graph of x against wavelength(lambda). Since in this case the graph is x against time(t), we can change the formula to => Phase angle = 2(pie)/T, where T is the period of oscillation and is the time difference.

It has been said that both waves have the same waveform, which implies same wavelength and same period. The only difference is that one would lead the other.

Take 2(pie) as 360^o since the phase angle has been given in degrees and not radians.
Hence 60^o = 2(180)/3 ---> =  0.5s

Therefore the new wave will lag behind by 0.5s, i.e it will have its first maximum at 0.5s and the first minimum at 2s.

Hope you get it.

Am sorry but I don't have any specific sites for this topic, but here are some links which a member has been so kind to look for us.