Physics: Post your doubts here!

[daredevil]

thanks a lot bro, Jazak Allah khairn

np

 

[Hassan Ali Abid]

for elastic collision momentum b4 = momentum after ......

assume that backward direction ie .ball moving on left hand side is +ve ....then momentum b4 collision is mu2 - (-mu1) and after collision is this : mv2 - mv1

as it is stated that it is elastic so both are equall u2-u1 = v2-v1 ........so ans is 'D'

ps do check its ans from ms .....idk whats its ans but this is the wayhow we solve this

 

[daredevil]

HI
does anyone know how 2 do this?
9702_s12_qp_12
question 14
thnx in advance!

keep in mind that the ground is frictionless and when the chair moves down the surface it is resting on will be reacting with forces in the opposite directions to the ones exerted by the ladder
so :
- the ladder when sliding down exerts a downward and 'on the wall force' so the wall exerts an upward force and a force perpendicular to the wall coming out so the resulting force P on the point will be towards top right.
- the weight W will be downwards
- the force exerted by the ground will only be an upward force in reaction to the downward force exerted by the ladder on the ground. no friction force will be acting here as in the start of the question they have mentioned that it is a FRICTIONLESS surface.

 

[Hassan Ali Abid]

how to solve q13 ..m/j12 (12) ?

 

[Scafalon40]

your absolute error should be correct to 1s.f. and then the value should be correct to the same no. od d.p. as ur resulting absolute error is.
like if u hav calculated a length to be 20.324 +/- 0.346cm

then ur correct absolute error should be +/- 0.3
and thus the resulting value should be 20.3 as the absolute error comes to 1 d.p.


for another case you have the speed to be 326.55 +/- 1 m/s
then ur answer should be:

327 m/s
because the absolute errror is correct to 0 s.f.

even if the value does not have a number after the decimal but the absolute error is correct to 1 d.p. or so u write abc.0 insead of just writing abc

hope u get it
1357913579 u might wanna check this out ... maybe it will help ^_^

Hey thanks dude! I'd totally forgotten about this stuff, even though I've already taken my AS papers. Thanks for the refresher! I owe you one

 

[Ahmedraza73]

HI
does anyone know how 2 do this?
9702_s12_qp_12
question 14
thnx in advance!

The Answer will be B that i think so,PLEASE do tell me if i am right?
A: In A weight is acting downward and the normal reaction force at the end of ladder is upward and the force P is the normal reaction force by the wall in straight direction.Which means the ladder is not sliding and on the equilibrium postion
B: In B weight is acting downward and the normal reaction force at the end of ladder is upward and the force P force P isacting slightly upward because the ladder is sliding .
C: In C weight is acting downward again and the normal reaction force at the end of ladder is now not acting against the direction of weight it is slighlt inward.
D : In D weight is acting downward again and the normal reaction force at the end of ladder is again not acting against the direction of weight it is slightly Outward
Thus the Answer will be B

 

[Hassan Ali Abid]

The Answer will be B that i think so,PLEASE do tell me if i am right?
A: In A weight is acting downward and the normal reaction force at the end of ladder is upward and the force P is the normal reaction force by the wall in straight direction.Which means the ladder is not sliding and on the equilibrium postion
B: In B weight is acting downward and the normal reaction force at the end of ladder is upward and the force P force P isacting slightly upward because the ladder is sliding .
C: In C weight is acting downward again and the normal reaction force at the end of ladder is now not acting against the direction of weight it is slighlt inward.
D : In D weight is acting downward again and the normal reaction force at the end of ladder is again not acting against the direction of weight it is slightly Outward
Thus the Answer will be B

how to solve Q13 of same ppr ????

 

[daredevil]

how to solve q13 ..m/j12 (12) ?

calculate the weight acting from the 2 kg block because that is the force moving the 8 kg block too:
2*9.81 = 19.62 N

the resultant force on the block would be this force minus the frictional force:
19.62 - 6 = 13.62 N

thus after calculating the resultant force, now calculate the acceleration with the formula F=ma :
a = F/m
= 13.62/8 = 1.7 m/s^2

is that the ryt answer ?? ^_^
if u don't get it then ask away

 

[Reh_456]

The Answer will be B that i think so,PLEASE do tell me if i am right?
A: In A weight is acting downward and the normal reaction force at the end of ladder is upward and the force P is the normal reaction force by the wall in straight direction.Which means the ladder is not sliding and on the equilibrium postion
B: In B weight is acting downward and the normal reaction force at the end of ladder is upward and the force P force P isacting slightly upward because the ladder is sliding .
C: In C weight is acting downward again and the normal reaction force at the end of ladder is now not acting against the direction of weight it is slighlt inward.
D : In D weight is acting downward again and the normal reaction force at the end of ladder is again not acting against the direction of weight it is slightly Outward
Thus the Answer will be B

yep it is

 

[Hassan Ali Abid]

calculate the weight acting from the 2 kg block because that is the force moving the 8 kg block too:
2*9.81 = 19.62 N
the resultant force on the block would be this force minus the frictional force:
19.62 - 6 = 13.62 N
a = F/m= 13.62/8 = 1.7 m/s^2

bro i also calculated like this but the ans is 1.4 ...option a .....

 

[daredevil]

bro i also calculated like this but the ans is 1.4 ...option a .....

what?? howww!!??!! :O

 

[Hassan Ali Abid]

what?? howww!!??!! :O

idk....just check its ms ......the ans is A

 

[daredevil]

idk....just check its ms ......the ans is A

yeah i just saw.... if anyone tells the answer plz tag me in it becuz i reaallllyy wanna know .....

 

[Hassan Ali Abid]

yeah i just saw.... if anyone tells the answer plz tag me in it becuz i reaallllyy wanna know .....

i wasted about 1 hr on this single question but was unable to calculate Ans A

 

[RadzMau]

Can someone pls help me with this question?
Question: A motorist travelling at 10 m/s can bring his car to rest in a distance of 10 m. If he had been travelling at 30 m/s, in what distance could he bring the car to rest using the same braking force?

 

[Ali_00921]

Distance between 2 successive heaps (minimum points) = λ/2

39cm covers 5 times of this distance, therefore:
5(λ/2) = 39 cm
λ = 15.6 cm = 15.6 x 10^-2 m

v = fλ
v = 2.14 x 10^3 x 15.6 x 10^-2 = 334 ms-1

Din't get the λ part..

 

[Dug]

how to solve q13 ..m/j12 (12) ?

T - 6 = 8a --- i
20 - T = 2a --- ii
We want to find acceleration so eliminate T:
8a + 6 = 20 - 2a
10a = 14
a = 1.4 ms-1

 

[Ahmedraza73]

bro i also calculated like this but the ans is 1.4 ...option a .....

Sorry dude i made a big mistake and not considered the Tension in my first answer: really apologize for tht:
here is the correct solution:
W=mg=2g(for the hanging part)
EQAUATION 1 : 2g-T=2a
Equa 2: T-6=8a
ADDING BOTH EQUATION WE GET:
2g-6=10a ---equation 3
puttin values in equation 3 we have:
2(9.81)-6/10=1.362 ROUNDED OFF TO 1.4m/s ^2

And thanks brood to helping me:

 

[Ahmedraza73]

yep it is

Did u understand?

 

[sweetjinnah]

plz i wanted explanations of the following mcqs:
o/n 2008 mcq 4 , 32 :
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf
m/j 2008 mcq 11 , 16 :
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_1.pdf
o/n 2007 mcq 23 , 25 :
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_1.pdf
o/n 2009 mcq 12 , 13 , 25 :
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_12.pdf