Physics /12 Discussion!

[Obsidian Fl1ght]

It was Potential energy I think I had done it before#repeat

For some reason I remmbred the graph of PE which was contantly increasing.
Now I realise it was a graph of PE to displacement!
And how in the world can momentum of the ball only be conserved!?
I mean law of momentum states:
Momentum of a system remains constant provided no external forces are applied.

But u know what?
Maybe the choice regarded the ball as the system?

happens well i still managed to get less then 5 wrong till now :/

Curiosity is curious about this statement of urs.
May I know what questions u did get wrong?

 

[Obsidian Fl1ght]

Again, if anyone remmbrs the values provided in that - "air density" (No other way to put it), question, kindly tell.

 

[Malik777]

If the ball would be the system then momentum and energy conserved were both Valid -_-

For some reason I remmbred the graph of PE which was contantly increasing.
Now I realise it was a graph of PE to displacement!
And how in the world can momentum of the ball only be conserved!?
I mean law of momentum states:
Momentum of a system remains constant provided no external forces are applied.

But u know what?
Maybe the choice regarded the ball as the system?


Curiosity is curious about this statement of urs.
May I know what questions u did get wrong?

 

[h4rriet]

I'd like to know what logic u applied for this choice...
Coz albeit I agree with the energy conservation bit... should not the energy of the whole system be being conserved?
Not of the ball only?

No, the ball's OVERALL energy will remain constant throughout. K.E. will turn into G.P.E. and G.P.E.

 

[Malik777]

Again, if anyone remmbrs the values provided in that - "air density" (No other way to put it), question, kindly tell.

I did 480 #Tukka

 

[h4rriet]

Graph of the resistors, Answer: D
Elastic collision, I chose speed unchanged

I chose relative speeds of approach = relative speeds of separation.

 

[Malik777]

No, the ball's OVERALL energy will remain constant throughout. K.E. will turn into G.P.E. and G.P.E.

was there no external force acting? :S

 

[Obsidian Fl1ght]

I did 480 #Tukka

I'm there with u man.
I stared some... a few minutes left I guess... hurriedly stared some more... do random calculations or something... some formula... anything... think.... use those grey cells....
End result... C
aka 480N.
Whoop

 

[Obsidian Fl1ght]

I chose relative speeds of approach = relative speeds of separation.

Ditto.

 

[h4rriet]

was there no external force acting? :S

Don't remember. I believe air resistance was neglected.

 

[Obsidian Fl1ght]

No, the ball's OVERALL energy will remain constant throughout. K.E. will turn into G.P.E. and G.P.E.

You might have something there... now that we've established that PE (which I chose) is NOT the answer... hmm.
But then why not momentum?

 

[h4rriet]

You might have something there... now that we've established that PE (which I chose) is NOT the answer... hmm.
But then why not momentum?

The speed changes since the K.E. changes, so the momentum changes too.

 

[Malik777]

You might have something there... now that we've established that PE (which I chose) is NOT the answer... hmm.
But then why not momentum?

Yup agreed with you If Energy conserved hay then momentum should be conserved :O

 

[Malik777]

what was the answer for that Graph :S ? #bulb I did C :/

 

[h4rriet]

what was the answer for that Graph :S ? #bulb I did C :/

Which graph?

 

[Malik777]

Which graph?

Non uniform scale? :S

 

[UziB]

The answer for the force exerted when the wind hits the wall was 16000N.
CALCULATIONS: Force exerted by air was supposed to be calculated from newton's second law of momentum which is:
F = ϼAv^2
F = 1.2 x 12 x (33)^2
F = 15681N ≈ 16000N hence option D.

The answers for h1 and h2 of the mercury columns were 4 and 2.
CALCULATIONS: They asked the possible values of h1 and h2, not EXACT values.
Density of mercury is 13,600 kg/m3
Pressure = ρgh
h1 = 16000/13,600 x 9.81 = 0.12m
h2 = 8000/13,600 x 9.81 = 0.06m
Ratio
h1 : h2 = 2 : 1
In option A the values were 4 : 2 which is 2 : 1 hence the answer was A.

In a perfectly elastic collision the relative speed of approach is equal to the relative speed of seperation.

For a ball moving upwards with air resistance being neglected, the total energy of the system remains conserved according to the law of conservation of energy.


In question 3 the lines of actions of forces met at the centre, hence resultant force was coming out from the centre towards the 4N force (as resultant always acts closer towards component of greater magnitude).

These are confirmed answers.

 

[afoo.666]

I dont remember any shit. Fml

 

[Malik777]

Q 3 was B confirmed Bro when the two forces perpendicular to each othr are acting on a same point away from the point the resultant would be by parallelogram rule hence by calculations it would b 5 N

The answer for the force exerted when the wind hits the wall was 16000N.
CALCULATIONS: Force exerted by air was supposed to be calculated from newton's second law of momentum which is:
F = ϼAv^2
F = 1.2 x 12 x (33)^2
F = 15681N ≈ 16000N hence option D.

The answers for h1 and h2 of the mercury columns were 4 and 2.
CALCULATIONS: They asked the possible values of h1 and h2, not EXACT values.
Density of mercury is 13,600 kg/m3
Pressure = ρgh
h1 = 16000/13,600 x 9.81 = 0.12m
h2 = 8000/13,600 x 9.81 = 0.06m
Ratio
h1 : h2 = 2 : 1
In option A the values were 4 : 2 which is 2 : 1 hence the answer was A.

In a perfectly elastic collision the relative speed of approach is equal to the relative speed of seperation.

For a ball moving upwards with air resistance being neglected, the total energy of the system remains conserved according to the law of conservation of energy.


In question 3 the lines of actions of forces met at the centre, hence resultant force was coming out from the centre towards the 4N force (as resultant always acts closer towards component of greater magnitude).

These are confirmed answers.

 

[Obsidian Fl1ght]

The answer for the force exerted when the wind hits the wall was 16000N.
CALCULATIONS: Force exerted by air was supposed to be calculated from newton's second law of momentum which is:
F = ϼAv^2
F = 1.2 x 12 x (33)^2
F = 15681N ≈ 16000N hence option D.

The answers for h1 and h2 of the mercury columns were 4 and 2.
CALCULATIONS: They asked the possible values of h1 and h2, not EXACT values.
Density of mercury is 13,600 kg/m3
Pressure = ρgh
h1 = 16000/13,600 x 9.81 = 0.12m
h2 = 8000/13,600 x 9.81 = 0.06m
Ratio
h1 : h2 = 2 : 1
In option A the values were 4 : 2 which is 2 : 1 hence the answer was A.

In a perfectly elastic collision the relative speed of approach is equal to the relative speed of seperation.

For a ball moving upwards with air resistance being neglected, the total energy of the system remains conserved according to the law of conservation of energy.


In question 3 the lines of actions of forces met at the centre, hence resultant force was coming out from the centre towards the 4N force (as resultant always acts closer towards component of greater magnitude).

These are confirmed answers.

Okay.
The formula used in the wind question... how did u get it from Newton's 2nd Law?