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Physics: Post your doubts here!

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Ahmed Ali Akbar

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may june 07 p1 mcq 18,and mcq 10..explain why in mcq 18 option D is incorrect....and why in mcq 10 D is is correct...
 
D

Doctor Nemo

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Ahmed Ali Akbar said:
may june 07 p1 mcq 18,and mcq 10..explain why in mcq 18 option D is incorrect....and why in mcq 10 D is is correct...
Click to expand...

Q 10
From the syllabus you are supposed to know that Newt´s second law is that the force on an object is the rate of change of momentum. When asked to define force this is the definition that you are supposed to give and not F = mass x acceleration. So the average force = average change of momentum is the (final momentum – initial momentum )/ (change in time).

So the answer is either B or D. Given the definition the answer is B. The final momentum is negative however when you substitute into a formula you don´t just substitute the magnitude you also substitute the sign. In the report this was listed as a difficult problem presumably because of confusion about the sign. So the lesson from this problem is that when dealing with letters and not actual numbers you do not change the sign in the formula even though you know that the value will be negative.

Q 18

D is not correct because work is not being done on the rubber while the load is removed. Perhaps you confused the word sample and thought it referred to the load and not the rubber. The wording is confusing, perhaps intentionally so. The lesson here, and it is a very important one, when you think that two of the choices are correct then go back and reread the question more carefully.
 
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Doctor Nemo

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Pals_1010 said:
http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9702 - Physics/&file=9702_s09_qp_1.pdf

Help with No 15 and No 18 please?
Click to expand...
15)

Two ways to look at problem.

The top half of the water on the left fell to the other side. The essence of solving this problem is to remember that the position and motion of any mass can be reduced to that of its centre of gravity.

The centre of gravity of the top half of the water is 1/4 h above the bottom half. When it falls to the other side the centre is now 1/4 h above the bottom. So the centre of gravity fell 1/2h. Since 1/2 of the mass fell 1/2 h the loss of gpe is 1/2h X 1/2 mass X g or 1/4mgh.

Alternatively consider that at the start the centre of mass of the entire mass is 1/2 h above the bottom. Afterwards it is 1/4h above the bottom. So this is the same as the entire mass falling 1/4 h so loss of gpe is 1/4h X mass X g.

18)

For manometers it is the difference in height between the two sides that is a measure of the pressure difference not just how much they move from the position when the pressure on the two sides are the same. Since the side on the right moved up h the side on the left moved down h. The difference in the two levels is 2h so the pressure difference is given by D.
 
Pals_1010

Pals_1010

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Doctor Nemo said:
15)

Two ways to look at problem.

The top half of the water on the left fell to the other side. The essence of solving this problem is to remember that the position and motion of any mass can be reduced to that of its centre of gravity.

The centre of gravity of the top half of the water is 1/4 h above the bottom half. When it falls to the other side the centre is now 1/4 h above the bottom. So the centre of gravity fell 1/2h. Since 1/2 of the mass fell 1/2 h the loss of gpe is 1/2h X 1/2 mass X g or 1/4mgh.

Alternatively consider that at the start the centre of mass of the entire mass is 1/2 h above the bottom. Afterwards it is 1/4h above the bottom. So this is the same as the entire mass falling 1/4 h so loss of gpe is 1/4h X mass X g.

18)

For manometers it is the difference in height between the two sides that is a measure of the pressure difference not just how much they move from the position when the pressure on the two sides are the same. Since the side on the right moved up h the side on the left moved down h. The difference in the two levels is 2h so the pressure difference is given by D.
Click to expand...

Thank you ^^
 
CaptainDanger

CaptainDanger

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The rotating blades of a hovering helicopter sweep out an area of radius 4.0 m imparting a downward velocity of 12 m/s to the air displaced. Find the mass of the helicopter.(density of air = 1.3 kg/m^3)?

Answer:
(9.4 x 10^2 kg)
 
D

Doctor Nemo

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CaptainDanger said:
The rotating blades of a hovering helicopter sweep out an area of radius 4.0 m imparting a downward velocity of 12 m/s to the air displaced. Find the mass of the helicopter.(density of air = 1.3 kg/m^3)?

Answer:
(9.4 x 10^2 kg)
Click to expand...

The weight of the helicopter must equal the downward force of the blades on air. So we need to calculate the force of the blades on air. We cannot use F = ma because we do not have acceleration and cannot calculate it from the information given in the problem. So we use instead F=rate of change of momentum where we assume that a mass of air equal to that of a cylinder with radius of 4 m and height of 12 m is given a velocity of 12 m/s every second.

(4^2 X pi X 12 X1.3) X12 = 9409 N. Now for this problem g = 10.0 and mass is equal to 9.4 X 10^2.

For reasons that I have not yet been able to figure out if you use the formula P = Fv assuming that the same mass is given a velocity of 12m/s every second and calculate power based on the resultant K.E you get one half the value. However in these types of problems momentum rules.
 
CaptainDanger

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Doctor Nemo said:
The weight of the helicopter must equal the downward force of the blades on air. So we need to calculate the force of the blades on air. We cannot use F = ma because we do not have acceleration and cannot calculate it from the information given in the problem. So we use instead F=rate of change of momentum where we assume that a mass of air equal to that of a cylinder with radius of 4 m and height of 12 m is given a velocity of 12 m/s every second.

(4^2 X pi X 12 X1.3) X12 = 9409 N. Now for this problem g = 10.0 and mass is equal to 9.4 X 10^2.

For reasons that I have not yet been able to figure out if you use the formula P = Fv assuming that the same mass is given a velocity of 12m/s every second and calculate power based on the resultant K.E you get one half the value. However in these types of problems momentum rules.
Click to expand...
Thank you! :)

Find the force exerted on each square meter of a wall which is at right angle to a wind blowing at 20 m/s. assuming that the air does not rebound. (density of air = 1.3 kg/m^3)? (5.2 x 102 N)
 
abcde

abcde

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CaptainDanger said:
Thank you! :)

Find the force exerted on each square meter of a wall which is at right angle to a wind blowing at 20 m/s. assuming that the air does not rebound. (density of air = 1.3 kg/m^3)? (5.2 x 102 N)
Click to expand...
AoA!
Here's my attempt. Maybe Doctor Nemo can improve it. :)
Force exerted per square meter = Pressure = Force / Area
Force is the rate of change of momentum so this translates into = m(v - u)/area
mass = density x volume.
=> Pressure = 1.3 x volume x (20 - 0)/ area
= 1.3 x area x 20 x 20 / area (based on volume = area x height, where 'height' is the distance covered in one second. Hope I'm right.)
= 5.2 x 10^2 Pa
So, per square meter, a force of 5.2 x 10^2 N is exerted.
 
L

leosco1995

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The rotating blades of a hovering helicopter sweep out an area of radius 4.0 m imparting a downward velocity of 12 m/s to the air displaced. Find the mass of the helicopter.(density of air = 1.3 kg/m^3)?
Click to expand...

Very interesting question.. requires a lot of thought.

Area = (4)²(π) = 16π
g = 9.81m/s²
v = 12m/s
Density of air = 1.3kg/m³
mass = ?

To find the mass, you will need the density and volume of air displaced. The density is already given, but you will need to find the volume.

You can do this simply by multiplying the area with the downward velocity. The volume of air displaced is (area x downward velocity) = 16π x 12 = 192π. Density is 1.3, so the mass of air here is 249.6π.

Now since momentum = rate of change of force, the force exerted on the air by the blades is 249.6π x 12 = 2995.2π.

Since the helicoptering is hovering, weight of the helicopter = upward force. We have the upward force, and W = mg so...

2995.2π = m x 9.81

My value of m comes out to be 959.2 kg, but if I assume g to be 10m/s² I get 941 kg which is probably the right answer.
 
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leosco1995

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Nibz said:
" When two objects collide and stick together, the collision is inelastic (perfectly inelastic is the term)." This is why D option is incorrect. As simple as that.
Click to expand...
...Oh, I didn't read the last part of the question. But anyway, would my reasoning be wrong?
 
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Nibz

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Only approximately. Because in elastic collision, both kinetic energy and momentum are conserved. But when snooker balls hit each other, there is some loss of kinetic energy.
 
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