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Physics: Post your doubts here!

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sagar65265

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Young Stunner said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf


Q25, 26, 33, 35
Click to expand...

Q25: The image of a wave has been given, and so has the direction it is traveling in.

Note that since this is a transverse wave and the disturbances are traveling from left to right, the particles at any point are vibrating in a direction perpendicular to that propagation.
In other words, if you focused on any one segment of the rope and that one segment alone (let's say by painting it a different color from that of the rope), then you would find it does not experience any sideways displacement; it only moves up, then down, then up, then down, etc.

At the same time, the waveform shifts towards the right, since that is the direction in which it is traveling.
So just imagine the waveform displayed in the question shift to the right; imagine each part of the waveform progress, and you would then see that P is moving downwards; since the trough just before P has to travel to the right, P has to "fall" into that trough, and if it "falls" into that trough, it is moving downwards.

Of course, this eliminates all the options in the question, leaving behind only A, the correct answer, but it is worth discussing the motion of Q:

At the instant of the displayed image, Q is a segment at the crest of the waveform; it is debatable that after some time, Q has to fall towards the equilibrium position, but taking a closer look at the wave motion (and seeing that we can estimate the motion of any one segment as Simple Harmonic Motion, SHM) it turns out that Q would be stationary.

It's just like the motion of a pendulum; at any extreme, the rate of change of it's position - it's velocity - is zero, but due to the forces acting on the pendulum (in THIS case, the force is the tension in the string) this changes slowly to make it change position. So that's most likely why the velocity of Q is zero at the instant shown.

Q26: There are two formulae we will need to use here; one relates Intensity of a wave to it's Amplitude, the other relates the Power of a wave to it's Intensity.

First, let's fix up an equation for the initial state/ situation (I = Intensity of Wave , A = Amplitude of Wave , k = arbitrary constant):

I = k * A^2

The other equation tells us (P = Power of Wave , S = Area component perpendicular to wave propagation direction):

P = I * S

Substituting the value of I from the first equation into the second,

P = k * A^2 * S

So all we have to do now is form two equations (one for each situation) and compare the two.
Initially,
P(i) = k * A^2 * S
Finally,
P(f) = k *4A^2 * 0.5S
Dividing the second equation by the first,
P(f) / P(i) = 2
Since P(i) = E,
P(f) = 2E =B

Q33: Seriously, this one is a doozy :) the wording is not very nice, to be honest, and it makes the question rather confusing.
What it actually asks for (as far as I can tell) are the variables required to find the different between OPEN circuit voltage and CLOSED circuit voltage. In other words, the difference between the potential difference measured when the circuit is OPEN and when it is fully connected with an external resistor.
(I thought it spoke about the normal decrease in voltage with running duration, which can be expected as a battery runs down. My mistake!)

A Google Images search for "internal external resistance" provides some very clear images, and i'm using the attached one for reference.
IMG635.JPG
When the circuit is open, no current flows through the battery, and the initial reading the equivalent to the EMF of the battery; no energy is lost per second to heat in the internal resistance, so it is equal to the EMF.
Let's write this down as V.

Once the circuit is connected as above, a current indeed does flow through all components, and a potential difference is maintained across each too. However, since the battery consists of an internal resistance (this internal resistance is NEVER separate from the battery; it can be considered as such, but the battery consists both of the cells and the internal resistance), there is a drop in potential across this internal resistance before the current even leaves the battery. Thus, the potential difference across the terminals of the battery drop.

Applying Kirchoff's Second Law while moving from B to A, what we get is (V is still the EMF, r(internal) is the internal resistance):

(Potential at A) + V - i * r(internal) = (Potential at B)

All this says is that the potential at A is one thing, it changes by so and so amount, as a result of which is becomes another value, the Potential at B.

So, since the Potential Difference is the (Potential at B) - (Potential at A), we have:

PD = V - i * r(internal)

Thus, the drop in potential is

V - (V- i * r(internal)) = i * r(internal)

The change is negative, but the decrease is positive, so the signs change there. Therefore, all you need to find the drop in potential are the final current, and the internal resistance of the battery = C

Q35: Assuming that the ammeter and the voltmeter are both ideal, the resistance of the ammeter is Zero and the resistance of the voltmeter is Infinity. Following on from there, the potential difference across the ammeter is Zero and the current through the Voltmeter is Zero.

So these don not affect the circuit in any way except to give us a reading. When the variable resistance is dropped, applying Kirchoff's Second Law gives us:

i = V/(∑R) where ∑R is the equivalent resistance of the circuit, in this case the algebraic sum of the two resistance values.

Since the overall resistance of the circuit decreases, the current automatically decreases and the answers are narrowed down to C or D.

Furthermore, since we have

V = I * R

and the current across the fixed resistor increases while the resistance remains the same,

V(f) = I(f) * R

Dividing the second equation by the first, we get V(f) = [I(f) / I] * V.
Since [I(f) / I] > 1, V(f) > V and so the reading on the voltmeter also increases, telling us the answer is D.
Note: If the voltmeter was across the Variable Resistor, our calculation would have been a little longer, since the increase in Current and the decrease in Resistance might *possibly* make up for each other and result in Zero Change in the voltmeter reading.

Hope this helped!
Good luck for all your exams!
 
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zonaali

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please someone help me with this one...answer is A. Examinar has mentioned that Both masses are being accelerated, so the total mass being accelerated here is 3 kg not 2 kg
 

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zonaali said:
please someone help me with this one...answer is A. Examinar has mentioned that Both masses are being accelerated, so the total mass being accelerated here is 3 kg not 2 kg
Click to expand...

The two most important things to note for this question are as follows:

i) Since the string is "mass-less", the tension in every part of the string is the same. In other words, the string pulls each block with the same force; it pulls the 2.0 kg block to the right with the same force as it tries to pull the 1.0 kg block upwards.

ii)Since the string is assumed to be in-extensible (in other words the length of the string cannot change, it cannot stretch at all) the acceleration magnitude of the two blocks are the same; if the 1.0 kg block moves a small distance downwards, the 2.0 kg block has to move the same distance in the horizontal direction. So, since both of them have to fall the same distance in the same time interval, they have the same velocity. From here, since any change in the speed of the 1.0 kg block during a particular time interval has to result in the SAME change in speed of the 2.0 kg block during the EXACT SAME time interval, the two accelerations at any point are the same.

Let's define the directions, first: for the 1.0 kg block the positive direction is downwards and the negative direction is upwards (it doesn't travel or accelerate sideways, so we don't need to worry about that direction). Also, for the 2.0 kg block, the positive direction is to the right and the negative direction is to the left.
We've chosen the axis so that the accelerations are all positive in sign, so that we don't have to juggle around any confusing negative values.

So, taking Newton's second law for the 1.0 kg block (Net Force = Mass * Acceleration):

(1.0 kg)(9.81 ms^-2) - T = (1.0 kg)a
9.81 - T = a

Alright, so that's the first equation. Considering the 2.0 kg block:

T = (2.0 kg)a
T = 2.0a

And there's the second one. Since there is no friction acting on the 2.0 kg block, that's it!

Now, since the two accelerations and the two tension forces are the same, we can substitute the second equation into the first as:

9.81 - 2.0a = 1.0a
9.81 = 3.0a

Therefore, a = 9.81/3 = 3.27 ms^-2

Since the acceleration here is constant and does not change with time, we can use the constant acceleration kinematics equations with:
u = 0 ms^-1
a = 3.27 ms^-2
s = 0.50 m (since both blocks travel this distance before the 1.0 kg block hits the ground and stops)
v = ?

v^2 = u^2 + 2 * a * s
v^2 = 0 + 2 * 3.27 * 0.5
v^2 = 3.27

Therefore, the final speed is √(3.27) = 1.8(08) ms^-1 = 1.8 ms^-1 = A.

Hope this helped!
Good Luck for all your exams!
 
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Salman917 said:
when will be the force of gravitation between 2 bodies will be zero n y?????????????
Click to expand...

It depends on the two bodies involved.

For example, if you have a circular ring (or a symmetrical one, it can even be an oval for that matter) somewhere in space and a particle is placed at it's center, the gravitational force on it is Zero. As that particle is moved away from the ring, the force of attraction increases at first, then decreases, and decreases until it reaches Zero at Infinity.

This is because, at the center of the ring, all the matter surrounding that point is symmetrically distributed; since segment of the ring exerts the same force on any particle at that center-point and the entire construct is symmetrical, you basically get pairs of forces cancelling each other out (segments of the ring at diametrically opposite points exert a net force of Zero; since the entire construct is like this, the net force there is also Zero).

As the particle moves away, the forces decrease due to increasing distance; the gravitational force follows an inverse-square law, so the force decreases with distance, and it decreasing very fast.

Another (very non-intuitive) example is a spherical shell; suppose you have a spherical shell (a hollowed out sphere would be a good way of describing it) and inside that shell you place a particle, it doesn't matter where you place that particle; as long as that particle is inside the hollow, the gravitational force on it will be Zero. There's very little physical explanation for that I can think of, aside from "the calculus works out in that way".
The beginning of this HyperPhysics page gives a lovely explanation that's really worth reading, and the calculus after that should give you a good idea of just how powerful the methods of calculus are, although that derivation is DEFINITELY NOT needed for CIE AS & A Levels:

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell2.html

And this last one applies for ANY two or more objects - as the distance between them increases, the gravitational forces exerted by each on the other also decrease, thus reaching (but never equaling) zero at a separation distance of Infinity. Usually it is so negligible at large values of separation distance that it has no tangible effect and can be stated as practically zero.
So, it's rather interesting to imagine that a star several dozens times heavier than our sun, in some far-away galaxy, is exerting a gravitational force on you and every particle around you, but is too weak to actually show!

Hope this helped!
Good Luck for all your exams!
 
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Thought blocker

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ZaqZainab said:
1 would be D because of the formula = Efficiency= Useful Output/Input
2 would be C from the formula work done= Pressure*change in volume this formula is also given in the data booklet
change in volume is (V1-V2)
Click to expand...
In 1) Efficiency = (output/input)*100 so, where is *100 ?? :(
 
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