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Physics: Post your doubts here!

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sagar65265

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unique111 said:
19) use the formula P=hdg where g=9.81 ... D comes out to be 1936.8=1900.
26) use dsin(theta)=n lambda. Where d=1/N and N=300,000 lines/m. Solve the equation 1=nlambda/d and n=7. This is the number of order, total maxima is double this plus 1=15 ( there is one more maxima in the zero order).
28) f=mg and f=Eq equate these two equations and charge/mass will be g/E. Don't know its polarity though.

* how did you do q.13? I got the torque, but what about the tension?
Click to expand...

Well, if the torque applied to Q is 3.0 Nm, and the only force applying the torque on Q is the upper part of the belt, then the force applying that torque should be the tension in the belt. So, since the torque = (Tension) * (Distance from rotation axis), we can say that

(Tension) = 3.0 Nm/(50/1000) = 60 Newtons tension.

This same tension force (since the tension in a massless string/ belt/ any other thread-like substance is constant throughout)pulls on P, and applies a torque there of magnitude

(Tension) * (Radius of the wheel) = (60 Newtons) * (75/1000) = 4500/1000 = 4.5 Nm = D.
Actually speaking, i'm not sure which part you didn't get - there's only one option with the right Tension, so were you asking concerning the answer or concerning the actual calculation to get the torque?

Good Luck for all your exams!
 
Zepudee

Zepudee

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sagar65265 said:
View attachment 45301

Since the pattern displayed as λ above repeats itself, that length is equivalent to the time period of the wave (since the horizontal direction is the time-base axis).
This is less that 1 centimeter, and actually measuring the length on the paper would give you an answer somewhere close to 0.6 cm, 0.7 cm or 0.8 cm - whichever one it is, we can use that for the calculation.

If 1 cm represents 10 ms = 10/1000 = 1/100 seconds, the 0.6, 0.7 or 0.8 centimeters represent 0.6/100, 0.7/100 or 0.8/100 seconds respectively.
So, if the time period of the wave is one of the above durations, then the frequency must be equal to 1/T, which turns out to be:

(100/0.6) = 166.6 Hertz, if the time period is 0.6/100 seconds.
(100/0.7) = 142.8 Hertz, if the time period is 0.7/100 seconds.
(100/0.8) = 125.0 Hertz, if the time period is 0.8/100 seconds.

So, whatever the answer time period is, the closest option appears to be 140 Hz, which is equal to B, our final answer.

Hope this helped!
Good Luck for all your exams!
Click to expand...

Thank you so so much sagar!!!! All the best to you too!
 
ZaqZainab

ZaqZainab

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Zepudee said:
But be careful here - all these changes are so simple because we are going in the direction of current. Suppose you go across a resistor with this method and you are going in the direction opposite to the current, then you will have an increase in potential, not a decrease.
About this, can you give me an example? :)
sagar65265 ZaqZainab
Click to expand...
I have a different way of doing that question Idk what he means by that
Heres how i do it
how many paths can be taken?
there can be
+LM-
+LNQ-
+PQ-

For +LM- 7+M=20
and so M=13
for +LNQ- 7+4+Q=20
and so Q=9
for +PQ- P+9=20
P=11
 
Venkata Gurrala

Venkata Gurrala

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sagar65265 said:
About your earlier query, I have no idea what you said "yeah" for, so here's an answer anyways ;).
Either ways, the book shows the graphs for different materials, and since different materials have different behavioral patterns under tension and compression, the graphs for different materials vary. So, the graph for an elastic band cannot be expected to conform to those patterns.

However, experience and assumption says that as you stretch a rubber band, it starts becoming more and more difficult to extend - in other words, after some time, a small extension will require a large force. In other words, after some time, the gradient of the graph (gradient = [change in applied force]/[change in extension]) increases.
The only graph that shows this kind of pattern (where after some extension the gradient of this graph increases) is A.

Q9)

First concept here - both energy and momentum are conserved in an elastic collision. So the kinetic energy and momentum and the colliding bodies before the collision are the same as the kinetic energy and momentum after the collision.

One more important characteristic of an elastic collision - the relative velocity between the interacting bodies is the same before the interaction as after. In other words, the rate of change of the distance between them is the same.

So, let's take D first, since it seems to be an outlier - it is an odd one out here, since it is the only one where both colliding objects move off together, so this is one we can easily eliminate as follows:

Initially, the ball moving right has kinetic energy 1/2 * (2m) * u² = mu².
The ball moving to the left has kinetic energy 1/2 * (m) * u² = 0.5 * mu².
Adding these up, the initial kinetic energy is equal to mu² + 0.5 * mu² = 1.5 * mu².

Now consider the situation after the collision. The mass of the moving object is 2m + m (since the velocities are the same and the objects stick together - you can take each separately and use the same speed to calculate their K.E., but it gives the same answer) = 3m. The option shows us a final velocity of (u/3).

So, the final K.E. = 1/2 * (3m) * (u/3)² = 1/2 * 3m * u²/9 = mu²/6.
This is not the value of the initial Kinetic Energy, so we can assume energy is lost somewhere or the other - so, this is not an elastic collision.

Let's do C next. The Initial Kinetic Energy, as calculated above, is 1.5 * mu².

In the final situation, the ball of mass (2m) has a velocity of (u/6), so it's K.E. = 0.5 * 2m * (u/6)² = m * u²/36 = mu²/36.
Also, the ball of mass (m) has a velocity of (2u/3), so it's K.E. = 0.5 * m * (2u/3)² = 0.5m * (4u²/9) = 2mu²/9 = 8mu²/36.
Adding these up, we get mu²/36 + 8mu²/36 = 9mu²/36 = mu²/4.

This is, again, not the same as before. So C is also out.

B is next - you can see immediately that this option is wrong, since it has the same speeds and masses as C - the (2m) mass is moving at (u/6) and the (m) mass is moving at (2u/3), which is the same combination we proved wrong before. Therefore, the only remaining option and our final answer is A.

Q14) When the projectile reaches it's highest point, it has a purely horizontal velocity and no vertical component of velocity.
Additionally, since there is no force on the projectile in the horizontal direction, it has the same horizontal velocity at it's highest point as it did when it was shot.

A little more on this - suppose the initial velocity has a magnitude of "v", the initial kinetic energy E = 1/2 * m * v² = 0.5mv².

Now, onto the focus of the question: since the initial velocity points at an angle 45° to the horizontal, the component of velocity in the horizontal direction should be

v(horizontal) = v * cos(45) = v/(√2)

This is, by our discussion above, the velocity of the projectile at it's maximum height. So, the kinetic energy at it's greatest height is

E(f) = (1/2) * m * (v/(√2))² = 0.5m * v²/2 = 0.25mv².

Let's divide E(f) by E, giving us

E(f)/E = (0.25mv²)/(0.5mv²) = 1/2
Therefore, E(f) = (1/2)E = 0.5E = A.

I'll post the rest after a while, let me know if you have any doubts.
Click to expand...
WOW! thank you so much. I understand it. Its kinda ironic how you still teach me stuff when we are so far away. Thanks again for your time!
 
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Venkata Gurrala said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf
Need help with Q9,14,15,19,26, and 28. Its pretty frustrating.
Click to expand...

Q15)

First things first, the law of conservation of momentum says:

"....the momentum of any system shall remain constant and unchanging if and only if no net external force acts on that system."

So, in most situations in physics, the easiest way to tackle a process is to isolate those parts of the process that, taken together, have no net force acting on them. This question serves a lovely example.

The floor is assumed to be frictionless - even otherwise, we are only concerned with the initial velocity, which friction does not necessarily influence directly. Regardless, let's continue with the assumption that the floor is frictionless (if it did exert frictional forces, we would have received some sort of information regarding that force).

Suppose we take the process to be the "moving apart of the two trolleys as the spring extends to it's natural length".
Taking the trolley on the left, there is a force acting on it - the force exerted by the spring when the system is released to it's natural action. This force is why the momentum of the trolley on the left changes, and so we should not take this trolley as the system if we want the simplest answer.

The same applies for the trolley on the right - the spring exerts a force on it in the right-hand direction, which causes an increase in momentum in the right-hand direction for the trolley on that side.

The spring, while it faces no net force (since it exerts equal forces on each trolley, and the trolleys exert the same forces by Newton's Third Law in the opposite directions, thus cancelling themselves out and leading to no net horizontal force) has no mass either, and doesn't have a kinetic energy associated with it.

So, no single object is fit to be the system. However, if we select both trolleys and the spring - yes, we can solve this, since there is no net force on this system! All forces here are between parts of the system, which do not contribute to the net momentum of the system itself. So, let's see what we can do!

Okay, so momentum of this system is conserved. Since the initial horizontal momentum is zero, the final horizontal momentum is also zero.

Suppose we take the right ward direction to be negative.
The trolley on the left moves to the left with a speed of 2 m/s. This means it has a positive leftward momentum of mv = (2 kg) * (2 m/s) = +4 kg m/s.
The trolley on the right moves to the right with an unknown speed, say "v". This means it has a negative rightward momentum of mv = (1 kg) * (- v m/s) =
- v kg m/s.

Adding these up, we get (4.0 kg m/s) + (- v kg m/s) = 0. Therefore, the magnitude of v = 4.0 m/s.
This makes sense - the lighter trolley moves faster than the heavier one with the same force on it. But that's not what we not, what we want is the energy in the spring.

What we can assume here is that there is no friction acting anywhere, and so the kinetic energy of the trolleys summed up is equal to the energy initially in the spring (alternatively, we can say that the energy of our "system" remains constant. Then, the initial energy is only in the spring, and the final energy is only in the trolleys, since the spring stretches all the way out and relaxes).

So, the final kinetic energy = 0.5 * (2 kg) * (2 m/s)² + 0.5 * (1 kg) * (4 m/s)² = 0.5 * 2 * 4 + 0.5 * 1 * 16 = 4 + 8 = 12 Joules = D.

Hope this helped!
Good Luck for all your exams!
 
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Zepudee said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_ms_12.pdf

Q18 (D)
ZaqZainab
Click to expand...

So the main point here is the specific nature of the formula for efficiency:

Efficiency = (Useful Energy Output)/(Total Energy Input) =(Useful Energy Output per second)/(Total Energy Input per second) =

(Useful Output Power)/(Total Input Power)

The only useful energy we get from this system is in the form of the electrical energy in the generator circuits - we do not want the kinetic energy the water gets, we do not want the change in potential energy, we do not want the sound energy resulting from the working of the machines involved and we do not want the heat generated from the collisions between water and wheel. The only Useful output energy is in the form of electrical energy.

Since we know from our studies in electric circuits, the Power in a circuit is given by P = IV, where V is the potential of that current and I is the magnitude or size of that current.

We are given both here, that the Voltage/Potential of the current is equal to 230 Volts, while the size of the current is 32 Amperes. Thus, the power is

P = (Useful Output Power) = IV = 32 * 230 = 7360 Watts.

This is the useful power output. Now to find the total power input.
When the input comes to the system, there are two forms of energy - the water turning the wheel has some kinetic energy, and a large amount of potential energy.

We are told in this question that we can ignore the small amount of kinetic energy the water has when it first hits the wheel, and that leaves us with the only form of input energy - the potential energy of the water.

So how much energy is put into the system per unit time? We are told that 200 kilograms of water pass the wheel per second. Therefore, we can assume that 200 kilograms of water fall the entire 8.0 meters height in 1 second. Therefore, the change in energy during this one second is

(Change in energy of water) = (Change in Potential Energy of water) = (Energy input to system) = mgh = 200 kg * 9.81 * 8.0 m = 15, 696 Joules.

Since this is delivered in 1 second, the (Power of falling water) = (Total Input Power) = 15, 696 Watts.

So, Efficiency = (Useful Output Power)/(Total Input Power) = 7,360/15,696 = 0.4689.
To convert to a percentage, we multiply by 100, so that Efficiency is equal to

0.4698 * 100 = 46.98 % = 47% = D.

Zepudee ZaqZainab hope this helps!

Good Luck for all your exams!
 
ZaqZainab

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sagar65265 said:
So the main point here is the specific nature of the formula for efficiency:

Efficiency = (Useful Energy Output)/(Total Energy Input) =(Useful Energy Output per second)/(Total Energy Input per second) =

(Useful Output Power)/(Total Input Power)

The only useful energy we get from this system is in the form of the electrical energy in the generator circuits - we do not want the kinetic energy the water gets, we do not want the change in potential energy, we do not want the sound energy resulting from the working of the machines involved and we do not want the heat generated from the collisions between water and wheel. The only Useful output energy is in the form of electrical energy.

Since we know from our studies in electric circuits, the Power in a circuit is given by P = IV, where V is the potential of that current and I is the magnitude or size of that current.

We are given both here, that the Voltage/Potential of the current is equal to 230 Volts, while the size of the current is 32 Amperes. Thus, the power is

P = (Useful Output Power) = IV = 32 * 230 = 7360 Watts.

This is the useful power output. Now to find the total power input.
When the input comes to the system, there are two forms of energy - the water turning the wheel has some kinetic energy, and a large amount of potential energy.

We are told in this question that we can ignore the small amount of kinetic energy the water has when it first hits the wheel, and that leaves us with the only form of input energy - the potential energy of the water.

So how much energy is put into the system per unit time? We are told that 200 kilograms of water pass the wheel per second. Therefore, we can assume that 200 kilograms of water fall the entire 8.0 meters height in 1 second. Therefore, the change in energy during this one second is

(Change in energy of water) = (Change in Potential Energy of water) = (Energy input to system) = mgh = 200 kg * 9.81 * 8.0 m = 15, 696 Joules.

Since this is delivered in 1 second, the (Power of falling water) = (Total Input Power) = 15, 696 Watts.

So, Efficiency = (Useful Output Power)/(Total Input Power) = 7,360/15,696 = 0.4689.
To convert to a percentage, we multiply by 100, so that Efficiency is equal to

0.4698 * 100 = 46.98 % = 47% = D.

Zepudee ZaqZainab hope this helps!

Good Luck for all your exams!
Click to expand...
Thank i got till 7360 and 15696 but didnt know what to do next
best of luck if you are giving your exam today
 
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sagar65265

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Zepudee said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_13.pdf

Q21 (B) Q10 (A)
Urgent!! Im sorry :(

ZaqZainab sagar65265
Click to expand...

Q10) This one has been discussed before, here are a few posts on that:

https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-318#post-692472

https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-567#post-828639

Q21) I'm pressed for time right now, i'll try posting it later.

Good Luck for all your exams!
 
Venkata Gurrala

Venkata Gurrala

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sagar65265 said:
Q15)

First things first, the law of conservation of momentum says:

"....the momentum of any system shall remain constant and unchanging if and only if no net external force acts on that system."

So, in most situations in physics, the easiest way to tackle a process is to isolate those parts of the process that, taken together, have no net force acting on them. This question serves a lovely example.

The floor is assumed to be frictionless - even otherwise, we are only concerned with the initial velocity, which friction does not necessarily influence directly. Regardless, let's continue with the assumption that the floor is frictionless (if it did exert frictional forces, we would have received some sort of information regarding that force).

Suppose we take the process to be the "moving apart of the two trolleys as the spring extends to it's natural length".
Taking the trolley on the left, there is a force acting on it - the force exerted by the spring when the system is released to it's natural action. This force is why the momentum of the trolley on the left changes, and so we should not take this trolley as the system if we want the simplest answer.

The same applies for the trolley on the right - the spring exerts a force on it in the right-hand direction, which causes an increase in momentum in the right-hand direction for the trolley on that side.

The spring, while it faces no net force (since it exerts equal forces on each trolley, and the trolleys exert the same forces by Newton's Third Law in the opposite directions, thus cancelling themselves out and leading to no net horizontal force) has no mass either, and doesn't have a kinetic energy associated with it.

So, no single object is fit to be the system. However, if we select both trolleys and the spring - yes, we can solve this, since there is no net force on this system! All forces here are between parts of the system, which do not contribute to the net momentum of the system itself. So, let's see what we can do!

Okay, so momentum of this system is conserved. Since the initial horizontal momentum is zero, the final horizontal momentum is also zero.

Suppose we take the right ward direction to be negative.
The trolley on the left moves to the left with a speed of 2 m/s. This means it has a positive leftward momentum of mv = (2 kg) * (2 m/s) = +4 kg m/s.
The trolley on the right moves to the right with an unknown speed, say "v". This means it has a negative rightward momentum of mv = (1 kg) * (- v m/s) =
- v kg m/s.

Adding these up, we get (4.0 kg m/s) + (- v kg m/s) = 0. Therefore, the magnitude of v = 4.0 m/s.
This makes sense - the lighter trolley moves faster than the heavier one with the same force on it. But that's not what we not, what we want is the energy in the spring.

What we can assume here is that there is no friction acting anywhere, and so the kinetic energy of the trolleys summed up is equal to the energy initially in the spring (alternatively, we can say that the energy of our "system" remains constant. Then, the initial energy is only in the spring, and the final energy is only in the trolleys, since the spring stretches all the way out and relaxes).

So, the final kinetic energy = 0.5 * (2 kg) * (2 m/s)² + 0.5 * (1 kg) * (4 m/s)² = 0.5 * 2 * 4 + 0.5 * 1 * 16 = 4 + 8 = 12 Joules = D.

Hope this helped!
Good Luck for all your exams![/
Great! thank you. tomorrow is my last exam. What brings you to the website anyway.
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