Physics: Post your doubts here!

[Muskan Achhpilia]

Thank you!!!

Can you please help me in the following question too...


Then the question is-
A second ball is thrown from point P with the same velocity as the ball in the diagram. For this ball, air resistance is not negligible. This ball hits the wall and rebounds.On Fig. 2.1, sketch the path of this ball between point P and the point where it first hits
the ground.

The answer given is
smooth curve with ball hitting wall below original
smooth curve showing rebound to ground with correct reflection at wall

I am confused whether it is rebounding then hitting wall or hitting the wall then rebounding...

Thanks a tonne!

 

[unique111]

*the bottom of the ladder rests on rough ground where there is friction. The top of the ladder is at a height h above the ground and the foot .....

 

[unique111]

Thank you!!!

Can you please help me in the following question too...
View attachment 38258

Then the question is-
A second ball is thrown from point P with the same velocity as the ball in the diagram. For this ball, air resistance is not negligible. This ball hits the wall and rebounds.On Fig. 2.1, sketch the path of this ball between point P and the point where it first hits
the ground.

The answer given is
smooth curve with ball hitting wall below original
smooth curve showing rebound to ground with correct reflection at wall

I am confused whether it is rebounding then hitting wall or hitting the wall then rebounding...

Thanks a tonne!

It hits the wall below the original point first.

 

[Muskan Achhpilia]

It hits the wall below the original point first.

Thank you!

 

[TheStallion-Reborn]

Saad Mughal usama321 AbbbbY asma tareen been stuck for over hours now!! Any help would be really appreciated !

 

[sagar65265]

Saad Mughal usama321 AbbbbY asma tareen been stuck for over hours now!! Any help would be really appreciatedView attachment 38272 !View attachment 38271

For Question 13, there are a few important points to be noticed in the question:

i) the string is light, so we can assume that the magnitude of the tension for in the string is the same at all points along the string;

Suppose the string were indeed heavy, each part of the string would have a particular mass, and since the string itself is accelerating, there would have to be a net force on each one of those parts. However, the only forces on the string would be gravity and the 2 tension forces, one from each side. So, for a horizontal part of the string, the tension would have to be different at each point (otherwise no individual part could accelerate, and so the rope on a whole won't accelerate either.

ii)all surfaces is frictionless, so we need not make any assumptions concerning the magnitude of any friction forces and so this simplifies the question a lot.

Now onto Newton's Laws:

Taking the 2.0 kg mass into account, we need only look at the horizontal forces , since there is no vertical acceleration of that mass.
The only horizontal force acting on the mass is the Tension in the string (which, as noted before, is constant throughout the length of the string).

Therefore, T = (2.0) * a

In other words, our positive direction for the 2.0 kg mass is towards the right, and the tension force is positive since it acts towards the right.

Taking the 1.0 kg mass into account, we need only look at the vertical forces, since there is no horizontal/ sideways acceleration of the 1.0 kg mass.
The only 2 horizontal forces acting on the mass are the Tension in the string (same value as in the previous equation) and it's weight (= 1.0 * 9.81 = 9.81 Newtons).

Therefore, since Tension acts in the direction opposite to the acceleration and the weight acts in the direction of the acceleration,

-T + 9.81 = (1.0) * a

In other words, for the 1.0 kg mass, our positive direction is downwards. Therefore, the tension is negative and the weight in positive (in terms of sign value).

The acceleration of each object is the same, since the string can be assumed to be rigid and in-extensible (cannot be stretched), so any movement of one mass will the mirrored in the movement of the other mass, the only difference being the direction.

Solving these equations, we get a = 9.81 / 3 = 3.27 ms^-2

Since this acceleration is constant, we can apply the constant acceleration equations for either object. Suppose we take the 2.0 kg mass, it will have traveled 0.5 meters when the other object hits the ground, it started from rest, and it moves with the acceleration calculated above, so

u = 0 ms^-1
a = 3.27 ms^-2
s = 0.5 m
v = ?

v^2 = u^2 + 2 * a * s
v^2 = 0 +3.27
So, the velocity v = 1.81 ms^-1 = A

For the next question, we can use several theories and laws, but lack of available information constricts our choices like so:

i) We can use Newton's Laws, but we do not know what the magnitudes of the forces that act between the two objects during the collision are; we also do not know how long the collision lasts, so we cannot answer this question using Newton's Laws.

ii) We can use the Work-Energy theorem, but there too we don't know the magnitudes of the forces acting during the collision and we do not know the distance the object move during the collision. So, we cannot use the Work-Energy theorem either.

iii) We can use the Principle of Conservation of Momentum, which does allow us to answer this question; the reason for this is that we do not need to know the forces acting between the objects involved if we take them both as our system.

So, taking both the pellet and the block as the system, we have:

Initial Momentum = (200 ms^-1) * (0.005 kg) = 1 kg ms^-1

Since no external forces act on the system during the collision (gravity is a rather negligible force compared to the forces acting between the two objects; it does cause some change, but gravity has rather little significance during the collision), momentum is assumed to be conserved. Not only that, since the pellet is embedded in the block, the two objects move as one mass, with a common velocity.

Therefore,

Final Momentum = Initial Momentum = [(0.095 + 0.005) kg] * [(final velocity) ms^-1)

0.1 * final velocity = 1

Therefore, final velocity = 10 ms^-1

During the collision we can assume that the objects don't move much upwards, and only start moving at 10 ms^-1 after the collision is complete. Therefore, since the only force acting on the system of the two objects is gravity = -0.981 Newtons, we can use the equations of constant acceleration motion:

u = 10 ms^-1
v = 0 ms^-1 (at the top of it's ascent)
a = -9.81 ms^-2

so v^2 = u^2 + 2as
0 = 100 - 2 * 9.81 * s
Therefore, s = 5.1 meters = A

Hope this helps!
Best of luck for your exams!

 

[sagar65265]

Help needed.

For the first image (the potential energy question):

Let's first try to find out whether the potential energy of the system increases or decreases:

The field shown can either correspond to a large, positively charged sheet on the left side of the image, or a large, negatively charged sheet on the right side of the image.

Since the charge is a positive charge, moving it to the left would bring it closer to the positive sheet and farther away from the negative sheet.

Note that just as we ignore the change in potential energy of the Earth when an object falls from a height (because that change is so small compared to the change in PE of the falling object), we ignore the change in potential energy of the charges producing the field, since that change is minuscule compared to the change in potential energy of the single charge. Therefore, the potential energy of the charge is in reality the potential energy of the charge + the potential energy of the system of charges that produce the field.

Since the force on the charge is to the right and will accelerate it in that direction, if we need to keep it's kinetic energy constant (i.e. keeping it's velocity constant, since mass remains unchanged), we need to apply a force to the left (if we applied a force to the right, then it would accelerate in that direction). Also note that the forces are all constant in magnitude since the field lines are equally spaced apart all the time, and so the electric force and the mechanical force involved remain the same.

So, if the force we apply is to the left, and the displacement is towards the right, and the kinetic energy remains constant, where does the work done by the force go?
The answer is simple: since the force does negative work, the internal forces of the system (where the system consists of the positive charge AND the charges producing the field) must do positive work, and thus they transfer energy OUT of the system. Therefore, the internal electric potential energy decreases when the charge is moved from X to Y.

Now to figure out how much the change is, i.e. the magnitude of the change. Since the kinetic energy of the system remains constant, the work done by all the forces adds up to zero. In other words,

(Work done by mechanical force) = (Work done by electric force) = (Decrease of Internal Energy of the system)

Therefore, using the dot product, we can say that the work done by the electric force in moving the charge from X to Y = F * r * cos(theta) = F * r * (s/r) - Fs

Therefore, the internal energy decreases by Fs = A.

For the second question, we can work with the fact that both wires are made of the same material, have the same length, and thus they will have the same resistivity (since resistivity is a property of the material, not the dimensions of the sample concerned).

Therefore, taking the formula for resistivity as ρ = RA/l ,

R(P) =R(Q) * A(Q)/A(P)

The Cross Sectional Area of Q= (pi) * (0.0005)^2 = A(Q)
The Cross Sectional Area of P =(pi) * (0.001)^2 = A(P)

So, taking these values, A(Q)/A(P) is equal to 0.25 = 1/4

Therefore, R(P) = R(Q)/4

Since the potential difference across the ends of the wires are equal and I = V/R,

I(P)/I(Q) = R(Q)/R(P) = 4/1 = D.

Hope this helped!

Best of luck for all your exams!

 

[unique111]

For the first image (the potential energy question):

Let's first try to find out whether the potential energy of the system increases or decreases:

The field shown can either correspond to a large, positively charged sheet on the left side of the image, or a large, negatively charged sheet on the right side of the image.

Since the charge is a positive charge, moving it to the left would bring it closer to the positive sheet and farther away from the negative sheet.

Note that just as we ignore the change in potential energy of the Earth when an object falls from a height (because that change is so small compared to the change in PE of the falling object), we ignore the change in potential energy of the charges producing the field, since that change is minuscule compared to the change in potential energy of the single charge. Therefore, the potential energy of the charge is in reality the potential energy of the charge + the potential energy of the system of charges that produce the field.

Since the force on the charge is to the right and will accelerate it in that direction, if we need to keep it's kinetic energy constant (i.e. keeping it's velocity constant, since mass remains unchanged), we need to apply a force to the left (if we applied a force to the right, then it would accelerate in that direction). Also note that the forces are all constant in magnitude since the field lines are equally spaced apart all the time, and so the electric force and the mechanical force involved remain the same.

So, if the force we apply is to the left, and the displacement is towards the right, and the kinetic energy remains constant, where does the work done by the force go?
The answer is simple: since the force does negative work, the internal forces of the system (where the system consists of the positive charge AND the charges producing the field) must do positive work, and thus they transfer energy OUT of the system. Therefore, the internal electric potential energy decreases when the charge is moved from X to Y.

Now to figure out how much the change is, i.e. the magnitude of the change. Since the kinetic energy of the system remains constant, the work done by all the forces adds up to zero. In other words,

(Work done by mechanical force) = (Work done by electric force) = (Decrease of Internal Energy of the system)

Therefore, using the dot product, we can say that the work done by the electric force in moving the charge from X to Y = F * r * cos(theta) = F * r * (s/r) - Fs

Therefore, the internal energy decreases by Fs = A.

For the second question, we can work with the fact that both wires are made of the same material, have the same length, and thus they will have the same resistivity (since resistivity is a property of the material, not the dimensions of the sample concerned).

Therefore, taking the formula for resistivity as ρ = RA/l ,

R(P) =R(Q) * A(Q)/A(P)

The Cross Sectional Area of Q= (pi) * (0.0005)^2 = A(Q)
The Cross Sectional Area of P =(pi) * (0.001)^2 = A(P)

So, taking these values, A(Q)/A(P) is equal to 0.25 = 1/4

Therefore, R(P) = R(Q)/4

Since the potential difference across the ends of the wires are equal and I = V/R,

I(P)/I(Q) = R(Q)/R(P) = 4/1 = D.

Hope this helped!

Best of luck for all your exams!

Really appreciate the effort you have put into helping! Thank you so much, bro. You are very smart! #respect
These are mcq questions, and yet take so much time to solve them, i always run out of time! How do i manage time? And bro, the first question, i didn't quite get the decreasing part. Why use an opposing force to the left? As the field lines indicate, left is positive and right side is negative, right? So, doesn't the positive charge placed at point X tend to move to the right?

 

[Thought blocker]

midha.ch Suchal Riaz

 

[sagar65265]

Really appreciate the effort you have put into helping! Thank you so much, bro. You are very smart! #respect
These are mcq questions, and yet take so much time to solve them, i always run out of time! How do i manage time? And bro, the first question, i didn't quite get the decreasing part. Why use an opposing force to the left? As the field lines indicate, left is positive and right side is negative, right? So, doesn't the positive charge placed at point X tend to move to the right?

First thing to look at is the Work-Energy Theorem in this form:

ΔW = ΔK.E

So, the change in kinetic energy of any object is equal to the work done on that object.

(Note: any object that can be modeled as a particle; for example, if two blocks connected by a spring were hurled through the air, this wouldn't apply on the system, since each part of the system can move with different speeds - since the spring can stretch and shorten, the blocks can have different velocities - but if you apply the theorem to one of the blocks, it would work, since every particle in the block travels with the same speed)

As you have pointed out, the positive charge tends to move towards the right due to the force exerted on it by the electric field displayed.
However, any displacement (if the particle were left alone) would mean that the electric force would do work that would automatically be converted to kinetic energy.
In this situation, all you would have to do is calculate the change in kinetic energy, and since the kinetic energy increases, the potential energy decreases by an equal amount, which can be calculated as the negative of the work done on the particle.

So yes, I guess I owe you an apology - I did take the long, drawn-out way, but hopefully it may add a little insight into something i'm about to say!

Thanks for the kind words, but to be fair i've done my A levels and have studied a little bit more, so the perspecties in that reply are mostly from those sources ;-) !

Concerning your query on the MCQs - yeah, i remember discussing with my classmates, Physics and Chem MCQs are seriously tough, sometimes even compared to the theory papers, but practice is no doubt the best way to approaching them.

The second best, I guess, should be taking time to learn, derive and understand all the unwritten laws;
taking the example above about the change in potential energy, there is a rather unspoken rule that says:

"The internal potential energy of any system at any point is the negative of the work done by the internal forces of that system in moving the parts of the system from infinity to that point."

Oh, dear, that's a horrible way of saying it. Sorry, I think the following page, Section 4.1.4 says it much better:

http://www.brown.edu/Departments/En...rticles_work_energy/particles_work_energy.htm

It says: "The potential energy of a conservative force is defined as the negative of the work done by the force in moving from some arbitrary initial position

to a new position in space," i.e. from infinity to any point in space.
That page is pretty complex, so don't worry if most of it looks formidable; there's just a lot of technical terms used there, and it probably isn't required knowledge for CIE AS and A Levels.

Since the Coulomb force is a conservative force, the above statement applies.

So, a little tinkering around with the above statement will eventually tell you that the change in potential energy of a system - within which acts a conservative force such as gravity or the spring force - is equal to the negative of the work done by those internal forces while moving from the initial to the final position.

In other words, since you can call any point a reference point and call the potential there 0, the potential at any other point is the negative of the work done by the internal force in moving from that reference point to the final point.
In this case, since the Coulomb force does positive work while moving from X to Y, the change in internal energy is automatically the negative of the work done by the Coulomb force. So all that remains is finding the magnitude of the work done by the Coulomb force.

Managing time....I guess it usually comes with practice, but the old advice always passes muster here:

Go through the entire paper, and complete at the earliest the questions you find easiest. Once you are done with those, find the next hardest questions, and spend a moment to think - what category does that question fall under(is it mechanics? statics? electric concepts? radioactivity?)? What does it ask? If it is asking for the rms voltage of an alternating current circuit, you can discard all the potentiometer and resistivity equations you know - unless those also figure in the question - and focus more on the correct equations, rather than meander around with all sorts of equations to end up with 0 = 0 (i did that once, and it broke my head ).

Hope it helps!
Best of luck for all your exams!

 

[unique111]

First thing to look at is the Work-Energy Theorem in this form:

ΔW = ΔK.E

So, the change in kinetic energy of any object is equal to the work done on that object.

(Note: any object that can be modeled as a particle; for example, if two blocks connected by a spring were hurled through the air, this wouldn't apply on the system, since each part of the system can move with different speeds - since the spring can stretch and shorten, the blocks can have different velocities - but if you apply the theorem to one of the blocks, it would work, since every particle in the block travels with the same speed)

As you have pointed out, the positive charge tends to move towards the right due to the force exerted on it by the electric field displayed.
However, any displacement (if the particle were left alone) would mean that the electric force would do work that would automatically be converted to kinetic energy.
In this situation, all you would have to do is calculate the change in kinetic energy, and since the kinetic energy increases, the potential energy decreases by an equal amount, which can be calculated as the negative of the work done on the particle.

So yes, I guess I owe you an apology - I did take the long, drawn-out way, but hopefully it may add a little insight into something i'm about to say!

Thanks for the kind words, but to be fair i've done my A levels and have studied a little bit more, so the perspecties in that reply are mostly from those sources ;-) !

Concerning your query on the MCQs - yeah, i remember discussing with my classmates, Physics and Chem MCQs are seriously tough, sometimes even compared to the theory papers, but practice is no doubt the best way to approaching them.

The second best, I guess, should be taking time to learn, derive and understand all the unwritten laws;
taking the example above about the change in potential energy, there is a rather unspoken rule that says:

"The internal potential energy of any system at any point is the negative of the work done by the internal forces of that system in moving the parts of the system from infinity to that point."

Oh, dear, that's a horrible way of saying it. Sorry, I think the following page, Section 4.1.4 says it much better:

http://www.brown.edu/Departments/En...rticles_work_energy/particles_work_energy.htm

It says: "The potential energy of a conservative force is defined as the negative of the work done by the force in moving from some arbitrary initial position

to a new position in space," i.e. from infinity to any point in space.
That page is pretty complex, so don't worry if most of it looks formidable; there's just a lot of technical terms used there, and it probably isn't required knowledge for CIE AS and A Levels.

Since the Coulomb force is a conservative force, the above statement applies.

So, a little tinkering around with the above statement will eventually tell you that the change in potential energy of a system - within which acts a conservative force such as gravity or the spring force - is equal to the negative of the work done by those internal forces while moving from the initial to the final position.

In other words, since you can call any point a reference point and call the potential there 0, the potential at any other point is the negative of the work done by the internal force in moving from that reference point to the final point.
In this case, since the Coulomb force does positive work while moving from X to Y, the change in internal energy is automatically the negative of the work done by the Coulomb force. So all that remains is finding the magnitude of the work done by the Coulomb force.

Managing time....I guess it usually comes with practice, but the old advice always passes muster here:

Go through the entire paper, and complete at the earliest the questions you find easiest. Once you are done with those, find the next hardest questions, and spend a moment to think - what category does that question fall under(is it mechanics? statics? electric concepts? radioactivity?)? What does it ask? If it is asking for the rms voltage of an alternating current circuit, you can discard all the potentiometer and resistivity equations you know - unless those also figure in the question - and focus more on the correct equations, rather than meander around with all sorts of equations to end up with 0 = 0 (i did that once, and it broke my head ).

Hope it helps!
Best of luck for all your exams!

Thank you for all the tips and advice you shared! I'm really happy to have received guidance from you. Though, I admit the parts of conservative energy just flew by my head, I get the gist of it. I owe you much gratitude for helping us (juniors).

Oh, btw Brown? OMG!

 

[unique111]

View attachment 38304midha.ch Suchal Riaz

Maths? why here, bro?

 

[Thought blocker]

Maths? why here, bro?

I was sleepy. So posted here.

 

[unique111]

I was sleepy. So posted here.

Bro, get some rest.

 

[Thought blocker]

Bro, get some rest.

I'am done. No more.
Btw, thanks for your concern.


You know the answers ?

 

[sitooon]

Where can i get examiner report for may jun 2013 physics , its not available on maxpapers

 

[Thought blocker]

Where can i get examiner report for may jun 2013 physics , its not available on maxpapers

I asked ♣♠ Magnanimous ♣♠, he is in search.

 

[sitooon]

I asked ♣♠ Magnanimous ♣♠, he is in search.

we hope he get them

 

[♣♠ Magnanimous ♣♠]

Search is ON =)

 

[umairch.]

Notes for possible sources of errors and improvements for Question 2 of paper 3...?