Physics: Post your doubts here!

[MYLORD]

Can someone help me with question 2 in paper 5

How to do all the above mentioned points.

 

[princess Anu]

http://maxpapers.com/wp-content/uploads/2012/11/9702_w14_qp_11.pdf
whats wrong with D ( Q7)
Q 11 how do we know if Upthrust is greater or Drag force??
Q12: aren't W & W and F & F making up couple of forces? :/ if I take pivot at the edge of the ladder near the ground level, I end up with Fh=Wa :/ but then,how to do it?
q20 why is D wrong
and also Q15,33,34..

HELP Pleaseee!

 

[princess Anu]

why can't force x be upthrust?

 

[Xylferion]

View attachment 51490 why can't force x be upthrust?

I believe since the ball reaches a constant velocity then there'd have to be a force resisting the one pulling it down. Don't quote me a 100% on this, but I get the feeling that upthrust is a force pushing an object up rather than resisting a force acting down. And if it's pushing the object up, then that doesn't necessary obtain a constant velocity does it?

That's just my two cents, hope it makes sense.

EDIT :- Here's something useful I happened to come across
http://lvp.lockyersmid.dorset.sch.u...0/content/html/portal_ks2science/pg000112.htm

 

[Bba321]

The papers are not opening.......Is anyone else having the same problem?

 

[The Godfather]

http://maxpapers.com/wp-content/uploads/2012/11/9702_w14_qp_11.pdf
whats wrong with D ( Q7)
Q 11 how do we know if Upthrust is greater or Drag force??
Q12: aren't W & W and F & F making up couple of forces? :/ if I take pivot at the edge of the ladder near the ground level, I end up with Fh=Wa :/ but then,how to do it?
q20 why is D wrong
and also Q15,33,34..

HELP Pleaseee!

Q7: Total momentum of the SYSTEM is always conserved, system of an isolated object may not be conserved.
Q12: Taking moments about anypoint its completly your discriction, answers will be same.
F and h are perpendicular,Weight and a are perpendicular, upward W is perpendicular to 2a
CLM = ACLM
Fh + Wa = 2Wa

 

[Llamas]

http://www.sheir.org/a-level-physics-43-june2010.pdf

Question 5, b2.

Can someone explain the method/concept, the mark scheme is useless.

 

[Xylferion]

http://maxpapers.com/wp-content/uploads/2012/11/9702_w14_qp_11.pdf
whats wrong with D ( Q7)
Q 11 how do we know if Upthrust is greater or Drag force??
Q12: aren't W & W and F & F making up couple of forces? :/ if I take pivot at the edge of the ladder near the ground level, I end up with Fh=Wa :/ but then,how to do it?
q20 why is D wrong
and also Q15,33,34..

HELP Pleaseee!

There are keywords in every question that you have to be watchful for.

Question 11 > The ball is falling at " terminal " speed in " still " air. Now the first thing to note is that if it's terminal, then the ball is not accelerating.
With that in mind, you look at the three forces they mention which are viscous drag, upthrust and weight.
Now viscous drag is sort of a retardation force that acts opposite to the direction of motion. Since the ball is falling, the motion of viscous drag will act upwards.
Upthrust will do the same as it uses the particles of the air to act upwards on the ball. The only force acting or you could say pulling the ball down is weight.

With this you from an equation Weight = Upthrust + Viscous drag.
This immediately tells us that the force with the greatest magnitude is Weight, so your answer can only be A or B. To eliminate the other option, you have to look at the other two forces. Like I mentioned above, viscous drag is a retardation force, so it will act opposite to the weight of the ball, slowing it down so that it stops accelerating due to free fall. That means that viscous drag is dependent on the weight of the ball.

Upthrust on the other hand deals with the fluid in which it resides. This is where you look at the key words " still air ". If the air is still, that hints that there is barely any motion with the air particles, therefore the upthrust will be very minimal when compared to viscous. Since viscous drag would have to be pretty large to counteract the acceleration due to gravity on the ball.
This shows us that viscous drag is > than upthrust in this question, but it's not always the case. It's only because we're dealing with still air, that this happens to be case.

So in the end, the only option available is A.


Question 15> Ok this question is all about calculations. They start of by giving you V, I and then mass and distance. They also tell you that the motor is 50% efficient.
You're asked to find the time. The key word here is efficiency. which immediately tells me we're dealing with power.

Power = Work / Time
Power = Voltage * Current ( VI )
Work = Force * Distance
Force = Weight = Mass * gravity

I first found the force, this is 200g/1000 to get kilograms and then multiplied by 9.81 to get the weight.
Work = Force(Weight in this case) * distance
Work = 1.962 * ( 90/100 m )
Work = 1.7658 ( 1.8 )

Power = VI ----> 6*0.5 = 3 watts, but it's 50% efficient so only 1.5 watts is being used.
Power = Work/Time, ----> 1.5 watts = 1.8 / Time
Time = 1.8/1.5 = 1.2 seconds, which is C, and that's your answer.

Hope that helped.

 

[princess Anu]

There are keywords in every question that you have to be watchful for.

Question 11 > The ball is falling at " terminal " speed in " still " air. Now the first thing to note is that if it's terminal, then the ball is not accelerating.
With that in mind, you look at the three forces they mention which are viscous drag, upthrust and weight.
Now viscous drag is sort of a retardation force that acts opposite to the direction of motion. Since the ball is falling, the motion of viscous drag will act upwards.
Upthrust will do the same as it uses the particles of the air to act upwards on the ball. The only force acting or you could say pulling the ball down is weight.

With this you from an equation Weight = Upthrust + Viscous drag.
This immediately tells us that the force with the greatest magnitude is Weight, so your answer can only be A or B. To eliminate the other option, you have to look at the other two forces. Like I mentioned above, viscous drag is a retardation force, so it will act opposite to the weight of the ball, slowing it down so that it stops accelerating due to free fall. That means that viscous drag is dependent on the weight of the ball.

Upthrust on the other hand deals with the fluid in which it resides. This is where you look at the key words " still air ". If the air is still, that hints that there is barely any motion with the air particles, therefore the upthrust will be very minimal when compared to viscous. Since viscous drag would have to be pretty large to counteract the acceleration due to gravity on the ball.
This shows us that viscous drag is > than upthrust in this question, but it's not always the case. It's only because we're dealing with still air, that this happens to be case.

So in the end, the only option available is A.


Question 15> Ok this question is all about calculations. They start of by giving you V, I and then mass and distance. They also tell you that the motor is 50% efficient.
You're asked to find the time. The key word here is efficiency. which immediately tells me we're dealing with power.

Power = Work / Time
Power = Voltage * Current ( VI )
Work = Force * Distance
Force = Weight = Mass * gravity

I first found the force, this is 200g/1000 to get kilograms and then multiplied by 9.81 to get the weight.
Work = Force(Weight in this case) * distance
Work = 1.962 * ( 90/100 m )
Work = 1.7658 ( 1.8 )

Power = VI ----> 6*0.5 = 3 watts, but it's 50% efficient so only 1.5 watts is being used.
Power = Work/Time, ----> 1.5 watts = 1.8 / Time
Time = 1.8/1.5 = 1.2 seconds, which is C, and that's your answer.

Hope that helped.

Thank you so much

 

[Xylferion]

Thank you so much

No problem ^_^. Can't help with 33 and 34 yet as I still have to go over my theory with those.

 

[princess Anu]

No problem ^_^. Can't help with 33 and 34 yet as I still have to go over my theory with those.

No problem..
If you can help with 15 or 12 or 20

 

[Xylferion]

No problem..
If you can help with 15 or 12 or 20

I already explained 15 :O

 

[manya]

can someone help me with this question

 

[Jenn]

How do we calculate the time in part (d)(1) and (2)? Help please.

 

[Xylferion]

How do we calculate the time in part (d)(1) and (2)? Help please.

Well firstly. you need to look at all the data you already have.
From (b), you've got the initial acceleration which is 0.570 ms^-2.
You also have the constant velocity which is 2.00 ms^-1.
You have the deceleration too, which you calculated to be 0.306 ms^-2.

You already have the formula written down, but re-arrange it to make t the subject.
t = v-u/a

For 1) t = 2.00 - 0 ( it starts from rest ) / 0.570, this gives you t = 3.5 seconds.
For 2) t = 0-2.00 / (-)0.306 ( since its decelerating ), this gives you t = 6.5 seconds.

Hope that helped! ^_^

 

[MYLORD]

No problem ^_^. Can't help with 33 and 34 yet as I still have to go over my theory with those.

can u help me with something ?
r u taking A level or AS Level ?
if u r taking a level, then how r u preparing for paper ?

 

[manya]

can someone help me with this question

thankyou so much

 

[waleed302]

http://www.sheir.org/a-level-physics-43-june2010.pdf

Question 5, b2.

Can someone explain the method/concept, the mark scheme is useless.

Let me try:
C=Q/V, and they tell you for a capacity of 30 micro Farads the max voltage is 6V. So substitute these values: 30=Q/6. Q=180. So now we've got the charge on one of the capacitors. In parallel the combined capacitance is 20, and they're asking for the max voltage, so plug these values into the formula: C=Q/V... 20=180/V. So V=9. Hope that helps.

 

[Saad ur rehman]

Salamz.
I got doubts in Paper 1 of May june 2010 variant 1. Questions are 3, 9, 13, 16, 27, 33 and 39. I would be glad if someone helps

 

[Saad ur rehman]

physicist, thanks fo that blog i cleared a few doubts but i couldnt get the rest.