Physics: Post your doubts here!

[Rizwan Javed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s15_qp_11.pdf
Can anyone please explain 6, 13, 19, 29 and 36?

6. In this question you're asked to calculate the percentage uncertainty.
First find the fractional uncertainties in k and x.
The fractional uncertainty in k is 2/100
The fractional uncertainty in x is 0.002/0.050
Now you can see that x is a squared term, so you'll multiply the fractional uncertainty of x with its power (that is 2 here): 2*0.002/0.050

Add up the fractional uncertainties and multiply with 100 to find the percentage uncertainy like this:

( 2/100 + 2*0.002/0.050 ) * 100 = 10%

So the answer is B.

 

[Rizwan Javed]

13. The force accelerating the box is actually the weight of the 2.0 kg mass. So first find this force (i.e. the weight of the 2.o kg mass).
It is 2 * 9.81 = 19.62 N
There's also a frictional force, so subtract that from the force calculated above.
19.62 - 6.0 = 13.62N

^ this force that you have calculated is accelerating two objects, the 8.0 kg box and the 2kg mass.

So use the equation F=ma (taking m = 8+2 = 10kg)

a = 13.62/10 = 1.362 ~ 1.4m/s^2

So the answer is A

 

[Rizwan Javed]

19. In this question, the sand is falling at a rate m kg/s. That means in 1 second, m kg undergoes a change in momentum which is mv. Force is the rate of change of momentum. So find the force, which is: mv/s.

Power is found by P = FV.
Substitute the value of F here. you get:
P = (mv/s)(v)
Since time (s) is 1 second, so the addition power will become
P = mv^2

So the answer is D.

 

[Rizwan Javed]

36.
Option A: The resistance of variable resistor is increasing, so the p.d. across it will also increase as R is directly proportional to V. So this statement is wrong.

Option B: It is a similar case. The resistance of variable resistor R is increasing, so p.d. will increase across R not r. So this statement is also wrong.

Option C & D are confusing here.

The power dissipated in R is (I^2)(R) and so it must be zero when R is zero, and must also be zero if R is infinite (as the current I will then be zero). The power must increase and then decrease as R increases from zero. As R increases, the current I decreases, the potential difference across R increases, the potential difference across r decreases and the power dissipated in r (given by I^2 * r) decreases.

So the correct option here is D.

 

[ashcull14]

need help to represent light gates or data logger /ticker tape to find the time of a falling body in nov 2003 paper 5 experiment
can anyone draw the diagram?

 

[funky brat]

13. The force accelerating the box is actually the weight of the 2.0 kg mass. So first find this force (i.e. the weight of the 2.o kg mass).
It is 2 * 9.81 = 19.62 N
There's also a frictional force, so subtract that from the force calculated above.
19.62 - 6.0 = 13.62N

^ this force that you have calculated is accelerating two objects, the 8.0 kg box and the 2kg mass.

So use the equation F=ma (taking m = 8+2 = 10kg)

a = 13.62/10 = 1.362 ~ 1.4m/s^2

So the answer is A

I get this but isn't the weight of box itself responsible for the acceleration too? Why isn't it?

 

[funky brat]

36.
Option A: The resistance of variable resistor is increasing, so the p.d. across it will also increase as R is directly proportional to V. So this statement is wrong.

Option B: It is a similar case. The resistance of variable resistor R is increasing, so p.d. will increase across R not r. So this statement is also wrong.

Option C & D are confusing here.

The power dissipated in R is (I^2)(R) and so it must be zero when R is zero, and must also be zero if R is infinite (as the current I will then be zero). The power must increase and then decrease as R increases from zero. As R increases, the current I decreases, the potential difference across R increases, the potential difference across r decreases and the power dissipated in r (given by I^2 * r) decreases.

So the correct option here is D.

And thanks a lot. It really did help.

 

[Rizwan Javed]

I get this but isn't the weight of box itself responsible for the acceleration too? Why isn't it?

No, the weight of the box (8kg) will not be responsible for acceleration, since the force that box exerts on the ground is balanced by the contact force. So it will not undergo any acceleration downwards. The only force that is acting upon the box is by that mass.

 

[funky brat]

No, the weight of the box (8kg) will not be responsible for acceleration, since the force that box exerts on the ground is balanced by the contact force. So it will not undergo any acceleration downwards. The only force that is acting upon the box is by that mass.

Alright. Thanks.

 

[ashcull14]

need help to represent light gates or data logger /ticker tape to find the time of a falling body in nov 2003 paper 5 experiment
can anyone draw the diagram?View attachment 59925

ANYONE PLEASE ANSWER THIS ?????????

 

[funky brat]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s15_qp_13.pdf
For question 13 why isn't this the right way to solve it?
5*200= 95*x
x= 10.53
then using the formula v^2= u^2=2as where v=0 This gives 5.6 as an answer and adding the 4 m gives 9.6. which is D. :/

 

[The Sarcastic Retard]

ANYONE PLEASE ANSWER THIS ?????????

SAME DOUBT!

 

[Rizwan Javed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s15_qp_13.pdf
For question 13 why isn't this the right way to solve it?
5*200= 95*x
x= 10.53
then using the formula v^2= u^2=2as where v=0 This gives 5.6 as an answer and adding the 4 m gives 9.6. which is D. :/

Your method is correct. But you are making a mistake. The mass after the collision is 100g instead of 95g because the bullets embeds itself in that block.

 

[funky brat]

Your method is correct. But you are making a mistake. The mass after the collision is 100g instead of 95g because the bullets embeds itself in that block.

Oh yeah.

 

[ashcull14]

SAME DOUBT!

what?? u didnt understand it either?

 

[funky brat]

Your method is correct. But you are making a mistake. The mass after the collision is 100g instead of 95g because the bullets embeds itself in that block.

Can you please explain 29, 26, 14 and 12 of the same paper. You don't need to solve it just explain the method.

 

[Rizwan Javed]

Can you please explain 29, 26, 14 and 12 of the same paper. You don't need to solve it just explain the method.

12. As the balls are making an elastic collision so the speed of approach will be equal to the speed of separation. So use this concept to find the final speed of ball X.

14. The upthrust depends on the pressure difference. Since they have the same dimensions, this means that pressure difference is also same. So all the cuboids will have the same upthrust.

26. Is the answer D btw? I'll tell the method, if my answer will be correct

29. Find the phase difference between the two waves arriving at point Z like this (34-24 = 10cm). As the amplitude of oscillation is zero at Z this shows that destructive interference is occuring. So use the formula for distructive interference: (1/2 +n)λ = phase differece. So λ = phase difference / (1/2 +n)

Keep substituting value for n (1, 2, 3 so on) until you get a number which matches with the answer options..

 

[funky brat]

12. As the balls are making an elastic collision so the speed of approach will be equal to the speed of separation. So use this concept to find the final speed of ball X.

14. The upthrust depends on the pressure difference. Since they have the same dimensions, this means that pressure difference is also same. So all the cuboids will have the same upthrust.

26. Is the answer D btw? I'll tell the method, if my answer will be correct

29. Find the phase difference between the two waves arriving at point Z like this (34-24 = 10cm). As the amplitude of oscillation is zero at Z this shows that destructive interference is occuring. So use the formula for distructive interference: (1/2 +n)λ = phase differece. So λ = phase difference / (1/2 +n)

Keep substituting value for n (1, 2, 3 so on) until you get a number which matches with the answer options..

For 12 why can't we use the momentum before collision and momentum after collision thing?
For 14 Why does an object with a greater density meaning more mass since the dimensions are the same not exert greater pressure?
Yeah it is D.

 

[Rizwan Javed]

For 12 why can't we use the momentum before collision and momentum after collision thing?
For 14 Why does an object with a greater density meaning more mass since the dimensions are the same not exert greater pressure?
Yeah it is D.

Because we don't know whether their masses are equal or not Or what exactly are their masses. We could have used that, if we had either known their masses or had known that they are equal in mass.

Upthrust depends on the volume of fluid that is displaced by the object. Similar volumes of fluid are displaced. So similar upthrusts. Also that pressure is exerted by the the water molecules (or water) not by the cuboid.

I'll show you the method for 26 with diagrams. Give me 5 minutes

 

[funky brat]

Because we don't know whether their masses are equal or not Or what exactly are their masses. We could have used that, if we had either known their masses or had known that they are equal in mass.

Upthrust depends on the volume of fluid that is displaced by the object. Similar volumes of fluid are displaced. So similar upthrusts. Also that pressure is exerted by the the water molecules (or water) not by the cuboid.

I'll show you the method for 26 with diagrams. Give me 5 minutes

It still does not make sense to me. Th water is displaced not only by the water molecules but also due to the object and a denser object will displace more or whatsoever.
And sure.