How did Physics P2 go?

[Farjad Ilyas]

I think two identical waves 180 degrees out of phase will completely cancel each other out and destructive interference occurs, no?
that's what I wrote. Plus the first one, with zero phase difference was the zero order maximab?. Correct me if I'm wrong.

Generally, that would be correct. However, in the b part they asked 'what is observed at P when the following vhanges are made to part a'
In part A the waves were in phase ast D1 and D2 but they were out of phase at P so destructive interference occured. The path diff was 4cm which is (4/1.6) equal to 2.5lambda. However in the B part if the wave are 180° out of phase at D1 and D2, they will be inphase at P since path diff will be either 2 or 3 lambda which is why constructive interference is seen.

 

[amina1300]

Are you talking about the young modulus question? Everyone I know got 0.13TPa for that one.

I got a really weird answer for that one weren't we suppose to plug in the values into equation ; gradient*L/(pi d^2/4) for the Young Modulus?

 

[techgeek]

Generally, that would be correct. However, in the b part they asked 'what is observed at P when the following vhanges are made to part a'
In part A the waves were in phase ast D1 and D2 but they were out of phase at P so destructive interference occured. The path diff was 4cm which is (4/1.6) equal to 2.5lambda. However in the B part if the wave are 180° out of phase at D1 and D2, they will be inphase at P since path diff will be either 2 or 3 lambda which is why constructive interference is seen.

I'm done with it then

 

[amina1300]

I think two identical waves 180 degrees out of phase will completely cancel each other out and destructive interference occurs, no?
that's what I wrote. Plus the first one, with zero phase difference was the zero order maximab?. Correct me if I'm wrong.

Yeah I wrote this too was this wrong?

 

[Farjad Ilyas]

I got a really weird answer for that one weren't we suppose to plug in the values into equation ; gradient*L/(pi d^2/4) for the Young Modulus?

Yeah thats what we had to do. Did you convert the extension and diameter to metres? Both were in millimetres I think

 

[techgeek]

Yeah I wrote this too was this wrong?

apparently so :/

 

[amina1300]

Yeah thats what we had to do. Did you convert the extension and diameter to metres? Both were in millimetres I think

yes i did but my answer was 4.01*10^-4 TPa

 

[amina1300]

apparently so :/

How many marks was that Q for?

 

[amina1300]

No we just had to calculate the loss in G.P.E and final K.E which was followed by the 1 mark question asking about the difference.

In the Kinematics Q how did we have to find the horizontal velocity?

 

[amina1300]

Yup 500hz was correct... For PD we had to find loss voltages by Ir and subtracting.it from 1.5 then

Could we use the E = I [ R + R + r] ?

 

[amina1300]

Termi

Terminal pd was 1.46

How did we have to find the current in the first part of the Electricity Q?

 

[Farjad Ilyas]

In the Kinematics Q how did we have to find the horizontal velocity?

you had the vertical distance, initial vertical velocity (0m/s) and vertical acceleration (due to gravity)= 9.81.
So you could use s= ut + 1/2at^2 and make time 't' the subject to find the time taken for the ball to reach the ground.
The ball travels a horizontal distance of 1.5metres during this time ans horizontal velocity is constant. So you can use Speed= Horizontal Dist travelled /Time to find the horizontal velocity 'v'

 

[Farjad Ilyas]

How did we have to find the current in the first part of the Electricity Q?

Don't you mean resistance? Didnt they give current as 0.18A?

 

[Farjad Ilyas]

Could we use the E = I [ R + R + r] ?

IRa + IRb= E-Ir - you get this if you rearrage your formula so if you took IRa + IRb as p.d across AB you should get 1.46V as the ans which is correct

 

[techgeek]

How many marks was that Q for?

I think that was worth 3 marks

 

[Omaima Uchiha]

4.5*10-7

Is that correct?

 

[amina1300]

Don't you mean resistance? Didnt they give current as 0.18A?

yesh the resistance. My bad I forgot.

 

[amina1300]

you had the vertical distance, initial vertical velocity (0m/s) and vertical acceleration (due to gravity)= 9.81.
So you could use s= ut + 1/2at^2 and make time 't' the subject to find the time taken for the ball to reach the ground.
The ball travels a horizontal distance of 1.5metres during this time ans horizontal velocity is constant. So you can use Speed= Horizontal Dist travelled /Time to find the horizontal velocity 'v'

Thankyouu!! I thought we had to do something else because I did this Q in the last 2 mins . Phew!

 

[Farjad Ilyas]

yesh the resistance. My bad I forgot.

The e.m.f was 1.5V and current was 0.18A. So you could find the total resistance using R=V/I. The gave said that the resistance of the variable resistor was set to 6 ohms i think and internal was 0.2ohms. So subtract the variable resistor's resistance and internal resistance from the total you calculated and you'd get the ans (because they were in series). The ans was 2.13 ohms i think

 

[amina1300]

The e.m.f was 1.5V and current was 0.18A. So you could find the total resistance using R=V/I. The gave said that the resistance of the variable resistor was set to 6 ohms i think and internal was 0.2ohms. So subtract the variable resistor's resistance and internal resistance from the total you calculated and you'd get the ans (because they were in series). The ans was 2.13 ohms i think

I got the same 2.13333333