WJEC AS Physics PH1 May 19th 2015

Thats what i wrote but what about current? Doesnt that also go down?

1. Horizontal v= 2
Vertical = 3.56
Magnitude = 4.1
Angle = 58
2.
3. Pd= 8.5
Pd = 3.75
Combined parallel resistance = 750
Variable resiator = 4500
Pd decreases
4. Experiment....
5.Inave?
6. Zip line..
7.Define displacement - mean velocity x time
Average speed = 48
Average velocity = 0
Thats all i can remember off the top of my head so pretty sure some of those are off

I circled the one that was pointing south east? it was like a tangent to the curve since that was the direction I thought the air resistance was acting?

Hahaha you can't define displacement as velocity x time! That's basically saying that displacement is defined as displacement/time x time XD.
Since velocity is defined as displacement/time, it wouldn't make sense to define displacement in terms of velocity.
I don't think your pds are correct.
Your vertical was my resultant in question 1. Not sure if I made a mistake or you remembered wrong...
The other answers I agree with.

I circled the one that was pointing south east? it was like a tangent to the curve since that was the direction I thought the air resistance was acting?

Hmm well the current of the overall circuit would increase, I can't fully remember the question but decreasing the resistance in the variable resistor would increase its share of the current. Without the numbers from the question I can't say whether the current in the other one goes up or down overall. I thought it only asked about pd not the voltage??

If you're on about the variable resistor question the voltage went down. Basically the R (parallel combo) of the R total went down so V(parallel combo) of the V total went down. The pd of the other resistor went up! Not sure if this explains it very well put if it had 3/4 of resistance it would dissipate 3/4 of pd. If resistance went down by 1 ohm it would have 2/3 of resistance so 2/3 of pd!

But if you change the resistance then that alters the current as well

I said that the pd stays the same, I mentioned the change in I and R cancelling out, and also mentioned the fact that variable resistor and the resistor in parallel must have the same Pd as they're parallel, and if their of changes then the pd of the resistor in series also changes therefore would have to same the same as before

You can take the circuit as a potential divider. Then since the resistance of the parallel combo goes down, the proportion of the emf that it gets decreases as well. Less pd in the combo means less pd across the 900 ohm resistor since pd is same in each branch.

Oh dear. You didn't know that pd is the same thing as voltage? I hope you didn't say/use that in the exam...

On question one how did everyone work out the initial velocity in the vertical competent? from what I remember I halved the time and put the final velocity as 0 then 0^2= u + -9.81 x half of time and rearranged to get U
I think I've done this completely wrong and then on the value of U resultant vector I used Pythagoras

I did v^2=u^2 + 2 x -9.81 x s, v was zero so u^2= 2 x 9.81 x s then took square root!

Typo was meant to say current!!!!!!

Do you think I just used the wrong equation? My answer came out like 2.94 I think 😖 and then U was 3. Something

Do you think I just used the wrong equation? My answer came out like 2.94 I think 😖 and then U was 3. Something

Im pretty sure 2.94 is right, your method should have worked using v=u+at however it looks like you did v^2 which fortunately still comes out as 0, whether or not an examiner will mark you down for that I have no idea seeing as it didn't affect your answer!

Im pretty sure 2.94 is right, your method should have worked using v=u+at however it looks like you did v^2 which fortunately still comes out as 0, whether or not an examiner will mark you down for that I have no idea seeing as it didn't affect your answer!