Higher physics 2014/15

Crap! I just realised the question said photodiode! I put inverting cause I thought it said op amp
:frown:

Also just realised I missed a question, only one mark though.

Other than that I thought it was relatively good

Reply 381

Btw how do you do the spacecraft question part A,
I did ft=m(v-u) and got the force upwards which is the same as the weight... XD

Reply 382

My multi-choice answers for NQ(old higher):

1E
2C
3B
4C
5B
6D (Got this one wrong, pretty sure it's B)
7C
8D
9B
10E
11E
12A
13A
14A
15D
16D
17A
18E
19D
20B

(edited 8 years ago)

Reply 383

Original post by muchensmile

I didn't put the max force value on my graph and I drew it in pen(says pencil) in the answer book:P
Load resistance q I also got 0.9V after a page of working:P
Also photoelectric one I added the energy and then did ek=1/2mv squared.
For explaining I said Ek doubles, however If you add the Ek from when electrons are released then total kinetic energy is not double, so there is no way speed can double...I didnt bother explaining about how its velocity squared:P

Oh I drew it in pen too :redface:

Honestly need to start reading questions properly...

Yeah that's what I did in the end :smile:

Yeah I was thinking about that too for the explanation - it was a strange question :confused:

Reply 384

how was the 21aii supposed to be done? i got some crazy answer and didn't have enough time to recheck it at the end. I thought the exam was very difficult, most difficult in many years personally ;(

Reply 385

i also done the first three questions in pencil! would i get the marks?

Reply 386

Original post by muchensmile

Btw how do you do the spacecraft question part A,
I did ft=m(v-u) and got the force upwards which is the same as the weight... XD

I calculated acceleration using a = v2-u2/2s
Then unbalanced force = ma
Which was like 7200N
Weight = mg = 4400N
So engine force = unbalanced force + weight = 11600N

Reply 387

Original post by steve909

how was the 21aii supposed to be done? i got some crazy answer and didn't have enough time to recheck it at the end. I thought the exam was very difficult, most difficult in many years personally ;(

I doubled 0.76s and then added it on to the time it takes for ball to fall 1.8m
thats for part A
Part b was d=vt and I got something like 21m I think

Reply 388

tomctutor

Original post by Omar717

My multi-choice answers:

1E
2C
3B
4C
5B
6D (Got this one wrong, pretty sure it's B)
7C
8D
9B
10E
11E
12A
13A
14A
15D
16D
17A
18E
19D
20B

JUST TO REMIND YOU ALL THERE ARE TWO DIFFERENT PAPERS BEING TALKED ABOUT- CONFUSION REIGNS -PLEASE TELL US IF YOU ARE TALKING ABOUT REVISED(CfE) or NQ (the older one) IN ANY RESPONSE!!!! :confused:

(edited 8 years ago)

Reply 389

Original post by chopinfan

I calculated acceleration using a = v2-u2/2s
Then unbalanced force = ma
Which was like 7200N
Weight = mg = 4400N
So engine force = unbalanced force + weight = 11600N

Ohh shoot I missed out the F=ma step :frown:

Probably get 1 mark out of 3 for getting acceleration:P

Reply 390

wordshark

If i helps there was a similar impulse question in the 2006 paper and you could still get two marks for drawing the graph without labels

Reply 391

Original post by tomctutor

JUST TO REMIND YOU ALL THERE ARE TWO DIFFERENT PAPERS BEING TALKED ABOUT- CONFUSION REIGNS -PLEASE TELL US IF YOU ARE TALKING ABOUT REVISED(CfE) or NQ (the older one) IN ANY RESPONSE!!!! :confused:

Oh sorry about that :tongue:

Reply 392

Original post by steve909

how was the 21aii supposed to be done? i got some crazy answer and didn't have enough time to recheck it at the end. I thought the exam was very difficult, most difficult in many years personally ;(

Calculate time for second half of parabola
Using s=ut+1/2at^2
u=0, so s=1/2at^2
at^2=2s
t=root2s/a=root2*4.7/9.8=0.98s
Therefore total time = 0.98 + 0.76 = 1.74s
Hope this is right :redface:

Reply 393

I swear that spacecraft question was the one I saw in my dreams :frown:

Reply 394

Original post by chopinfan

I calculated acceleration using a = v2-u2/2s
Then unbalanced force = ma
Which was like 7200N
Weight = mg = 4400N
So engine force = unbalanced force + weight = 11600N

Forgot to calculate engine force :frown:

2/3 marks I guess

Reply 395

Original post by chopinfan

Calculate time for second half of parabola
Using s=ut+1/2at^2
u=0, so s=1/2at^2
at^2=2s
t=root2s/a=root2*4.7/9.8=0.98s
Therefore total time = 0.98 + 0.76 = 1.74s
Hope this is right :redface:

thanks!

Reply 396

Original post by muchensmile

Ohh shoot I missed out the F=ma step :frown:

Probably get 1 mark out of 3 for getting acceleration:P

Ahh yeah you'll probably get like 2 marks! :smile:

What did you get for the final answer?

Reply 397

tomctutor

Original post by steve909

i also done the first three questions in pencil! would i get the marks?

All answers must be written clearly and legibly in ink.

but I suppose you won't be penaliaed as you changed to ink- but for future ref read the instructions first in any paper!

Reply 398

wordshark

The bit that confused me about the spacecraft question was that it talked about average gravitational strength which implies that the acceleration due to gravity is not constant. Don't you need constant acceleration to be able to use newtons equations? Am i just looking far too much into it?

Reply 399