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AQA Physics B Unit 4 Discussion (PHB4)

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jas0nuk

Also did this paper today. It wasn't too bad. Surprised about the lack of circular motion!

What did everyone get for the magnitude of the frictional force against the piston in the SHM question?
They said 0.76J of energy is removed in one complete cycle.
I worked it out as

Work done = force * distance
0.76 = force * (0.15 * 4) <-- x4 because in one cycle it goes from +0.15 to 0 to -0.15 back to 0 and then to 0.15 again
.: Force = 1.2N

But other people put different answers. Someone said 9.2N and someone else 2.4N :s-smilie:

The velocity of the alpha particle after hitting the gold nucleus was -1.44 x 10^7 metres per sec (opposite direction but they asked for speed not velocity)

the way i did that one was KE=0.5xmxv^2
to find the velocity then
F=(mv^2)/r
so F=m x omega x velocity
omega = 2 x pi x f
so
F= m x 2 x pi x f x v

cant see this being the correct way tell ya the truth but it was alll i could think of during the exam

Reply 21

02sagooa

Thoughtless

you cant use F= (mv^2)/r because it wasnt circular motion

Reply 22

Ydde

I simply couldn't work out how to rearrange the formula that needed the sq root of KE. But for some reason I just squared the momentum and went from there. Was pretty straight forward after that but boy did that scare me lol.

Reply 23

memenme_1

jas0nuk

The velocity of the alpha particle after hitting the gold nucleus was -1.44 x 10^7 metres per sec (opposite direction but they asked for speed not velocity)

Ditto! that is what i got for the two answers...though instead of minus i put to the left!

Reply 24

memenme_1

By they way, For T and A , what values did you guys get in question 2? Oh and the for question where it asked how the electron shows wave properties and particle properties in the apparatus shown? I put that, the electron produces interference patterns ( Wave) and it can be accelerated using high potential difference or electric field ( particle!)

Reply 25

02sagooa

memenme_1

i put the same thing

Reply 26

jas0nuk

For Q2, T and A were read off the graph allowing f to be determined.
T = 0.25sec .: f = 4Hz
A = 0.15m

edit: memenme_1, I put exactly the same things for the last question xD

Reply 27

alan910_2

jas0nuk

The velocity of the alpha particle after hitting the gold nucleus was -1.44 x 10^7 metres per sec (opposite direction but they asked for speed not velocity)

Yeah I also got 1.2 N but I used F=P/v

It's pretty unorthodox but it seems to have worked.

Your speed value however... I got something to the power 5, not 7. (i mean like, something times 10^5)

I was pretty confident for that one... What was your method?

Reply 28

nigel_s

Ydde

Yeah basically that's what I did. Actually I think I multiplied both sides of the KE equation so you had

mKE=0.5m^2v^2

Reply 29

nigel_s

jas0nuk

The velocity of the alpha particle after hitting the gold nucleus was -1.44 x 10^7 metres per sec (opposite direction but they asked for speed not velocity)

Got the same as you in all :smile:

Reply 30

Thoughtless

KE=0.5mv^2
(2KE)/m = v^2

v=sqroot((2KE)/m)

p=mv

p=m x root((2KE)/m)

p=root(2KEm)

p=root(KE) x root (2m)

p=1.35something... x root(KE)

thats how i did it
if anyone would like to put that into latex form that would be great - i dont know how to use it lol

Reply 31

liamo1

alan910_2

I got the speed to be something times 10 to the power 5. Also, it was negative because it is rebounding in the opposite direction.

Speed negative? - I don't think so. If it asks for speed it wants magnitude not direction hence you put the positive value. You will only lose 1 mark. :frown:

Reply 32

liamo1

I thought it was a very generous exam. Did you see the small section on question 5 I think:

Work function: The minimum energy input required to liberate and electron from a surface.

2nd part was E=hf.

3rd part was basically:

ekmax = hf - work function

Took about 2 minutes worth 8! marks.

I believe I done well. I think I got an A and I think to get 100 UMS marks you need 65/75. So i'm hoping I got 65.

Reply 33

nigel_s

liamo1

Is this paper worth 100 UMS? The AS papers were worth 105 each.

Reply 34

memenme_1

nigel_s

Is this paper worth 100 UMS? The AS papers were worth 105 each.

they are 105 UMS i think

Reply 35

ginge133

Module 4's worth 90 UMS I think. http://www.aqa.org.uk/qual/pdf/AQA-5456-6456-W-SP-09.PDF

Page 6

Reply 36

alan910_2

liamo1

Speed negative? - I don't think so. If it asks for speed it wants magnitude not direction hence you put the positive value. You will only lose 1 mark. :frown:

I don't think it said speed actually, the equation for momentum uses velocity.

Don't know why I said 'speed' in my last comment.

So no, I won't lose that mark :smile:

Reply 37

jas0nuk

It did ask for the speed, not the velocity :P

Reply 38

alan910_2

jas0nuk

It did ask for the speed, not the velocity :P

Oh bloody hell, what a stupid question. Momentum uses velocity, so the answer should be velocity.

Regardless, I still don't think they'll dock the mark for the minus sign. They'll expect many to put it I think...