A2 (Edexcel) Physics Unit 4 & 5 6PH08 June 2012

No ur right I had a question like this in my m2 book and had to calculate loss in height for every bounce it followed a series pattern

It all depends on coefficient of restitution of the ball

Ie snooker balls are more elastic in collisions than tennis balls

But when bouncing them on the ground it's other way round

Ugh things like this bother me so much lol
Most of the marks these days depend whether you can interpret their writing in the way they want you to :/

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Original post by iesians

WHAT ?!
almost every oscillation is damped !

I researched this answer for you.

Conditions for SHM:
1. Fixed time period
2. Acceleration ∝ - displacement

For bouncing ball:
1. Time period is NOT constant. The amplitude is 1. very large 2. decreases by 1/3 or 2/3 each bounce I think - so it is heavily damped. The oscillations you are considering are of small amplitude (with 10') and with negligible damping by air.

2. Acceleration is NOT ∝ - displacement. In fact acceleration is constant = g regardless of displacement.

Another thing to note: In SHM, maximum velocity is at the middle of the oscillation. But in bouncing ball, the maximum velocity is NOT when the ball is at middle height, but just before it hits the ground.

Reply 121

d_94

7

Is it just me or did time go by reallly fastt! I mean we have like 2 days left for this exam!! Ughhh

Reply 122

meraphox

Original post by d_94

Is it just me or did time go by reallly fastt! I mean we have like 2 days left for this exam!! Ughhh

Lol I have that feeling for everything
Once the day comes it seems as if everything is just summed up in 90min =|
100's of hours of studying come down to 1 exam :frown:

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Reply 123

meraphox

Does anyone know how many raw marks qualify as 120ums?

Eg if you get 60/80 and 60/80 is an A*
That's 108 ums then?
Like how many raw marks are 120ums

Can someone explain this using last years grade boundaries?

I need 225 ums in unit 4/5 to meet my offer :frown:

Edit:
Can some1 confirm this?

Last year 58 raw marks was A*
=> 58 = 108 ums
So (58/108) x 120 = 64.4 = 65

Is 65/80 then 120 ums?

Thanks
This was posted from The Student Room's iPhone/iPad App

(edited 11 years ago)

Reply 124

iesians

6

Original post by Parthenon93

I researched this answer for you.

Conditions for SHM:
1. Fixed time period
2. Acceleration ∝ - displacement

For bouncing ball:
1. Time period is NOT constant. The amplitude is 1. very large 2. decreases by 1/3 or 2/3 each bounce I think - so it is heavily damped. The oscillations you are considering are of small amplitude (with 10') and with negligible damping by air.

2. Acceleration is NOT ∝ - displacement. In fact acceleration is constant = g regardless of displacement.

Another thing to note: In SHM, maximum velocity is at the middle of the oscillation. But in bouncing ball, the maximum velocity is NOT when the ball is at middle height, but just before it hits the ground.

hey thanks very much !!
i think your conditions for SHM are wrong
in MS it is written that :-

-acceleration is proportional to displacement
-acc is in opposite direction to the displacement.

for something to exhibit SHM, theres no point for constant time period....

Reply 125

Parthenon93

7

Original post by iesians

hey thanks very much !!
i think your conditions for SHM are wrong
in MS it is written that :-

-acceleration is proportional to displacement
-acc is in opposite direction to the displacement.

for something to exhibit SHM, theres no point for constant time period....

I get what you are saying. I found the conditions in the book.

Simple harmonic motion is a special type of periodic motion, so to be simple harmonic, it has to be periodic motion first.

And you the MS points sum up in acceleration ∝ - displacement, which I gave as the second condition.

Reply 126

d_94

7

Original post by meraphox

Lol I have that feeling for everything
Once the day comes it seems as if everything is just summed up in 90min =|
100's of hours of studying come down to 1 exam :frown:

This was posted from The Student Room's iPhone/iPad App

Soooooo freakking truee. CAnt wait until I'm done with this *****

Reply 127

rkh300

does anyone have unit 5 notes Q_Q i'm dying here

Reply 128

584385

Original post by rkh300

does anyone have unit 5 notes Q_Q i'm dying here

unit 5 notes

Original post by iesians

hey thanks very much !!
i think your conditions for SHM are wrong
in MS it is written that :-

-acceleration is proportional to displacement
-acc is in opposite direction to the displacement.

for something to exhibit SHM, theres no point for constant time period....

Constant time period isnt a condition for shm its a characteristic of it, if it is undamped (which is assumed in shm) then it will always have a constant time period.

Reply 130

Parthenon93

7

Does anyone have the chapterwise answers (not examzone) to Unit 4 of Mile Hudson book?

Reply 131

d_94

7

Original post by Parthenon93

Does anyone have the chapterwise answers (not examzone) to Unit 4 of Mile Hudson book?

http://www.scribd.com/mohd_jamal_35/d/80536475-a2-Physics-Text-Book-Answers

Reply 132

Dissh.M

For unit 4
How do you know which particle is entering the bubble chamber by its track and does it necessarily depend on the magnetic field ?(that is out of page oor into )

Please help

Reply 133

astaraeal

1

Original post by Dissh.M

For unit 4
How do you know which particle is entering the bubble chamber by its track and does it necessarily depend on the magnetic field ?(that is out of page oor into )

Please help

They will give you information in the question - the only thing you can work out from a track is the direction of motion of the particle, so they will tell you either the direction of the magnetic field or the charge on the particle (ie whether it is positively or negatively charged.)

You can then use Fleming's LHR to work out the piece of information that isn't given (magnetic field direction/charge on particle.)

The direction of motion of the particle will depend on the magnetic field direction (by Fleming's LHR).

If they want you to work out more specific information (magnitude of charge/momentum/mass/velocity of particle) they will give you enough information to be able to use r = p/BQ. They may give you a scale and get you to measure the radius of the particle in the photograph so you can use this to work out r.

Reply 134

username287755

how do you do question 2, jan 2012?

Physics 4 Jan 2012 QP.pdf727.9KB

Reply 135

astaraeal

1

Original post by Killerstorm2

how do you do question 2, jan 2012?

You place the momentum vectors before nose-to-tail, and the momentum vectors after nose-to-tail. The resultant momentum vector before is the missing side of the triangle, as is the resultant momentum vector after. As momentum is conserved, the resultant momentum vector should be the same before and after.

So the answer is C :smile:

Reply 136

mtynan1

Hi, has anyone done the specimen paper for unit 4? 15 b ii is really screwing me up :s the mark scheme says 1.6/16, but where's the 1.6 come from? why isn't it 12/16 :frown:

Reply 137

astaraeal

1

Original post by mtynan1

Hi, has anyone done the specimen paper for unit 4? 15 b ii is really screwing me up :s the mark scheme says 1.6/16, but where's the 1.6 come from? why isn't it 12/16 :frown:

The paper is wrong - it's a specimen paper so wasn't checked properly :smile:

Either re-label the axis or pick your own value for initial voltage/capacitance so that your values fit on the graph!

Reply 138

Alexok

8

Original post by Dissh.M

For unit 4
How do you know which particle is entering the bubble chamber by its track and does it necessarily depend on the magnetic field ?(that is out of page oor into )

Please help

You can deduce the direction of the particle by looking at the radius of its curve. If the radius gets smaller, or the circles get smaller than its going in that direction.

So you will see straight lines that slowly curve and get more curved which means its losing kinetic energy. it cant go from curved line to straight line because that would mean its gaining energy.

Also we know its losing energy because of equation F = mv^2/r.... r = mv^2/F .... F, m is constant. As Velocity decreases due to loss in kinetic energy due to ionisation of bubble chamber you will see r(radius) decrease as its directly proportional

The magnetic field only changes the direction of the curves so it doesn't really make a difference

(edited 11 years ago)

Reply 139

poopypie

Hey does anyone know what can be concluded if a positron is found to have a mass of 1.8 x 10^-29 kg?

the answer is that it is travelling close to the speed of light. Why??

A2 (Edexcel) Physics Unit 4 & 5 6PH08 June 2012

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