EDEXCEL A2 Physics EXAM Unit 4 Physics On The Move 20th June 2016 (NOT I-A-L)

Original post by target21859

Did you have to mention charge?

I think this is what the question was about, but my memory is absolutely terrible so don't take my word for it.

Reply 182

qq1102141262

I would say the speedometer will be overestimated.

if the tyre is under inflated, r will be less as well as v will be less.
As v=(2pi*r)/t. The r for the speedometer is constant for that specific bikeso actual speedometer will display higher v.

Reply 183

Original post by qq1102141262

I would say the speedometer will be overestimated.

if the tyre is under inflated, r will be less as well as v will be less.
As v=(2pi*r)/t. The r for the speedometer is constant for that specific bikeso actual speedometer will display higher v.

Same as I did. I said would be exaggerated due to lower radius. I also mentioned that the value itself wouldn't change as the angular velocity is constant.

Reply 184

habiba_9

Original post by oShahpo

Same as I did. I said would be exaggerated due to lower radius. I also mentioned that the value itself wouldn't change as the angular velocity is constant.

Wasnt this question related to something with pulses of emf?

Reply 185

Original post by habiba_9

Wasnt this question related to something with pulses of emf?

Was it? I don't remember the question exactly. Do you remember the wording of the question?

Reply 186

habiba_9

Original post by oShahpo

Was it? I don't remember the question exactly. Do you remember the wording of the question?

Nop but i wrote my answer in terms of magnitude of emf anyone else who did that aint sure about my answer tho

Reply 187

dfgfd

Original post by oShahpo

Yea I said the same thing, but I also that the value itself wouldn't change as the angular velocity stays the same, does that negate the first bit I said?

The angular velocity of the wheel remains constant and hence so does the speed reading.

Reply 188

akiwicat

Original post by oShahpo

Was it? I don't remember the question exactly. Do you remember the wording of the question?

I said that for a fixed speed V, the frequency of rotation of the wheel is V/2*pi*r and so for a smaller radius at a given speed the frequency of rotation would be greater and so the frequency of EMF pulses would be greater causing the speedometer to overestimate the speed.

Reply 189

Salty carpet

Original post by jtebbbs

Resolve vertically: F*sin22 = 80 * 9.81 gives F = 846

Set horizontal component equal to mv^2/r, rearrange to get

r = (80 * 9^2) / (846 * cos22) which gives you 20.4m

I thought I didn't round but got 20.5m, did you use exact values in your calc?

Reply 190

Original post by akiwicat

I said that for a fixed speed V, the frequency of rotation of the wheel is V/2*pi*r and so for a smaller radius at a given speed the frequency of rotation would be greater and so the frequency of EMF pulses would be greater causing the speedometer to overestimate the speed.

V isn't constant, in fact, V is reduced when the tire is not inflated properly. It would overestimate the speed, but not because the frequency of emf pulses is greater, rather because of the calibration of the device.

Reply 191

Martin010

Guess I was the only one who somehow got 40.5% for max decrease... oops :s-smilie:

Reply 192

AtomicAnal

Question 3 was obviously D, Newtons aren't mass, can't believe you all got that wrong lol. My teacher said it was D too lol

Reply 193

josephinemar25

Original post by AtomicAnal

Question 3 was obviously D, Newtons aren't mass, can't believe you all got that wrong lol. My teacher said it was D too lol

What was question 3 again???

Reply 194

jtebbbs

Original post by PhysicsIP2016

Surely the speed is directly proportional? v=2pir/T?

Yeah you're right, looking back I'm not sure what I meant!

I think what I'm saying is as the radius decreases, the speed on the speedo relative to the true speed increases, but yeah, the true speed is directly proportional to the radius. Hoping I didn't make the same mistake in the exam, can't remember!

Reply 195

tkjoption

Original post by jjjzzz

Thank you!!!!!!!!! I insisted on 33.3% until I saw your comment. 👍

I agree on 33.3%
In fact I am pretty sure that 33.3 is more likely to be correct cause
This is what u get from the result of
{Max-Min} /Max *100%=33.3%

Reply 196

jtebbbs

Original post by AtomicAnal

Question 3 was obviously D, Newtons aren't mass, can't believe you all got that wrong lol. My teacher said it was D too lol

MeV was option C, and is a unit of energy, so it was C. D was Nm^-1s^2, in other words Newtons divided by metres times seconds squared. A newton is kgms^-2, or kilograms times metres divided by seconds squared, so the metre and second parts cancel out, leaving mass.

Reply 197

Qcomber

3

Original post by AtomicAnal

Question 3 was obviously D, Newtons aren't mass, can't believe you all got that wrong lol. My teacher said it was D too lol

I put MeV was not a unit of mass. I think D was Nm^-1s^2 which is a unit of mass.

Reply 198

jtebbbs

Original post by tkjoption

I agree on 33.3%
In fact I am pretty sure that 33.3 is more likely to be correct cause
This is what u get from the result of
{Max-Min} /Max *100%=33.3%

I think the question asking for "max decrease" implies that you can either increase or decrease from the starting point. If you use 2.8mm to find your starting point, you're already at the max so you can't increase further, therefore 3.5mm must be the 'starting point'.

Reply 199

AtomicAnal