AQA PHYA4 ~ 13th June 2013 ~ A2 Physics

Hi anyone chance someone could explain question 12 to me?? http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JAN11.PDF

Is the answer A? GPE decreases (become more negative) as the comet approaches the source of gravity. The GPE lost by the comet is converted to kinetic energy, hence the increase in kinetic energy :smile:

Reply 161

laurenb11

Original post by posthumus

Is the answer A? GPE decreases (become more negative) as the comet approaches the source of gravity. The GPE lost by the comet is converted to kinetic energy, hence the increase in kinetic energy :smile:

Yeah, it is, thanks! I didnt think about the GPE becoming more negative, hence decreasing. Thankyou :smile:

Reply 162

masryboy94

8

Original post by laurenb11

Hi anyone chance someone could explain question 12 to me?? http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JAN11.PDF

ok what they are basically trying to say it is following an elliptic orbit (oval shaped orbit) rather than a circular orbit, this means height and speed is changing rather than staying constant throughout the orbit. you should also know that E_k + E_p = constant, therefore if E_kincreased that means E_phas to decreases so that constant is fulfilled. Therefore when it is at the minimum distance, it feels the greatest gravitation force because F obeys the inverse square law, so when distance reduces, F increases dramatically, and to stop it from crashing into the sun, velocity would HAVE to increase dramatically, and since

EK = \frac{1}{2}mv^2

and we agreed that E_k + E_p = constant that means since E_kincreases, then E_p decreased, so your answer would be A because they said E_p would increase. hope that clears it up :biggrin:

(edited 10 years ago)

Reply 163

Jack93o

15

can someone explain this question for me please?

Reply 164

posthumus

20

Original post by Jack93o

can someone explain this question for me please?
x

mv^2 / r = mg

v^2 / r = g

v= root(rg)

this is the speed you need...

since T = 2pi r / v

You can find the time period :smile:

Reply 165

masryboy94

8

Original post by Jack93o

can someone explain this question for me please?

so if the object is weightless this means that the centripetal force must equal the gravitational force. therefore

\frac{mv^2}{r} = \frac{GMm}{r^2}

do your cancelling out, then rearrange V to get

V = \sqrt \frac{GM}{r}

we also know that

v = \frac{d}{t}

and the distance around a whole circle is

2\pi r

so you get

v = \frac{2\pi r}{T}

, equate the 2 v's and make T the subject, for this question dont bother simplifying and square rooting ... etc but do it for good practice if you want, anyways then when you get the answer, it will be interms of seconds, they want answer interms of hours, so you would have to divide by 3600 (divide by 60 to get to minutes and another 60 to get to hours) and then you get the answer they are looking for which is B :biggrin:

(edited 10 years ago)

Reply 166

Jack93o

15

Original post by posthumus

mv^2 / r = mg

v^2 / r = g

v= root(rg)

this is the speed you need...

since T = 2pi r / v

You can find the time period :smile:

Original post by masryboy94

so if the object is weightless this means that the centripetal force must equal the gravitational force. therefore

\frac{mv^2}{r} = \frac{GMm}{r^2}

do your cancelling out, then rearrange V to get

V = \sqrt \frac{GM}{r}

we also know that

v = \frac{d}{t}

and the distance around a whole circle is

2\pi r

so you get

v = \frac{2\pi r}{T}

, equate the 2 v's and make T the subject, for this question dont bother simplifying and square rooting ... etc but do it for good practice if you want, anyways then when you get the answer, it will be interms of seconds, they want answer interms of hours, so you would have to divide by 3600 (divide by 60 to get to minutes and another 60 to get to hours) and then you get the answer they are looking for which is B :biggrin:

centripetal acceleration being equal to 'g' is what I'm a bit confused about

I thought centripetal acceleration is always equal to 'g' at the equator (ground) level. Why does it only equal 'g' when the earth rotates fast enough for objects there to appear weightless?

and what is the centripetal acceleration normally equal to then? would it be less than or more than 'g'?

Reply 167

masryboy94

8

Original post by Jack93o

centripetal acceleration being equal to 'g' is what I'm a bit confused about

I thought centripetal acceleration is always equal to 'g' at the equator (ground) level. Why does it only equal 'g' when the earth rotates fast enough for objects there to appear weightless?

and what is the centripetal acceleration normally equal to then? would it be less than or more than 'g'?

nono dont get the wrong idea, when it says weightless it doesn't mean it isn't at ground level, but the fact you begin to feel weightless yet touching the ground means that they must equal each other exactly !!

Reply 168

Jack93o

15

Original post by masryboy94

nono dont get the wrong idea, when it says weightless it doesn't mean it isn't at ground level, but the fact you begin to feel weightless yet touching the ground means that they must equal each other exactly !!

yeah I know the question is referring to ground level, what I don't get is, why isn't centripetal acceleration at ground level always equal to 'g' (= 9.81)? I thought it is....

right now i'm sitting here, I'm still experiencing a centripetal acceleration of 'g' due to my weight (mg) at this ground level that I'm at, yet I'm not feeling weightless/free fall

so i don't understand how subbing in g = 9.81 for centripetal acceleration would solve this question

(edited 10 years ago)

Reply 169

The H

1

-SHM
-Transformers
-Magnetic fields
-Gravitational force/field strength

some likely topics

Reply 170

cooldudeman

18

Just to confirm, grav potential is infinity when object is closer than earth's radius right? Looking at the graph, it seems to be so...

Posted from TSR Mobile

Reply 171

fuzzybear

12

magnetic field line is always directed north to south, right?

Reply 172

StudentAnon

How do you all approach the 'which of the following cannot be a unit of..' questions? I find there's too many equations to look at etc

Reply 173

laurenb11

Original post by cooldudeman

Just to confirm, grav potential is infinity when object is closer than earth's radius right? Looking at the graph, it seems to be so...

Posted from TSR Mobile

Yeah, i think so as in deep space its zero right? and grav potential is the amount of energy needed to pull a small test mass from deep space to that point?

Reply 174

masryboy94

8

ok i have a question for all of you, i know the answer to it and know how to do it, but im curious to how everyone answers it:

A discharging capacitor loses 70% of its initial charge, Q₀, in 20 seconds.

a) Write down the value of Q/Q₀ after 20 seconds.

b) Find the time constant of the capacitor - resistor circuit

Reply 175

posthumus

20

Original post by masryboy94

ok i have a question for all of you, i know the answer to it and know how to do it, but im curious to how everyone answers it:

A discharging capacitor loses 70% of its initial charge, Q₀, in 20 seconds.

a) Write down the value of Q/Q₀ after 20 seconds.

b) Find the time constant of the capacitor - resistor circuit

a) 0.3 ??

b) e^-20/RC = 0.7
20/RC = -ln(0.7)
RC = 20/-ln(0.7)

Reply 176

masryboy94

8

Original post by fuzzybear

magnetic field line is always directed north to south, right?

yep always !!

Reply 177

masryboy94

8

Original post by posthumus

a) 0.3 ??

b) e^-20/RC = 0.7
20/RC = -ln(0.7)
RC = 20/-ln(0.7)

for part a) are you sure its 0.3 or are you hesitant about it? and your part b) is not correct, you made a small mistake somewhere.

Reply 178

posthumus

20

Original post by masryboy94

for part a) are you sure its 0.3 or are you hesitant about it? and your part b) is not correct, you made a small mistake somewhere.

Yh I'm quite sure of a) though I think that was an odd question if it's from a past paper... but makes me more convinced since I should have used it in the second part :tongue:

should have made it equal to 0.3 not 0.7

Reply 179

masryboy94

8

Original post by posthumus

Yh I'm quite sure of a) though I think that was an odd question if it's from a past paper... but makes me more convinced since I should have used it in the second part :tongue:

should have made it equal to 0.3 not 0.7

ok good, so you obviously know why its 0.3 and 0.7, like you can explain why yh? and lool nope its from the book, but was in an old paper, so it can come up. haha yeh thats right :biggrin:

AQA PHYA4 ~ 13th June 2013 ~ A2 Physics

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