PAT Solutions.
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Original post by samkanwoods
I get 700seconds and 23C. How did you get 49s. I used E=mc(delta)T. The frying pan receives 70 Joules per second and I worked out E=2*490*(70-20)=49000J. Therefore Time=49000/70=700s.
Oh dear, I messed up and calculated it as though the area of the pan was 1m^2, not 0.07m^2. Redid it and got 700s
Original post by cameron95
For question 17 on the 2010 paper, what is meant by the 'line of intersection of the two mirrors'?
I didn't understand that either. In the end i presumed that it was the line that would connect the two mirrors to form a right angled triangle...
On a separate note, could someone post a solution to 2010 Q.22??
Original post by Miles434
I didn't understand that either. In the end i presumed that it was the line that would connect the two mirrors to form a right angled triangle...
On a separate note, could someone post a solution to 2010 Q.22??
On a separate note, could someone post a solution to 2010 Q.22??
Thank you, Ive noticed a number of the questions are badly worded which can be confusing. I don't know how to write my solutions here, but I calculated the masses as;
green-24g and red-81g.
Original post by joemkcarr
Has anyone got answers to 2011 24, or the long question about the archer? Cheersss
24 I got:
alpha:beta:gamma
30:5:4
26 I got:
a)60ms^-1
b)50ms^-1
c)1s
d)5m above centre
e)5000N
f)0.2ms^-1
Original post by Miles434
I didn't understand that either. In the end i presumed that it was the line that would connect the two mirrors to form a right angled triangle...
On a separate note, could someone post a solution to 2010 Q.22??
On a separate note, could someone post a solution to 2010 Q.22??
These are my answers (by no means correct):
I'm leaving them as velocity squared so it looks less messy, but obviously for the answer you need to root them...
a) v^2=[k(x-l)^2]/m
b) v^=2[(x-l)(k(x-l)-20m)]/m
c) h=[(x-l)(k(x-l)-20m)]/20m
Original post by fayled
24 I got:
alpha:beta:gamma
30:5:4
26 I got:
a)60ms^-1
b)50ms^-1
c)1s
d)5m above centre
e)5000N
f)0.2ms^-1
alpha:beta:gamma
30:5:4
26 I got:
a)60ms^-1
b)50ms^-1
c)1s
d)5m above centre
e)5000N
f)0.2ms^-1
How did you get 60m/s I got 85m/s by saying that 0.5mv^2=120*0.6, therefore to get v=root(72000)=60root(2)=85m/s
Original post by samkanwoods
How did you get 60m/s I got 85m/s by saying that 0.5mv^2=120*0.6, therefore to get v=root(72000)=60root(2)=85m/s
Its a bowstring so it stores elastic potential energy=0.5FL
It is non-calculator so getting root3600 which can be done mentally makes sense.
Getting 36J energy also makes sense for part 2 as 25/36x36=25.
(edited 11 years ago)
Original post by fayled
Its a bowstring so it stores elastic potential energy=0.5FL
It is non-calculator so getting root3600 which can be done mentally makes sense.
It is non-calculator so getting root3600 which can be done mentally makes sense.
Oh **** I used worked done W=Fl, but I dont see why using W=KE wouldnt work either.
(edited 11 years ago)
Original post by samkanwoods
Oh **** I used worked done W=Fl, but I dont see why using W=KE wouldnt work either.
I originally thought to use Work done but it clearly doesn't work on the non calculator paper - if you get numbers you can't compute you're normally going wrong.
It is a bowstring so stores energy elastically hence you must use the EPE formula.
Original post by fayled
I originally thought to use Work done but it clearly doesn't work on the non calculator paper - if you get numbers you can't compute you're normally going wrong.
It is a bowstring so stores energy elastically hence you must use the EPE formula.
It is a bowstring so stores energy elastically hence you must use the EPE formula.
Makes sense I suppose. thanks
2011 question 24:
A radioactive source emits a parallel beam of alpha, beta and gamma radiation
and is placed 10 cm from a detector, which is sensitive to all forms of radiation
and receives 100 counts/sec.
When a sheet of aluminium of thickness 1 cm is placed in front of the detector,
the radiation level is seen to fall to 50 counts/sec.
When the source is taken away completely, the radiation levels are seen to
fall to 10 counts/sec.
When the source is placed 1 cm from the detector, the radiation levels are
seen to increase to 400 counts/sec.
In what proportion is the source emitting alpha:beta:gamma particles?
Have seen this type of questions in other past papers. Thought I could skip it every time but this time its a 4 mark question instead of one in the multiple choice part. Slightly worried now since the test is in a day's time.
A radioactive source emits a parallel beam of alpha, beta and gamma radiation
and is placed 10 cm from a detector, which is sensitive to all forms of radiation
and receives 100 counts/sec.
When a sheet of aluminium of thickness 1 cm is placed in front of the detector,
the radiation level is seen to fall to 50 counts/sec.
When the source is taken away completely, the radiation levels are seen to
fall to 10 counts/sec.
When the source is placed 1 cm from the detector, the radiation levels are
seen to increase to 400 counts/sec.
In what proportion is the source emitting alpha:beta:gamma particles?
Have seen this type of questions in other past papers. Thought I could skip it every time but this time its a 4 mark question instead of one in the multiple choice part. Slightly worried now since the test is in a day's time.
Original post by fayled
Its a bowstring so it stores elastic potential energy=0.5FL
It is non-calculator so getting root3600 which can be done mentally makes sense.
Getting 36J energy also makes sense for part 2 as 25/36x36=25.
It is non-calculator so getting root3600 which can be done mentally makes sense.
Getting 36J energy also makes sense for part 2 as 25/36x36=25.
pulled back a distance of 0.6m doesn't necessarily mean the extension of the string = 0.6m does it?
The numbers shown by two dice are labelled d1 and d2; a score is constructed
from these by the expression: S = Ad1 + Bd2 + C, where A, B and C are
constants. Determine the values of A, B and C such that the range of possible
values for S covers all integers from 0 to 35, with an equal probability of each
score.
This one..I found C= -7 but I have no idea what I could do next
from these by the expression: S = Ad1 + Bd2 + C, where A, B and C are
constants. Determine the values of A, B and C such that the range of possible
values for S covers all integers from 0 to 35, with an equal probability of each
score.
This one..I found C= -7 but I have no idea what I could do next
Original post by freakynerdlol
pulled back a distance of 0.6m doesn't necessarily mean the extension of the string = 0.6m does it?
yes...
Original post by fayled
yes...
does it?? :s
Original post by freakynerdlol
does it?? :s
If the string is pulled back a distance 0.6m then it must be extended by 0.6m. What else could it mean exactly?
Original post by fayled
These are my answers (by no means correct):
I'm leaving them as velocity squared so it looks less messy, but obviously for the answer you need to root them...
a) v^2=[k(x-l)^2]/m
b) v^=2[(x-l)(k(x-l)-20m)]/m
c) h=[(x-l)(k(x-l)-20m)]/20m
I'm leaving them as velocity squared so it looks less messy, but obviously for the answer you need to root them...
a) v^2=[k(x-l)^2]/m
b) v^=2[(x-l)(k(x-l)-20m)]/m
c) h=[(x-l)(k(x-l)-20m)]/20m
Wrong question.....
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