# Edexcel IAL Physics Unit 1 WPH01 May June 2016 DiscussionWatch

Announcements Poll: How was your exam ?
Easy ... piece of cake xD (6)
8.33%
emmm not that bad :D (39)
54.17%
I got screwed :( (19)
26.39%
It was a NIGHTMARE !! :0 (8)
11.11%
3 years ago
#141
(Original post by Mustafalord99)
I wrote plot a graph of velocity against radius^2 then find the gradient of the graph, i conapred the eqution of v=2gr^2(density of solid-density of fluod)/9n then compare it with y=mx+c so the gradient will be equal to 2g(Ps-Pf) then substistute values and get viscosity.
Do you think I'll get the mark if I say r squared against velocity. And then I wrote the gradient accordingly. I know Indepedant is supposed to be on the x axis but still?
0
reply
3 years ago
#142
I wonder how did you guys get that answer of 500J question? Cos the line is curved, I didn't even know how to get the area under the graph. 0
reply
3 years ago
#143
(Original post by louisasia)
I wonder how did you guys get that answer of 500J question? Cos the line is curved, I didn't even know how to get the area under the graph. I used trapizium rule and divided the graph into eight shapes 😂i got like 554 then i said since the answer is an overestimte , we could subtract 55 so i got finaly answer as 497 i guess
0
reply
3 years ago
#144
(Original post by louisasia)
I wonder how did you guys get that answer of 500J question? Cos the line is curved, I didn't even know how to get the area under the graph. I just counted the number of boxes and multiplied the number of boxes by the area of one box.
0
reply
3 years ago
#145
(Original post by Ayesssshaaa)
Do you think I'll get the mark if I say r squared against velocity. And then I wrote the gradient accordingly. I know Indepedant is supposed to be on the x axis but still?
yeah you should get the mark.. I also said plot a graph of v against r^2.. then compare the equation for viscosity with y=mx+c.. then simply substitute m with eta( n )
0
reply
3 years ago
#146
(Original post by Einsteinj.n.r)
yeah you should get the mark.. I also said plot a graph of v against r^2.. then compare the equation for viscosity with y=mx+c.. then simply substitute m with eta( n )
I said plot a graph for difference in weight and upthurst in y axis and 6pi r v on x axis
Gradient will give us viscosity
0
reply
3 years ago
#147
(Original post by louisasia)
I wonder how did you guys get that answer of 500J question? Cos the line is curved, I didn't even know how to get the area under the graph. I got 519J by make rectangles and triangles in it
0
reply
3 years ago
#148
(Original post by OnTheHorizon)
The ball question 2nd part.Horizontal velocity.
Btw for time did u multiply by 2?
0
reply
3 years ago
#149
(Original post by MrBioBombastic)
Btw for time did u multiply by 2?
Why?
0
reply
3 years ago
#150
(Original post by OnTheHorizon)
Why?
Time we got was from x to y for the previous motion part it would be same so total will be 2xt
0
reply
3 years ago
#151
(Original post by jannz)
How did u guys show the distance is about 0.4 m in that bouncing ball ques....
I used equations of motion
0
reply
3 years ago
#152
(Original post by MrBioBombastic)
Time we got was from x to y for the previous motion part it would be same so total will be 2xt
I did something else,Though I made a mistake I believe I took the time of 24 bounces instead of 20.24/20=1.2s.If I recall correctly You shouldn't be able to use that time between X and Y multiplied by 2.
EDIT:But I guess you've taken a certain distance? I just took the whole horizontal distance for 24 images.
0
reply
3 years ago
#153
(Original post by OnTheHorizon)
I did something else,Though I made a mistake I believe I took the time of 24 bounces instead of 20.24/20=1.2s.If I recall correctly You shouldn't be able to use that time between X and Y multiplied by 2.
EDIT:But I guess you've taken a certain distance? I just took the whole horizontal distance for 24 images.
I took distance between 2 balls on floor from ruler and used s=ut+1/2at2 where u was zero as from x to y and s was 0.44m so I fot time between x and y so I multiplied by 2 for whole projectile motion then I used speed=d/t where d was horizontal distance and I got 0.38ms-1 is it correct
and the distance I took b/w two balls wasmy range i.e horizontal distance
0
reply
3 years ago
#154
(Original post by MrBioBombastic)
I took distance between 2 balls on floor from ruler and used s=ut+1/2at2 where u was zero as from x to y and s was 0.44m so I fot time between x and y so I multiplied by 2 for whole projectile motion then I used speed=d/t where d was horizontal distance and I got 0.38ms-1 is it correct
Honestly,I don't know I barely remember the values I took.But it does make sense.
0
reply
3 years ago
#155
(Original post by OnTheHorizon)
Honestly,I don't know I barely remember the values I took.
lol btw method is correct?
it was lengthy af
0
reply
3 years ago
#156
For everyone asking about the ball question, here is what i did: For the first question to find the maximum height, the time was 6/20 because 20 was 1 second. And then use of equation s=ut+1/2at^2 because initial is zero as it falls from maximum height, all gpe no ke. You should get around 0.44m. Then for the second part, use the picture, if 0.44 was actual height, 6 cm was height on picture, that means total length(horizontal) was 9.3 cm. Convert both cm to meter. Use: If 6cm is 0.44m what is 9.3cm. You should get horizontal distance is 0.682m. Then use simple distance/time. Time for total should be 20/20 is 1 second and + 4/20 as it was 24 pictures and 20 was one second, so you need to add the extra 4, time should be 1.2. So velocity is 0.568m/s, atleast this is what i got, hopefully, it's correct.
0
reply
3 years ago
#157
From where does this multiplying the time by 2 comes from?..we were dealing with just the fall of the ball i.e (from x to y)..its not like the ball moving up and coming down so there is no point of doing anything with the flight time..just imagine it as if the ball has fallen from a roof and we are told to workout now..
0
reply
Thread starter 3 years ago
#158
for that horiz velocity question, it said "use the photograph". so you had to measure the horiz distance between X and y and it came to be like 1.5cm. then use the scale from the prev question where 5.5 cm on pic was 0.44m
5.5 cm ----- 0.44m
1.5 cm ------0.12m

horiz velocity= 0.12/0.3= 0.4 ms-1
0
reply
Thread starter 3 years ago
#159
(Original post by RotiKebab)
For everyone asking about the ball question, here is what i did: For the first question to find the maximum height, the time was 6/20 because 20 was 1 second. And then use of equation s=ut+1/2at^2 because initial is zero as it falls from maximum height, all gpe no ke. You should get around 0.44m. Then for the second part, use the picture, if 0.44 was actual height, 6 cm was height on picture, that means total length(horizontal) was 9.3 cm. Convert both cm to meter. Use: If 6cm is 0.44m what is 9.3cm. You should get horizontal distance is 0.682m. Then use simple distance/time. Time for total should be 20/20 is 1 second and + 4/20 as it was 24 pictures and 20 was one second, so you need to add the extra 4, time should be 1.2. So velocity is 0.568m/s, atleast this is what i got, hopefully, it's correct.
the whole calc had to be done between x and y im afraid. those are just 6 pics
0
reply
Thread starter 3 years ago
#160
(Original post by Zainat0987)
From where does this multiplying the time by 2 comes from?..we were dealing with just the fall of the ball i.e (from x to y)..its not like the ball moving up and coming down so there is no point of doing anything with the flight time..just imagine it as if the ball has fallen from a roof and we are told to workout now..
exactly. we had to take those 6 pics in account between x and y not all
0
reply
X

Write a reply...
Reply
new posts

### 1,763

people online now

### 225,530

students helped last year
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of Dundee
Undergraduate Open Day Undergraduate
Mon, 26 Aug '19
• University of Aberdeen
General Open Day Undergraduate
Tue, 27 Aug '19
• Norwich University of the Arts
Postgraduate (MA) Open Day Postgraduate
Sat, 31 Aug '19

### Poll

Join the discussion

#### Do you have grade requirements for your sixth form/college?

At least 5 GCSEs at grade 4 (65)
14.16%
At least 5 GCSEs at grade 5 (70)
15.25%
At least 5 GCSEs at grade 6 (90)
19.61%
Higher than 5 GCSEs at grade 6 (185)
40.31%
Pass in English and Maths GCSE (21)
4.58%
No particular grades needed (28)
6.1%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.