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# Edexcel IAL Physics Unit 1 - January 12, 2017 watch

1. (Original post by JeSuisAhmedN)
I do remember that it was something like that,
Thank god I didn't exactly know what to do,i think i took the max value for delta x from the graph of the linear region and to find delta L or something and then put it all in the formula for Y.M
2. (Original post by Waheed1998)
Good luck for the full UMS! Do youremember what you got for the frictional force the last part of that question!Was it around 17 or 18?i think i got it wrong.
Thanks man you too. I saw that question in the last 2 minutes of my exam, I didn't see it before. I remember the resultant force being similar to 18 or something, lemme just go through what I did, I used F = ma and that's it. So 18 is really the resultant force, I shouldn't say it's the frictional force, the friction could've been found by 18 = Force of box on man - frictional force, but that's as far as I got. So I think if you got 18, you'll probably get 1 or 2 marks out of the 3 marks, I think.
3. (Original post by Waheed1998)
Thank god I didn't exactly know what to do,i think i took the max value for delta x from the graph of the linear region and to find delta L or something and then put it all in the formula for Y.M
Yeah I think that's exactly what I did as well
4. I got 1.13x10^11 for the youngmodulus and 64KG for the first question what did you get?
5. (Original post by Waheed1998)
Was the points 90.06?
Was the angle of arrival of the skier 54?
What was the Youngs Modulus of the wire?
What was the mass of air in the first question?
Did you get 64kg for the mass of the air?
6. Is the force of pull around 371 Newtons?

This year's exam was kinda weird. So many 4 mark questions for calculations. I tried the best as I could.
7. (Original post by tarqaq)
Is the force of pull around 371 Newtons?

This year's exam was kinda weird. So many 4 mark questions for calculations. I tried the best as I could.
Yeah I think that sounds about right. We needed to show it was about 400 Newtons right?
8. (Original post by JeSuisAhmedN)
Thanks man you too. I saw that question in the last 2 minutes of my exam, I didn't see it before. I remember the resultant force being similar to 18 or something, lemme just go through what I did, I used F = ma and that's it. So 18 is really the resultant force, I shouldn't say it's the frictional force, the friction could've been found by 18 = Force of box on man - frictional force, but that's as far as I got. So I think if you got 18, you'll probably get 1 or 2 marks out of the 3 marks, I think.
But wasn't the question out of 4 marks? So what did u get as a frictional force on the PERSON? I got like 304.5N
9. (Original post by MrRJ)
But wasn't the question out of 4 marks? So what did u get as a frictional force on the PERSON? I got like 304.5N
could you inform me on how you did the first part of that question
i can kind of remember the question
the pull was 35 degrees to the horizontal and the mass of the box was 85 kilos
thats all

how could you get the magnitude of the force with only as much information
10. (Original post by ethat)
could you inform me on how you did the first part of that question
i can kind of remember the question
the pull was 35 degrees to the horizontal and the mass of the box was 85 kilos
thats all

how could you get the magnitude of the force with only as much information
Mass of box=85kg
Mass of person=90kg
Friction on the box (as calculated in b(ii))~305.5N
Acceleration of both mass and box=0.2ms-2
F=ma
F(box)=85*0.2
=17N
17N is resultant of pull (being the greatest) and friction
Therefore the total pull force is friction + resultant= 305.5+17N=325.5
But F(person)=90*0.2
= 18N
The considering that the pull force on the person is same 325.5 - 18 =304.5N
11. (Original post by MrRJ)
Mass of box=85kg
Mass of person=90kg
Friction on the box (as calculated in b(ii))~305.5N
Acceleration of both mass and box=0.2ms-2
F=ma
F(box)=85*0.2
=17N
17N is resultant of pull (being the greatest) and friction
Therefore the total pull force is friction + resultant= 305.5+17N=325.5
But F(person)=90*0.2
= 18N
The considering that the pull force on the person is same 325.5 - 18 =304.5N
But didn't the question said that the man increased his pull on the object? If I remember correctly, I think it did which is why I didn't consider the pull force to be the same and left it at 18 N because I was confused m
12. (Original post by JeSuisAhmedN)
But didn't the question said that the man increased his pull on the object? If I remember correctly, I think it did which is why I didn't consider the pull force to be the same and left it at 18 N because I was confused m
I can certainly say that 18N would probably get u 1 or 2 marks, but the last two marks involved of using the calculated answers from part i and ii. Because F is resultant of two or more forces in the equation F=ma, u have to either add or deduce it from another force that u have found, in this case being the force of friction in the box as in part ii.
I'm not saying the answer that I got was correct, coz I doubt my logic as well, but that was what I made sense out the question.
13. (Original post by MrRJ)
Mass of box=85kg
Mass of person=90kg
Friction on the box (as calculated in b(ii))~305.5N
Acceleration of both mass and box=0.2ms-2
F=ma
F(box)=85*0.2
=17N
17N is resultant of pull (being the greatest) and friction
Therefore the total pull force is friction + resultant= 305.5+17N=325.5
But F(person)=90*0.2
= 18N
The considering that the pull force on the person is same 325.5 - 18 =304.5N

I am sorry but i suppose you did not answer my question

I need to know about the first part of that question which said show that the pulling force is about 400 newtons

mass of box=85 kg
the pulling force was acting at an angle of 35 degrees

with only this much information provided in the question paper how could it be deduced that the pulling force is about 400 newtons

(Was there any other info that was provided and i missed to see? if so please include it and provide the working for the question)

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