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# GCSE Edexcel physics - 12th June 2014 watch

1. What did everyone think of the paper? I wasn't sure about the ice rink question or the 14000N rope question when the car has a weight of 12000N ( can't remember the exact numbers)
Thanks
2. Easiest paper I have done in my life, I will get 100% raw marks, as per usual.
3. (Original post by TheSweatyNerd)
Easiest paper I have done in my life, I will get 100% raw marks, as per usual.
Ikr! What did you get for current? Calculation. .

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4. (Original post by huwgerard)
What did everyone think of the paper? I wasn't sure about the ice rink question or the 14000N rope question when the car has a weight of 12000N ( can't remember the exact numbers)
Thanks
The rope has 12000N

Whilst the car has 14000N.

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5. For the car and rope one I said it wouldn't snap because weight is a vertical force, the only thing which would affect the tension of the rope would be a horizontal force like the drag of the car.
Ikr! What did you get for current? Calculation. .

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5.83 or 7.83 I think.
7. (Original post by jhjamie)
For the car and rope one I said it wouldn't snap because weight is a vertical force, the only thing which would affect the tension of the rope would be a horizontal force like the drag of the car.
I wrote it will snap because as the velocity is constant, the car is pulling the truck back by its drag by 14000N. Thus the truck will have exert a equal but opposite force and thus the truck will have 14000N to opposite direction. The rope will break as tension is applied because it has a force of 12000N.
8. (Original post by TheSweatyNerd)
5.83 or 7.83 I think.
It wad 7.7.... round it up to 7 8

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I wrote it will snap because as the velocity is constant, the car is pushing the truck back by its drag by 14000N. Thus the truck will have exert a equal but opposite force (<--) thus will be 14000N. This will mean that tension will applied to the rope and will split as it has less force then both the directions of the car (drag) and truck.

Dunno I guessed that one.

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I'm not sure, I asked my teacher about it and told him my answer and he said I was correct? I could be wrong though.
It wad 7.7.... round it up to 7 8

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Yeh 7.8
11. If the velocity is constant and all forces are in balance then surely the rope can't snap? It will only snap if there's a resultant force, no?
12. The rope will not snap. It will be able to tow the car. I had a 5.83 answer somewhere.
13. (Original post by Akashi)
The rope will not snap. It will be able to tow the car. I had a 5.83 answer somewhere.
5.8? For what. .

It was 7.8

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5.8? For what. .

It was 7.8

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I can't remember if the question was how long did it take for the acceleration to take. It was something like 20-13/something. What was the current question?
15. (Original post by Akashi)
I can't remember if the question was how long did it take for the acceleration to take. It was something like 20-13/something. What was the current question?
It was rearranging the acceleration formula

T= 20-13 × (... I forgot.

Current =power/voltage

It was 7.8
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16. 20-13/T = 1.2 right?
17. (Original post by Akashi)
20-13/T = 1.2 right?

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Yep. 1.2 = 20-13/T so 1.2 x t = 20-13. 20-13/1.2 = T. T = 5.83
19. (Original post by Akashi)
Yep. 1.2 = 20-13/T so 1.2 x t = 20-13. 20-13/1.2 = T. T = 5.83
I don't remember doing that or getting that answer.

Most people rearranged the acceleration formula (v-u/t) to make t the object.

T= v-u × A

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I don't remember doing that or getting that answer.

Most people rearranged the acceleration formula (v-u/t) to make t the object.

T= v-u × A

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A = v-u/T. You can't just do A x v-u because v-u isn't dividing - it is being divided. Therefore, its reciprocal is not to just multiply it out.
The numbers were 1.2 = 20-13/T. Using your method T would = 8.4. Insert it into the first equation (A = v-u/t). 1.2=7/8.4. It doesn't work.

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