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SL Physics M04 paper2 question, help

Hey guyz
i have problem with question A2, part a and d from May04 Physics Paper 2 SL Time Zone 1

The markscheme has this as the way they got their answer:

horizontal component = R sin14;
= 8500 tan14;
= 2119 N --> which approx is 2100 N


my area of concern is how they arrived at the value for R, here is my calculation

x = 14 degrees, R = Reaction force of the road on the car, F = Horizontal component of the force R

tanx = F/R
F = Rtan14

and R = 8500cos14

therefore,

F = Rtan14 = 8500cos14tan14 = 2056.3N --> which approx equal 2100

its hard to see thigns the way im describign it without looking at the quesiton itself, but if anyoen can spare a bit of time that'd b awesome

as for part d, i thought that if velocity increase, F will increase (i.e. for example.. instead of 2100 it become 3000 but it is still towards center of circle), therefore the car should slide DOWN the ramp?, they say that
"friction must supply larger force2ward center, car tends to slide UP he ramp"

sorry i couldnt ask my own physics teacher.. he's a wonderful bludger.

Thanks in advance
Reply 1
Here is the actual question for those of you who can help plat4m6

http://img494.imageshack.us/img494/4303/motionofcar5xa.png
I had the same problem with the same question. It's hard to see where the Rtan14 came from, and i still dont get that. I solved the problem the way i would have done if it was friction and motion up/down an inclined plane, ie that the horizontal component was Wsinx, which gives the 2056 answer or something like that. I also thought the same thing for question d and havent gotten around to asking a teacher yet. I'll ask tomorrow and post whatever info my teacher gives me. Btw, if ur standard level ur lucky, the next question is about a chicken, egg and entropy. IB really has some creative people in the wings. (little bit of sarcasm there....)
plat4m6

tanx = F/R
F = Rtan14

and R = 8500cos14

therefore,

F = Rtan14 = 8500cos14tan14 = 2056.3N --> which approx equal 2100

as for part d, i thought that if velocity increase, F will increase (i.e. for example.. instead of 2100 it become 3000 but it is still towards center of circle), therefore the car should slide DOWN the ramp?, they say that
"friction must supply larger force2ward center, car tends to slide UP he ramp"



are you sure tan x = F/R ? I get the exact answer as the markschemes..
i dunno how to explain, but to be simple ... let's play with vector .. draw a triangle on the R line, with R as the hypotenus, 8500 N as the vertical line, and F as the horizontal force .. The angle between the R and 8500 will be 14 degree .. so F/8500 = tan 14 .. F=8500tan14 ≈2100 N ..

For (d), i think as velocity increases, the car would move upwards .. let's say you spin something in a horizontal direction. The faster you spin, it will move higher, isn't it ? I think so ..

I'm not really good in physics .. so correct me if i'm wrong ..
plat4m6
Hey guyz
as for part d, i thought that if velocity increase, F will increase (i.e. for example.. instead of 2100 it become 3000 but it is still towards center of circle), therefore the car should slide DOWN the ramp?, they say that
"friction must supply larger force2ward center, car tends to slide UP he ramp"


another thing, as velocity increases, the F will increase .. that's true. But bear in mind that the force is rely on the R... The frictional force = µR. So, the frictional force increases as the F increases .. but in order to make the frictional force to be constant, the F can not increase .. F must remains constant. How ? by increasing the radius ..
Reply 5
thanks heaps guyz 4 ur help,
umm spellbinder, ur method works perfeclty well - atleast now i know how markscheme got to their answer-, however, What's wrong with my logic there?, i put my reasoning here, in graphical form:

http://www.rahmahwear.com/mymethod.doc

also, how did you deduce that the angle between the weight force (8500) and R was 14 degrees?,

normally when we're looking at stuff sliding up and down ramps, we take weight force to be the hypotenuse and the normal to be adjacent, which is why the Force in direction of ramp is taken to be 'mgsinx'

In ur case, you took 8500 to be the adjacent and R to be hypotenuse

Thanks once again!

PS: Thanks IB@KKICaus 4 posting the page from the past paper great help!
plat4m6
how did you deduce that the angle between the weight force (8500) and R was 14 degrees?,

normally when we're looking at stuff sliding up and down ramps, we take weight force to be the hypotenuse and the normal to be adjacent, which is why the Force in direction of ramp is taken to be 'mgsinx'



the 14 degrees is a math problem .. not physics concern anymore .. i dunno how to explain to you .. first, the angle between the horizontal line(basement) to the inclined plane is given as 14 degrees. Then, using Z angle, since the horizontal line and the horizontal force are parallel; the angle between the horizontal force and the inclined plane is 14°; agree with me ? followed up by the angle between the 8500N and the R is 14°. So, i conclude that the angle between the 8500N and the R = 14°. Just play around with vector and its angle .. that's all.

I understand your problem, and I do get the same answer when i first do it. But you see, if you take the weight force as hypotenuse, it can't resolve into a horizontal component since itself is a vertical component. Do you understand what i mean ? You see, when you're sitting down, your weight, let's say 65g N can't simply be resolved into horizontal component. isn't it ? it will be 8500cos(90) .. right ? The same case here, you resolve 8500N into an inclined force, which is 8500sin(14). And from there, you resolve again this force into horizontal component, which i think it's not logic .. a vertical force can never resolve and resolve until you get the horizontal force.

clear now ?
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in your case, you're taking 8500N as hypotenuse. Then after you obtained 8500sin(14), you use this as your hypotenuse.. what is the logic here ? if you take 8500sin(14) as the hypotenuse, you won't get 8500N as your vertical component if you reverse and find the vertical component back .. isn't it ? from your diagram, 8500sin(14) is the hypotenuse, horizontal force is the adjacent, while the vertical is the weight .. using these info in ur diagram, you wouldn't get 8500N as the vertical force .. which is contradict to the weight force given.
Reply 7
heya spillbidner, guess what mate, i figured out my error

look closely, i said

R = 8500cos14 that's a big statement
it was supose to be
R = 8500/cos14

and then you can take it from there and it comes out fairly easily
Horiz Force = 8500tan14

but ofcourse.. the method in markscheme much easier, i was just gettign excited about the thought of me being the first to discvoer a flaw in maths lol, till math made me discover a flaw in myself!

thanks mate
Ahmed