Edexcel IAL Physics Unit 1 14 May 2019 Watch

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i face the same problems...its confusing which value to use for the next eq.
S.U.V.A.T !!!
Name:  21.png
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Size:  31.8 KB know what values are there in the 4 equations...u dont kneed to have all the data to plug in at once..sometimes u gotta find a value and then plug it into another Eq
remember to divide the velocities into 2 components...sometimes the object might not have any vertical initial velocity when thrown from a height...
sometimes u have to find the total horizontal distance traveled by someth.... using only vertical measurements find the time..
if there no air resistance, horizontal velocity remains same throughout ! and use speed=d/t for its calculations...
learn howto find the final velocity of the object when its about to hit the ground (resultant of horizontal and vertical components) and the angle its gonna strike with(trig) ..post Q's u find tricky here...and just be careful while handling the data..
i cant help u if i dont know which part do u go wrong in...so watch vids and so some practice Q's..best
and it happens...it is, a little tricky..

(Original post by usman200186)
My only problems are of projectile motions. I always mess up in them. I dunno sometimes the answer is so easy but I can't figure it out. Can anyone help me with this please.
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(Original post by : ).)
i face the same problems...its confusing which value to use for the next eq.
S.U.V.A.T !!!
Name:  21.png
Views: 17
Size:  31.8 KB know what values are there in the 4 equations...u dont kneed to have all the data to plug in at once..sometimes u gotta find a value and then plug it into another Eq
remember to divide the velocities into 2 components...sometimes the object might not have any vertical initial velocity when thrown from a height...
sometimes u have to find the total horizontal distance traveled by someth.... using only vertical measurements find the time..
if there no air resistance, horizontal velocity remains same throughout ! and use speed=d/t for its calculations...
learn howto find the final velocity of the object when its about to hit the ground (resultant of horizontal and vertical components) and the angle its gonna strike with(trig) ..post Q's u find tricky here...and just be careful while handling the data..
i cant help u if i dont know which part do u go wrong in...so watch vids and so some practice Q's..best
and it happens...it is, a little tricky..
Yeah, I know all this stuff but just sometimes you know. I can't find a specific question now but sometimes in markscheme it is give to divide the time by 2 and they change +ve value for g to -ve value and it makes no sense to me. These two are my only problems that why we need to divide time by 2 sometimes
and why a is taken as negative when its supposed to be positive i.e the object is moving down. But thanks for explaining and clearing my concepts tho
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alright so ..when the ball is going up..its working against gravity..so take it as -ve..when the ball is falling down..its usually taken as +ve .u can interchange it tho....signs are there to denote direction..sometimes even the value of height obtained can be -ve...it all depends on the direction of the movement of the object..i usually take up as +ve and down as -ve...and i apply it to every quantity
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Size:  54.0 KB for the time (/2)...ill haveto see the question ..try to find the Q ..itll be helpful
in some questions u might take the initial vertical velocity and final vertical velocity (this is only for the first half of the projection)..to find the total time ...after using v=u+at .u get the time for the first half of the projection...so u need to multiply by 2 to find the total time.....so the question u came across might have something similar to this concept...or the object must have been thrown from a cliff because of which u hadto div. by 2 ..im just guessing tho
(Original post by usman200186)
Yeah, I know all this stuff but just sometimes you know. I can't find a specific question now but sometimes in markscheme it is give to divide the time by 2 and they change +ve value for g to -ve value and it makes no sense to me. These two are my only problems that why we need to divide time by 2 sometimes
and why a is taken as negative when its supposed to be positive i.e the object is moving down. But thanks for explaining and clearing my concepts tho
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(Original post by : ).)
alright so ..when the ball is going up..its working against gravity..so take it as -ve..when the ball is falling down..its usually taken as +ve .u can interchange it tho....signs are there to denote direction..sometimes even the value of height obtained can be -ve...it all depends on the direction of the movement of the object..i usually take up as +ve and down as -ve...and i apply it to every quantity
Name:  21il.png
Views: 13
Size:  54.0 KB for the time (/2)...ill haveto see the question ..try to find the Q ..itll be helpful
in some questions u might take the initial vertical velocity and final vertical velocity (this is only for the first half of the projection)..to find the total time ...after using v=u+at .u get the time for the first half of the projection...so u need to multiply by 2 to find the total time.....so the question u came across might have something similar to this concept...or the object must have been thrown from a cliff because of which u hadto div. by 2 ..im just guessing tho
Its not the same question, but still I cannot solve this one. Can you helo me with this? I really appreciate!Name:  IMG_20190513_111025-compressed.jpg.jpeg
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Here's the mark scheme Name:  IMG_20190513_111159-compressed.jpg.jpeg
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Thanks!
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(Original post by : ).)
no worries...i myself couldnt do it..i took help from the link i sent u
if ur confused about why we didnt take 10 minutes into consideration took at Brown panther's comment and reply (after uve opened the link)
Yeah i was using 10minutes
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after finding the 2 components of velocity ...find the duration when the ball is in the air by...
using the H distance and H initial velocity ...after finding the time ...calculate the V distance (height) the ball reaches ...using s=ut+0.5 at^2 (calculate with V velocity).....if that max height is greater than 2.4 ..the ball wont go in...keeping in mind that u have to keep the margin for the ball(diameter of the ball)..the mid point of the ball might hit the edge causing it to bounce off ...

(Original post by usman200186)
Its not the same question, but still I cannot solve this one. Can you helo me with this? I really appreciate!Name:  IMG_20190513_111025-compressed.jpg.jpeg
Views: 16
Size:  90.3 KB

Here's the mark scheme Name:  IMG_20190513_111159-compressed.jpg.jpeg
Views: 16
Size:  62.3 KB

Thanks!
u might be thinking that after reaching the max height the ball ..while coming back down..may enter the goal...
to argue, there's also a possibility that the ball can enter before or even at its max height...max height should be 2.4..below that everything is fine ...were not bothered..thats why they have not encouraged us in the marking scheme to find what H distance the ball traveled.
my Q: when i did this i started finding if the ball crossed 11 m or not..mightbe that the ball just traveled 10 m and stopped ..why didnt they calculate to find whether the ball makes it to the goal horizontally... ?
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why is the answer c?
why not A
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(Original post by : ).)
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why is the answer c?
why not A
Because the A is in cm which is not standard units for lenght
All other answers are correct, but the most precise and accurate one is the C. Because in B and D the next figure after 5 may have been something other than 0 like 2-3-4
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(Original post by : ).)
my Q: when i did this i started finding if the ball crossed 11 m or not..mightbe that the ball just traveled 10 m and stopped ..why didnt they calculate to find whether the ball makes it to the goal horizontally... ?
Read the question, it says *Ball is kicked towards the goal "from a distance of 11m"* So it means 11 m is from the goal. Btw thanks for explaining, I got the answer after 2 tries.
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yea..but it doesn't say that the ball is guaranteed to go 11m or more
should i think logically here...after all..11 meters is enough to get the ball in the goal if applied sufficient force
(Original post by usman200186)
Read the question, it says *Ball is kicked towards the goal "from a distance of 11m"* So it means 11 m is from the goal. Btw thanks for explaining, I got the answer after 2 tries.
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(Original post by : ).)
yea..but it doesn't say that the ball is guaranteed to go 11m or more
should i think logically here...after all..11 meters is enough to get the ball in the goal if applied sufficient force
You cannot calculate how far the ball travels horizontally with the information given. If you assume it does not travel 11m then the question cannot be done. Only if you assume that it travels exactly 11m can you even attempt the question.
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oh..owkay
(Original post by AsithU)
You cannot calculate how far the ball travels horizontally with the information given. If you assume it does not travel 11m then the question cannot be done. Only if you assume that it travels exactly 11m can you even attempt the question.
but how will i know in the exam....like what further info is needed to calculate the horizontal distance
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(Original post by : ).)
oh..owkay
but how will i know in the exam....like what further info is needed to calculate the horizontal distance
If you had the time you could do it by horizontal velocity x time.
Here you have to calculate the time yourself and you can only calculate the time for exactly 11m because that's the only other value you have
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thankyou!
(Original post by AsithU)
If you had the time you could do it by horizontal velocity x time.
Here you have to calculate the time yourself and you can only calculate the time for exactly 11m because that's the only other value you have
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how do i approach this question?
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how do i approach this question?
With sadness
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Which paper is this
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I think for this ull have to prove its not the average speed
So according to me it should be written somewhat as
It is the instantaneous speed as it gives the speed at a particular moment whereas the average speed = total distance /total time which equals to 6.843/(13 divided by 60)
So average speed= 32kmh-1 while the speed shown is only 28
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(Original post by Nighthunter67)
With sadness
applies for every question really
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Why is the answer of 7 D??
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