AS Mechanics in 30 minutes
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- เผยแพร่เมื่อ 30 ก.ย. 2024
- AS Mechanics revision video. Quick last minute revision in less than 30 minutes. Covers suvat, motion under gravity, connected particles with pulley and with inclined plane, vectors, variable acceleration.
i subscribed and liked! You're a life saver. About to do a mock and im terrified because im aiming for an A but ive been doing so bad- your videos are so very helpful. Thank you for taking the time out of your day to help strangers :)
How did u do?
@@crave4654failed
Hey I'm in your exact same position now, how did you do, did your a levels go well? I hope so 🤞🏻🤞🏻
Thank you Tony, this was well structured video has helped my son to understand the topic. Keep up the good work!
Alsalam Alaykoum brother (Hello dear brother). Can you send me the slides' link?
Thank you so much.
Allah bless you.
Wa-Alaykum-Salaam.
docs.google.com/presentation/d/1ozUFikQvwgm6unqcs3wZ4FMgv3ycnKTD/edit?usp=sharing&ouid=113090209927709126638&rtpof=true&sd=true
Or for a pdf drive.google.com/file/d/1lJQVionh3zFAT9aQrhdTY9YkojsVbB1d/view?usp=sharing
Thank you very much for this video. Whenever my teachers taught mechanics i just never seemed to understand it as easily as everyone else, and the lessons where so fast paced. Thank you for helping me understand what is actually going on. Its much easier to do now that i understand
My pleasure ☺
Thank you for the video, i have at least general knowledge for tomorrows test :D
BLESS YOU! IHAVE MY IGCSE TOMORROW AND YOU ARE AN ANGEL
A bit late but this is AS
@@brunocarrascosa632 Brooooo😭😭 Good Luck
TH-cam had to recommend me this video AFTER my mech exam today..... thanks youtube :I
The video might be 4 years old but it is still so relevant. You've managed to get me to (mostly) understand these topics in not even half an hour better then my teacher in weeks of lessons. Thank you!
you sound a bit like David Attenborough, thanks for the amazing vid!
Ha ha, flattery indeed!
hello fellow procrastinators
🫣
i think i love you
hi tony, at 22:50 how do i work out the distance travelled for each value of t??
Substitute the value of the into the equation for x, namely x =1/2t^2(t^2 - 2t +1) which is given in the question.
To be more precise x is the displacement, which means if it starts at zero, becomes 1/32, and then goes back to zero the distance travelled is 1/32 + 1/32 = 1/16, then from x=0 at t=1 it becomes x=2 at t=2, so it travels a further distance of 2, giving a total distance travelled of 2 1/16, whilst the displacement at this point in time is still 2. So it is important to understand the difference between distance travelled and displacement.
@@tonydebling no problem thank you. wasnt using my brain very well for my question👍 thanks again.
Good lad tony love that from you xx
agreed, thanks tony xoxo
Thank youu so much.....it really helped me..😃😃😃
Hi is this video enough for mechanic?:)
at 14:47 why the acceleration of Q is not -5g/7 ???
You are asked to find the magnitude which us always positive. Whether the acceleration is positive or negative depends on you choice of positive (displacement, velocity, acceleration and force) up or down.
Will this make me pass tomorrow's exam, or shall I ask you one on sports instead?😂😂
same bruh
For part B of the variable acceleration question, instead of plugging in all the different values, could you do a definite integral of the velocity equation since the area under the curve of a velocity time graph is displacement? You would still need to split it up of course but would it still work? its probably the most complicated way to do it but its the first thing that came to my mind 😅
Just remember distance travelled is different from displacement. Areas below the axis represent negative displacement but still positive distances travelled.
Please do a video on variable acceleration!
You save lives🙏
why is it for finding T on q1 you didnt use area of the triangles
For the ball question at 8:22 did you take the upwards direction as positive and is that why 9 was positive cause the ball was moving up? And why wouldn't 9 be negative cause the ball moves down too
Yep. Positive was chosen to be up. So it's +9 on the way up and -9 on the way down 😀
@@tonydebling ah okay thanks, but then how come you only used +9 when substituting it in and not -9?
@@ratofalady8125 that was the u in the formula, initial velocity
14
What is bro waffling
Harsh?
only issue is you dont really explain why youre doing what youre doing which is a bit confusing :/
An interesting philosophical point. In general the answer is to choose some laws of mechanics to allow us to write some equations, and then to solve the equations to find out the missing information that we are asked for.
Great vid! Got me through my mocks!
How do you get T=8.4N at 16:48?
Take g=9.81, rearrange equation 1,29 -T = 1.2x2.81, T = 1.2(9.81-2.81)=1.2x7=8.4N
@@tonydebling ohh, thanks
Thank you so much!!
God bless america
thank u so much!!!
No problem 😊
hi sawaiz
lol
Had to Google it but thank you.
5:35 answer for acceleration how does it equal 0.8? isn't it 2.5? or am I tripping?
Too many mushrooms in the omelette I think. Are you happy with 40=50- 1/2 a 5^2? -10 = - 1/2 a x 25 so 20 = 25 a so a = 20/25 = 0.8 :)
@@tonydebling where the -10 come from?
40 - 50 (subtract 50 from both sides of the equation)
@@tonydebling oh yup, thank you I was just tripping
Thank you!
Cheers cuz
Hi Toney, shouldn't the value of acceleration due to gravity be 9.8. Is 10 acceptable?
You will normally be told what approximation to use in the question.
Hi, for 6:13, why couldn't you use s=vt-0.5at^2? Why did you use s=ut+0.5at^2? Wouldn't it be easier to use the first than having to find the value for u to use the second equation
could you have not done 20=10t-0.5(0.8)t^2 I came to the answer t=22.8 t=2.2 (to 1dp)
(I also hope you're having a great day)
ahhhhh wait is it because v is the final velocity at the end of the journey at AB but we don't know what it is at the midpoint of AB/ it will be different at the midpoint so we cant use 10ms^-1 for the second part
wait no what but you used 10ms^-1 to find out u but I thought (well came to the conclusion that) we couldnt use v for any other point other than the end at AB
Are you procrastinating?
@@tonydebling but can you help me and answer why because I have no idea and I can't move on if I don't understand
@@yaya_04ad do you get it yet? lmaooo i saw this and had to ask
does not matter i am gonna fail anyway
good luck!
do your best
we've got 6 hours
Bruh
why is t=-2 for the suvat equation @7:55
t = +2 as a root ot the quadratic.
s = -2 as when particle hits ground displacement is 2 in negative direction
Thank you for the video! At 8:12, why did you say that the acceleration is -10? Shouldn't it be positive 10 since the force of gravity downwards into the ground is positive?
Up direction has been chosen to be positive.
@@tonydebling er sorry if im wrong but shouldn't gravity be 9.8, and not 10
@@mattyhalsall2820 depends on accuracy used. Question will usually tell you 10, 9.8, 9.81 and I'm sure there are more accurate versions, which would also depend on altitude!!!
@@tonydebling but in my exams what degree of accuracy is expected, thanks
@@mattyhalsall2820 you will usually be told which value to use for g in the question or at the start of the paper.
thank you so so much!!
which formula are you rearranging at 23:20
in the question you are given x=1/2t^2(t^2 -2t +1) which can be factorised/rearranged to give x=1/2t^2(t-1)^2
Thankyou very much, this was so concise and helpful for revision :)
well done u did all corect
9:51 why do you +\- 0.458 ?
Values of t as particle when it is at height are roots of the quadratic 5t^2 - 9t + 3 = 0 using the quadratic formula you get the roots as 0.9 +/- 0.458. The first part is -b/2a second part is the discriminant divided by 2a. Sqrt(9^2 - 4(5)(3))/2(5). Hope this helps.
@@tonydebling tysm for the quick reply! i used quadratic but kept getting different answers compared to the vid, maybe there was an error in my calculation
@@jasmine9215 I think you made a mistake in your calculation. B^2 -4ac is 21, sqrt(21) is 4.582
Tony I hava q quostion any chance you could help me?
Sure
@@tonydebling hey I know this sounds Ridiculous but I have a test next week and I might need some help moment any chance you could contact me at my email Ariyan26f@gmail.com so I can send u what I need help with if not I completely understand
@@Aryan-fy6cq so how did the test go, were you prepared enough 👍
@@Aryan-fy6cq did he answer ???
Great video, thank you 👍🏽
Hi Tony, could you explain why there are two Rs in 18:06? I did not understand that part x
Hi Kiara, there is an equal and opposite reactions between the lift and the woman. The blue R reprsents the lift pushing the woman up, and the red R represents the woman pushing the lift down. These are equal and opposite. The colours were used to separate the forces acting on the woman and the forces acting on the lift.
@@tonydebling thank you!
How did you rearrange the formula at 5:37
40 = 50- 1/2a x 25, so 1/2a x 25 = 10, so 25a = 20, and a = 20/25 [was that your question?]
Thanks for the video. Exam in less than 24 hours👍🏽
Pleasure
hi tony, just wanted to ask how you got t= 1000(g-2) at 19:47
Just rearrange the equation above it and factorise.
-T + 1000g = 2000, 1000g = 2000 + T, T= 1000g - 2000 = 1000(g - 2)
@@tonydebling ohh so 7800 N, thanks alot for clearing that up
For 19:01 is it meant to be (R - T + 950g = 950 x 2) or is it (R + 950g - T = 950 x 2)
nevermind its the same thing lol
They're equivalent?