AS Mechanics in 30 minutes

[@marvellousadenuga1239]

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

[@tonydebling]

My pleasure ☺

[@nanaa428]

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 :)

[@crave4654]

How did u do?

[@brandonpotter7574]

@@crave4654failed

[@irismagic4667]

Hey I'm in your exact same position now, how did you do, did your a levels go well? I hope so 🤞🏻🤞🏻

[@Nafisalhanafi]

@@brandonpotter7574 Yeah man I learnt not to do this the hard way, I'm exactly in the same position as you right now. I just plan to finish this video as well as the one he did for statistics and do exam questions. I blame the teachers for only giving us two days to revise.

[@JoeMama-kl8mj]

Thank you Tony, this was well structured video has helped my son to understand the topic. Keep up the good work!

[@Josh-zd9pq]

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!

[@SamiKhan-pb8nz]

BLESS YOU! IHAVE MY IGCSE TOMORROW AND YOU ARE AN ANGEL

[@brunocarrascosa632]

A bit late but this is AS

[@aimminjale3032]

@@brunocarrascosa632 Brooooo😭😭 Good Luck

[@eggy22222]

KZbin had to recommend me this video AFTER my mech exam today..... thanks youtube :I

[@Vic4life]

Thank you for the video, i have at least general knowledge for tomorrows test :D

[@mr.samfouad28]

Alsalam Alaykoum brother (Hello dear brother). Can you send me the slides' link? Thank you so much. Allah bless you.

[@tonydebling]

Wa-Alaykum-Salaam. docs.google.com/presentation/d/1ozUFikQvwgm6unqcs3wZ4FMgv3ycnKTD/edit?usp=sharing&ouid=113090209927709126638&rtpof=true&sd=true

[@tonydebling]

Or for a pdf drive.google.com/file/d/1lJQVionh3zFAT9aQrhdTY9YkojsVbB1d/view?usp=sharing

[@adityatekriwal1693]

you sound a bit like David Attenborough, thanks for the amazing vid!

[@tonydebling]

Ha ha, flattery indeed!

[@siennathompson2969]

Please do a video on variable acceleration!

[@adamj4346]

Good lad tony love that from you xx

[@eleanorachurch5279]

agreed, thanks tony xoxo

[@amybaldwin8552]

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 😅

[@tonydebling]

Just remember distance travelled is different from displacement. Areas below the axis represent negative displacement but still positive distances travelled.

[@tasniazafarqarina2162]

Thank youu so much.....it really helped me..😃😃😃

[@The.Brother.in.Christ]

You save lives🙏

[@yuna6934]

at 14:47 why the acceleration of Q is not -5g/7 ???

[@tonydebling]

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.

[@crsilvajrr970]

why is it for finding T on q1 you didnt use area of the triangles

[@leiamartin6930]

Great vid! Got me through my mocks!

[@nubla11]

Hi is this video enough for mechanic?:)

[@elliothosie3206]

hi tony, at 22:50 how do i work out the distance travelled for each value of t??

[@tonydebling]

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.

[@tonydebling]

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.

[@elliothosie3206]

@@tonydebling no problem thank you. wasnt using my brain very well for my question👍 thanks again.

[@ratofalady8125]

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

[@tonydebling]

Yep. Positive was chosen to be up. So it's +9 on the way up and -9 on the way down 😀

[@ratofalady8125]

@@tonydebling ah okay thanks, but then how come you only used +9 when substituting it in and not -9?

[@tonydebling]

@@ratofalady8125 that was the u in the formula, initial velocity

[@elerimiles1946]

Thankyou very much, this was so concise and helpful for revision :)

[@tendaijakutindi4175]

Thank you!

[@swipez_yt1741]

5:35 answer for acceleration how does it equal 0.8? isn't it 2.5? or am I tripping?

[@tonydebling]

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 :)

[@swipez_yt1741]

@@tonydebling where the -10 come from?

[@tonydebling]

40 - 50 (subtract 50 from both sides of the equation)

[@swipez_yt1741]

@@tonydebling oh yup, thank you I was just tripping

[@syeed7659]

How do you get T=8.4N at 16:48?

[@tonydebling]

Take g=9.81, rearrange equation 1,29 -T = 1.2x2.81, T = 1.2(9.81-2.81)=1.2x7=8.4N

[@syeed7659]

@@tonydebling ohh, thanks

[@mehdih396]

Thank you so much!!

[@heyitsadelinaaa]

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?

[@tonydebling]

Up direction has been chosen to be positive.

[@mattyhalsall2820]

@@tonydebling er sorry if im wrong but shouldn't gravity be 9.8, and not 10

[@tonydebling]

@@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!!!

[@mattyhalsall2820]

@@tonydebling but in my exams what degree of accuracy is expected, thanks

[@tonydebling]

@@mattyhalsall2820 you will usually be told which value to use for g in the question or at the start of the paper.

[@yaya_04ad]

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)

[@yaya_04ad]

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

[@yaya_04ad]

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

[@tonydebling]

Are you procrastinating?

[@yaya_04ad]

@@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

[@LucyLucyLucyLucyLucy99]

@@yaya_04ad do you get it yet? lmaooo i saw this and had to ask

[@stilllmkp]

why is t=-2 for the suvat equation @7:55

[@tonydebling]

t = +2 as a root ot the quadratic. s = -2 as when particle hits ground displacement is 2 in negative direction

[@ad-dinkashfi4962]

Hi Toney, shouldn't the value of acceleration due to gravity be 9.8. Is 10 acceptable?

[@tonydebling]

You will normally be told what approximation to use in the question.

[@07jas]

thank u so much!!!

[@tonydebling]

No problem 😊

[@aryaslay715]

hello fellow procrastinators

[@Sxlanki._x]

🫣

[@azp_ad4m277]

well done u did all corect

[@ultimateneon3867]

which formula are you rearranging at 23:20

[@tonydebling]

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

[@williamsussex4422]

Cheers cuz

[@jasmine9215]

9:51 why do you +\- 0.458 ?

[@tonydebling]

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.

[@jasmine9215]

@@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

[@tonydebling]

@@jasmine9215 I think you made a mistake in your calculation. B^2 -4ac is 21, sqrt(21) is 4.582

[@Kiarasartt]

Hi Tony, could you explain why there are two Rs in 18:06? I did not understand that part x

[@tonydebling]

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.

[@Kiarasartt]

@@tonydebling thank you!

[@krypton1142]

i think i love you

[@ajeanvi11e]

Great video, thank you 👍🏽

[@infinitegamer2632]

Thanks for the video. Exam in less than 24 hours👍🏽

[@tonydebling]

Pleasure

[@Aryan-fy6cq]

Tony I hava q quostion any chance you could help me?

[@tonydebling]

Sure

[@Aryan-fy6cq]

@@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 [e-mail] so I can send u what I need help with if not I completely understand

[@Zen_Zen_Zense]

@@Aryan-fy6cq so how did the test go, were you prepared enough 👍

[@Vic4life]

@@Aryan-fy6cq did he answer ???

[@mxria1447]

only issue is you dont really explain why youre doing what youre doing which is a bit confusing :/

[@tonydebling]

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.

[@mo.sharif1]

Will this make me pass tomorrow's exam, or shall I ask you one on sports instead?😂😂

[@yahya5022]

same bruh

[@DopeAK7]

For 19:01 is it meant to be (R - T + 950g = 950 x 2) or is it (R + 950g - T = 950 x 2)

[@DopeAK7]

nevermind its the same thing lol

[@tonydebling]

They're equivalent?

[@ellasara9631]

How did you rearrange the formula at 5:37

[@tonydebling]

40 = 50- 1/2a x 25, so 1/2a x 25 = 10, so 25a = 20, and a = 20/25 [was that your question?]

[@rhysgibbs6337]

God bless america

[@ahmadferozali9297]

hi sawaiz

[@epicshibexd5049]

lol

[@tonydebling]

Had to Google it but thank you.

[@condoriano4217]

hi tony, just wanted to ask how you got t= 1000(g-2) at 19:47

[@tonydebling]

Just rearrange the equation above it and factorise. -T + 1000g = 2000, 1000g = 2000 + T, T= 1000g - 2000 = 1000(g - 2)

[@condoriano4217]

@@tonydebling ohh so 7800 N, thanks alot for clearing that up

[@bluelight628]

14

[@JVpiano]

What is bro waffling

[@tonydebling]

Harsh?

[@TalhaTaskin-o2c]

does not matter i am gonna fail anyway

[@haniakhakwani]

good luck! do your best we've got 6 hours

[@K1NGBoSS1595]

Bruh

[@charlottedrnovsky]

thank you so so much!!