2022 Edexcel Maths A Level Mechanics Paper 3 Walkthrough

[@jakewakeling2381]

any danger of less ads

[@aashiralitaj291]

You're a life saver sir cheers 🎉🎉🎉

[@MrAstburyMaths]

You’re most welcome! Be sure to subscribe and share - I have more A level content on the way 👍

[@ahmedpatel4237]

@@MrAstburyMaths do you think you could go though questions that are most likely going t come up?

[@Rua1557]

51:40 are these refinements made based on the modelling assumption it’s a particle?

[@kausarlolz]

this video is really helpful! but for question 3(a) at 22:32 why do we equate the forces with the speed its moving at? Also for question 4 29:02 why did you resolve the Tension force into the horizontal and vertical components? I thought it was perpendicular to the hypotenuse so you don't resolve it?

[@MrAstburyMaths]

Hi, the object was at rest and then a force applied in a particular direction. Afterwards the object would have speed in that same direction. Just as if someone pushed you forward you would move in the forward direction. That’s why we can make the direction of the speed the object the same as the direction of the force. Regarding the components of the Tension… I like to draw a very detailed force diagram for these types of questions as you might have noticed! The components won’t be useful for taking moments you are right but they will be useful when equating forces horizontally and vertically which I believe I had to do at some point in this question. I hope that’s all clear and makes sense! Thanks

[@kausarlolz]

@@MrAstburyMaths that makes perfect sense now!! thank you so much!

[@MissRM-n8i]

Thank you so so much for these videos and explaining them💛

[@hundbudguhdcbdu2638]

Question 4 d, how comes u dont take into account the horizontal components of mg and 2mg?

[@chatterbot___]

It's because it's the coefficient of friction and the friction is horizontal while the mg and 2mg is vertical, not on the X and y axis but in actual space yk

[@charliemccarthy3676]

Really great video! Just wondering as I did mine slightly different, For the first part of question five, could you have rearranged 0=usinaT-4.9T^2 to 4.9T^2 = UsinaT, divided through by T then divided by 4.9 to get T=Usina/4.9, then substituted this value into 120=UcosaT to eventually get the same answer? Both methods seem to work, just wanted to check.

[@MrAstburyMaths]

Yes that would work out 👍 good effort, that was a particularly tricky question!

[@kimberlygillian9230]

Thank you so much. Understood every bit👍👍😍😍

[@MrAstburyMaths]

You’re welcome. Be sure to sub, like and share for more content! 👍🙏

[@mine-rb4hy]

Mr astbury pls help before tommorow, why on question 5b can you not use the time from 5a as 60/cos(a)

[@AlevelStudent-c1y]

Hi Sir, For Q4 wouldn’t the frictional force be in the opposite direction. I thought that as the string “pulls the rod” it causes the rod to move to the right, causing friction to act to the left?

[@AlevelStudent-c1y]

I understand that the question explicitly says to the right but i just can’t visualise why .

[@tacct1kk715]

The component of the tension actually pulls the rod to the left causing the friction to act to the right

[@davidjenner5665]

Projectile question, part (b): K.E. lost = P.E. gained. Initially: K.E. 1/2 mU^2 and P.E. = 0. At max. ht: K.E. = 0 and P.E. = mg.10 = 980m. Therefore 1/2mU^2=980m., giving the required answer: U^2 = 1960 (m cancels). A far more appropriate method, that your students should be made aware of!

[@kausarlolz]

and for question 4 30:35 why dont we draw a reaction force at C? sorry for asking too many questions !

[@MrAstburyMaths]

No problem! Ask away. A reaction force is only present when a object comes into contact with a surface. In this case C is hanging from a rod not resting on a surface

[@kausarlolz]

@@MrAstburyMaths ah that makes sense! so there would be a reaction force on objects that are below or above the rod but not 'on' the rod.

[@salma5345]

Hi sir for question 4 part a you initially took the forces mgcos and 2mgcos but when you got to T you used T and not Tcos . I am a bit confused ? Thanks:)

[@ranjotthiara8581]

casue T was the one that was perpendicular not the other one. The other one was straight up

[@MrAstburyMaths]

I was looking only at forces perpendicular to the slope.

[@lilsels]

I’ve lost my like I think I dropped it, have you found it?? Great video! ❤

[@shirdong]

for q2, why is Miu 0.5?

[@Zen_AlSeed]

its the cofficient of friction mentioned in the question

[@MissRM-n8i]

Can you please do integration question walkthroughs from year 13 maths please?

[@MrAstburyMaths]

You could go through the past paper videos, I’ve labelled every question. Skip to the integration ones. Edexcel A-Level Maths Pure Walkthrough Full Papers kzbin.info/aero/PLrHSLsLY6wii0MUgH083i_e9gx-lbNlHQ

[@lucyparks1263]

Mr Astbury, please help, on Q4 why is there no reaction force at point C?

[@MrAstburyMaths]

Because at point C the rod is not in contact with a surface. A reaction force is when a surface, like the ground or a wall, push back on an object

[@vek5661]

how many adsss

[@archie8459]

hi why is there no normal reaction at c ?

[@kerenkoleoso3423]

fr🥲

[@brunoklotz]

Do not the weight of the rod and the other weight also have a component in the direction of the horizontal ground?

[@remikarim562]

Its horizontal components are zero as weight acts only vertically. Hope that makes sense.

[@ahmedpatel4237]

is there another way apart from guessing friction in exam :)

[@ahmedpatel4237]

for q2

[@OGGY365]

At 32:41 why is there normal reaction at point C?

[@Shindrã_yãn_lucid]

there is none do you mean why is there no normal reaction at c?

[@v75tan]

2023 mechanics ??

[@user-vv6lb5rc6r]

the paper is locked on the edexcel website

[@Infinitespam]

@@user-vv6lb5rc6rSo was the 2022 paper last year March

[@user-vv6lb5rc6r]

@@Infinitespam last year march was 2023 right? usually, papers don’t get released on the website until it’s been quite well over a year since the exam happened. So you wouldn’t be able to access a June 2022 paper in March 2023; you would have to wait until June/July

[@TheAppleWork]

this video was made in march so he has access to the papers

[@KungFuWizardOfJesus]

@ravjayakodi2746 Could get the channel taken down if he does those papers.

[@shirdong]

for question 2, it says the mass of block b is 5kg, but when we are drawing the diagram we used 5g, why is that

[@mariosspanos1542]

This is how we normally do it. The force acting down from gravity is the mass of the particle in kg x g (9.8 normally). As the weight of the object is relative to its mass. Going forward just know whenever you do these typed of questions the arrows going down are in terms of mass in kg x g. Sam for pulleys/ lifts and all those styles of force diagram qs. Then you can just take components of the forces.

[@riley2476]

g just stands for the gravitational constant 9.8

[@Eldennoob786]

Ty 🎉🎉🎉🎉🎉🎉