A Level Physics Revision: All of Newton's Laws, Impulse and Momentum

[@Wartix01]

Grateful for all your lessons. Was really stressed because I couldn’t understand these topics. But you made them much easier to understand

[@zhelyo_physics]

Wow, thank you so much for your comment!

[@arctichoundgamer2105]

Personally, I would have preferred if there wasn't the background music. However, it was still a great video so thank you very much!

[@zhelyo_physics]

Interesting! I was actually wondering recently about that! I'll experiment with no background with the next one and see the result. Thanks for the comment!

[@arctichoundgamer2105]

@@zhelyo_physics No problem :) maybe just making it a tad quieter would work too, I’ll watch them either way!

[@TaibaRaza-k3d]

I’m so grateful for these video. I have A levels in about two days. Really covered all me revision. Thank you.

[@zhelyo_physics]

thanks a lot for the comment! Good luck!

[@OmarAntar-er3kl]

My head will blow up 💥

[@major8409]

11:38 music decided to go hard for a moment 😂

[@aakashs5420]

i want you to solve using components of velocity in x direction

[@Aksa-y8y]

For the momentum question on 18:00, I tried getting my answer using horizontal components but didn't get the same answer. I assumed that the second object was stationary and used the equation M(a)V(a) + M(b)V(b) = M(a)V(a') + M(b)V(b'), which gave me ((1)(15)+(1)(0) = (1)(10cos(33)) + (1)(vcos(50)), why or how would this method be wrong??

[@zhelyo_physics]

Hi, I think I covered this in some of the older comments but I think set up the problem wrong with the initial angles. Typically your method would work though.

[@anonymous99923]

For the last question, does that mean 10cos(33) + 7.1cos(50) = 15 as the total momentum in the x direction before = total momentum in x direction after? But total momentum before is 15 Nm and total momentum after is 12.95 Nm? Where have I gone wrong?

[@zhelyo_physics]

This would be correct, I made up the example to showcase the method. I think the input angle of 50 in the question is wrong. The method you describe would be absolutely correct!

[@rajvardhanpatankar3547]

Hi I had question on the momentum in 2d example as when I did it with momentum in the x axis i got a different value for v my calculation was mVcos(50)+m10cos(33)=15. Sir can you explain where I went wrong

[@zhelyo_physics]

Hi, have a look at the previous comments where I have answered this in detail. So when I made this question years ago the angles should have added up to 90 degrees which leads to inconsistency in x and y answers. Your calculations are okay! : ) Good work!

[@stelsie922]

this was beautiful cheers

[@zhelyo_physics]

thank you for the comment!

[@story.shifters]

Very useful lessons! Thank you so much! May you, please, post your notes? (whole screens of lessons) It would make our life much more easier ;) Thaaanks!!!

[@zhelyo_physics]

Thanks for the comment! Sorry only the videos are available. Glad they are useful though! : )

[@nikolayyordanov8986]

Hello Sir, in an exam question would we be asked to find the momentum in the vertical or horizontal direction for 2D momentum, or do we have to know which one they're looking for? It confuses me how you know which one you have to find. Thank you!

[@zhelyo_physics]

Hello, you can normally find a problem using both. I tend to just go for the easier way. E.g. if the total momentum in the y direction is 0 initially, use the y momentum after as they will add up to 0. Hope this helps! : )

[@nikolayyordanov8986]

@@zhelyo_physics Thank you!

[@sh_legendsh_legend6754]

About momentum in 2d you only found rebound speed via equation momentum y before= momentum y after. When would you use momentum x before= momentum x after. Or can you do either to find rebound speed? Not sure how you would find the rebound speed by using momentum x equation if if you can use either equation.

[@zhelyo_physics]

Funny I have just published a brand new video on this topic. Hope this helps!

[@fariataxnim]

shouldnt it be 0=10sin33 - vsin50 because the two momentum in different direction?

[@zhelyo_physics]

if added up they should give 0, you can write it as +vsin50 but when you solve for v this would give a negative answer. Hope this is helpful!

[@As-iy2ki]

At 16:36 why is it sin instead of cos? I thought the vertical components cancel out so we use cos. Correct me if I am wrong.

[@zhelyo_physics]

You can do these problems with either sin or cos, if the vertical components cancel out you can directly set them equal to one another. Hope this helps!

[@PavanYK]

For the momentum in 2d question, why would you look at the vertical component rather than the horizontal component; would it matter?

[@zhelyo_physics]

Nope it will not (please note that there is an inaccuracy in this question though which I mention in other comments which leads in difference in answers)

[@pratyasha279]

3:15 How is R electrostatic in nature in this example?

[@zhelyo_physics]

so the any surface is made out of electrons that cannot occupy the same space at the same time. Hope this helps!

[@AmL-L-L]

random question but is there a brief explanation as to why R (reaction force) is electrostatic?

[@zhelyo_physics]

a good analogy (warning just an analogy) is that it works a bit like a trampoline, you push down, displacing electrons and changing their balance point, pushing back up. Hope this helps!

[@Aakash-yy4jf]

i think you have made a mistake in momentum in 2d collision. when using momentum in y direction the momentum after the collision in y direction of the other ball is in opposite direction so you have to include the negative sign...btw great video

[@aaronkibs8353]

Thanks for the video. For the last question why is mvsin50 defined as positive rather than negative, as you know the velocity is travelling down .

[@zhelyo_physics]

excellent question, it is not defined as positive per say, but when you do the algebra it comes out as negative, essentially both y components need to add up to 0. Hope this helps!

[@uncloned69]

why is the momentum before zero? 16:19 explain a bit more to understand pls

[@zhelyo_physics]

As there is no motion in that direction

[@iSuperMC]

but why do you get a different solution for v when you resolve in the x axis ( 15 = 10cos(33)+vcos(50) ) because the answer you get that way is v = 10.29

[@zhelyo_physics]

interesting, I'll investigate that!

[@indiantechsupport546]

@@zhelyo_physics I think your angle 50 is wrong. I got it to be an angle of about 39.5 (39.47322373) degrees. If you use that, you get a velocity of about 8.6ms-2 (8.567311691) for v. Keep up the great work :)

[@zhelyo_physics]

@@indiantechsupport546 I think you are absolutely right! I actually remember just coming up with an example to showcase the method. Thanks!

[@persowa8940]

Thanks bro I thought I was going insane cuz I had the wrong solution turns out I was right all along

[@devakp8832]

If you use the x component would you get the same velocities?

[@zhelyo_physics]

actually not in these case due to the initial conditions, the principle is the same though. I will make a further video addressing this.

[@Azeem8166]

Hi sir how do tou know to use the horizontal instead of the vertical for the sohcahtoa

[@zhelyo_physics]

A good way of thinking of is the total momentum in the y direction was 0 before, so it will be 0 after which makes the calculations easier. Hope this helps

[@abdi5383]

was wondering if you had the midmaps that you create in a colection

[@zhelyo_physics]

thanks for the comment, sorry only the videos are available.

[@kausarlolz]

hello sir, if i wanted to find the area under a non linear curve. do i find the area of one box and multiply it by the total number of boxes? what if we have some incomplete boxes like a quarter box in the area or a half box?

[@zhelyo_physics]

I think it's a question of approximating in that case. Typically the answers are not exact in those types of questions.

[@benjaminfox3761]

Hi Sir. I was doing an exam question that asked about working out the change in momentum after an object rebounds off of a wall, and it was -2mv. Do you happen to have a video explaining how you get this? Thank you for all your videos btw, they are an absolute lifesaver!

[@zhelyo_physics]

I do have an old video on this actually: kzbin.info/www/bejne/pKjYepaqiryirdk enjoy!

[@benjaminfox3761]

@@zhelyo_physics brilliant thanks you!

[@teacupcakes2739]

i literally learnt this in fmaths today. since change in momentum = mv - mu you know mv = -mu since it rebounds at same speed. So momentum = -mu -mu = -2mu

[@justblitz1566]

For the last question, woulndt the momentum before be 15? Since MaUa + MbUb gives 15 + 0? With that im getting v as 12.47 m/s

[@justblitz1566]

Sorry I just saw the clarification in the other comments, thanks for the amazing video though!

[@zhelyo_physics]

thank you for the comment!

[@enderman4514]

youre awesome ❤

[@zhelyo_physics]

thanks a lot for the kind comment!

[@S_H-R]

Very helpful sir

[@zhelyo_physics]

thank you so much!

[@tayyibjamil8573]

Hi is momentum in 2d covered in the AS spec for AQA physics

[@zhelyo_physics]

I can't remember off the top of my head sorry but a quick Google of the spec will give you an answer and even better you can use it as a check list. Good luck!

[@clevereagle1911]

Does it work for international Alevels Edexcel ?

[@zhelyo_physics]

Yes

[@teacupcakes2739]

I like the silly music in the background

[@indiraflo6770]

Thank you

[@zhelyo_physics]

Anytime! : )

[@m4rzb4rz-qq3yq]

2x speed bgm isnt the best sound in the world 😅

[@lukienure5007]

Hi sir , is this aqa ?

[@zhelyo_physics]

applicable to AQA but for all exam boards, the last part on 2d Momentum is not part of the AQA spec, but worth double checking.

[@gautammorkhandikar394]

how do you identify that a question will use newtons laws?

[@gautammorkhandikar394]

the shower head question is a bit confusing when you first see it

[@zhelyo_physics]

anytime when there is acceleration. If the mass is constant, you can use F=ma, if not e.g. a fluid flow, the rate of change of momentum as described : )

[@dureadan-p9k]

M¹v¹ = m²v² M¹v¹ +m²v²= m²v¹ + m²v² Big confusion in thisss in which cases we will apply them Or maybe the second formula doesnt even exist😅 im very bad at phy

[@zhelyo_physics]

Depends on the initial velocity of the objects, we are always just conserving mass x velocity for each object. Hope this helps!

[@asyadallykhankhodabocus6278]

No music again plss

[@zhelyo_physics]

Definitely, this was amongst my first batch of videos. There has been no music since 2020 : )

[@nadiab7747]

cant watch the vidoe with thaat music

[@zhelyo_physics]

I agree, this is one of my first videos from years ago