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

Chapters:
00:00 Intro
00:10 Newton's First Law
00:38 Newton's Second Law
01:46 Newton's Third Law
02:21 Weight and Normal Reaction
03:29 Non-constant mass
07:09 Impulse
08:04 Force against Time graph
09:33 Momentum
10:00 Conservation of Momentum
12:31 Elastic and Inelastic Collisions
14:00 Momentum in 2D
This is excellent A Level Physics revision for all exam boards including OCR A Level Physics, AQA A level Physics, Edexcel A Level Physics, CIA Cambridge International A Level Physics, Eduqas etc.

[arctichoundgamer]

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

[Wartix]

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

[TaibaRaza-kd]

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

[major]

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

[aakashs]

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

[anonymous]

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?

[Aakash-yyjf]

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

[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!!!

[fariataxnim]

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

[rajvardhanpatankar]

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

[PavanYK]

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

[sh_legendsh_legend]

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.

[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

[devakp]

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

[nikolayyordanov]

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!

[Azeem]

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

[abdi]

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

[benjaminfox]

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!

[aaronkibs]

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 .

[teacupcakes]

I like the silly music in the background