5 Momentum Conservation & Impulse Momentum Theorem | Centre of Mass Class 11 | JEE Main & Advanced

[@ganeshchaudhari203]

Sir your are the best teacher of physics in whole KZbin.👍👍😃

[@shivangipriyadarshini1409]

Right

[@annubarthwal9607]

Your are right bro 💯👌

[@sumitnaiyaa]

I saw this same comment in physics wallah's video🤦‍♂️

[@Arnab_Chakraborty_IITKGP]

@@sumitnaiyaa lol..its advance bro

[@SwayamKrishnaartsandcrafts]

bhai aapka kitna rank aya mai 2025 aspirant hun

[@bharatbarnwal8540]

Bina halla kiye free me padhane wale shikshako ko sat sat naman

[@kartikgoyal4815]

I understand your sarcasm 😂😂

[@adarshdixit7568]

?? Halla kaun teacher karta hai ??

[@bharatbarnwal8540]

@@adarshdixit7568 unacademy, vedantu, go prep, bla bla

[@saswatmishra5644]

@@bharatbarnwal8540 exactly 😂😂😂

[@anuragkumar8335]

@@adarshdixit7568 jake ek baar bahr ki duniya dekho🤪🤪😂😂

[@Ndiedddd]

In the first question, the motion of the bodies is very interesting. 3 Separate oscillatory motions at the same time. But what is more interesting is abj sirs explanation

[@kishangopal1424]

ABJ sir God Of physics

[@ayushuttarwar5744]

I think no one skips hard question although they understand it or not..😅😂

[@jaygondaliya6245]

true bro

[@tripurikumar8500]

yes.peoples like you

[@MashoomAyush]

Bhai hua iit

[@AparnaJEE2025]

​@@MashoomAyush🌝

[@Jeeeee2025]

Bat to sahi hai 😂😂

[@sarvdadevi378]

Sir apki video dekhne par concept crystal clear ho jata hai

[@sst940]

1st que is very good que!!

[@iitkarenge]

U can make anyone fall in love with physics .... even difficult questions feel easier with u

[@aryanpal8600]

Thanks sir .... The best physics teacher

[@atharva3490]

This 1st quescame in my aits(mains) that's beauty of Abj sir here again to revise this 💕🙏🏻

[@Parth-dc8mp]

@aryan4679 nope , he is talking about test series

[@VkKingdom-j7u]

Which coaching bro fiitjee

[@nikola6tesla]

i remember the 1st question which i struggled understanding in 11th. And now because of ABJ sir, it seems like piece of cake.

[@VkKingdom-j7u]

Bhai mujhe abhi bhi usme struggle Karna pad rha hai 😢😢 samjh nhi aa rha bead ke respect me vo circular motion kaise kar rha

[@NilanjanHalder-lp9vd]

​​@@VkKingdom-j7uabbe wo perpendicular velocity hai wrt bead hai na, normal circular motion ko visualize karo to center ke about jo circumference pr motion hota hai uske ek velocity perpendicular to radius hota hai, yaha wo center ke jagah bead hai bss

[@PadhleBSDK-164]

30:18 Equations: 1. Relative velocity of bead ko lekr ke centripetal force wali equation bana denge. Pseudo force consider krna hoga is baar. 2. Momentum conserve kr denge along horizontal. (GROUND FRAME se velocity lenge is baar 2m ki, jo ki string ke perpendicular nhi h) 3. Potential drop = kinetic obtained, firse, ground frame se hi velocity lena hoga 2m ki. mera 1.2188 mg aaya, shayad kahi calculation mistake hui ho

[@yashwanth2837]

1:10:00 similar question was asked in adv 2021.

[@hypocratus7341]

18:30 guy named "momentum conserve kar do" : relatable

[@jayeshverma6755]

Thanks so much sir i was struggling in those hinge problems....i was not able to understand that how v along string becomes zero ...love u sir ❤️

[@jaskaransinghsodhi3152]

Thank you so much sir. Please MECHANICS ki frequency increase kar dijiye at least rotational tak. Thank you sir

[@kashyaptandel5212]

Oye papaji

[@amay96]

Oye paaji

[@shrekek]

Oye papaji

[@PANDITDADDY-kq9uf]

Oye papaji

[@vedanthumbe4649]

Oye papaji

[@shivanshuagrawal9532]

May God bless you 🙏

[@sashrikgupta9708]

Ohhh my god kal mera isshi kabtest hai aur mujhe ye aajh nahi araha tha I can't believe on my luck😁😁😁😁😁😁

[@divyakrity198]

30:00 beat problem ❤😊

[@prakharmishra9185]

The last question was too good!

[@hemanggadhvi7505]

1:10;19 similiar question jee advance 2021

[@shashvatpandey3028]

](1944*L+150)*mg]/1225 --- Ans.30:00

[@eagl-3]

What is the answer of that question of tension in string while making an angle 37° with horizontal?

[@swatisharma5230]

34mg/7

[@lavanyabalivada4773]

THANK YOU SO MUCH SIR..!!!😊😊😊

[@a2motivationfan495]

00:00 to 31:37

[@18harshraj61]

Impulse new topic starts after 30 minutes of this lecture from starting

[@Adityasinghm26v]

Tension at 37° is 1.91mg

[@sameerkushwaha584]

hw : 258 mg/125 .

[@harsh9023]

1:22:37 sir can we find velocity by TMEC on system as there are no external forces on system. if yes, then i am not getting the correct answer from that. what is the problem. i mean; 0 + (1/2)*(2)*(10^2) = (1/2)*1*(v^2) + (1/2)(2)((5^2 + V^2 )^(1/2)) 150 = 3(v^2) v = 10/2^(1/2) but v = 10/2^(1/2)

[@kishangopal1424]

SIR PLEASE PLEASE LEAKING OF DIELECTRIC CAPACITIOR BI KARA DENA SIR

[@aadarshkumar6398]

Fun fact:-Motion of 2m mass particle will be an ellipse wrt ground (in the 1st question),it can be easily proved...

[@shashvatpandey3028]

how?

[@JaswanthSai_25]

@@shashvatpandey3028 Do some math and you fill figure out. Find the equation of motion using given info, I just drew 6 phase diagrams in one place, can easily prove its a part of an ellipse.

[@NK-bo7lt]

​@@JaswanthSai_25bhai ham topper nhi hai Bata Dena kaise

i knew it had a locus from one of the conic section

[@vedanthumbe4649]

30:17 478/125xmg

[@peace-f4n1m]

bhai aap kabse prep kar rhe ho aapka ek comment bohot pehle dekha tha

[@aloksisodiya30]

Sir On competishun app when we give AITS pratham we do not have option to go back on previous section once we done it. Sir could this issue be resolved.

[@sagarbhardwaj2438]

Sir answer to the given problem is (690/361)mg?

[@AshishSingh-jc9tx]

Yes

[@terabaap5422]

Sir aapne momentum conserve kaha se kiya? M1 se toh kar nahi sakte kyuki usme horizontal force lagray agr aap system bolke conserve kar rahe ho toh fir X com ya fir ground se karna hoga naaa? Aur fir vohi reference se energy conserve karna hoga naa? Please reply sir i am maxed confused. Ground aur com se toh fir velocities bhu alag hojayenge naaa

[@sports_boy329]

Best physics teacher ever I saw on KZbin ❤..it 2023 I complete my jee preparation by these lectures

[@harivatsaparameshwaran4174]

Sir in the first question last part while conserving mechanical energy shouldnt we take length of centre of mass? You have taken l , but shouldnt it be 2/3l ?

[@aayushjoshi3681]

We are applying circular motion equation on 2m ,not on COM, so we will take l.

[@bharatbarnwal8540]

Sir puri syllabus upload kariye please

[@RedFrost6]

explained well

[@MohitTyagi]

Keep working hard, student! Wishing you all the best.

[@vishwajeetchoudhary8264]

30:00 My answer - T = 966/305 mg

[@NeerajKumar-oj8uv]

*SIR PLEASE TELL US ABOUT YOUR TEST SERIES FOR JEE 2021 DROPPERS with plan... EAGERLY WAITING FOR YOUR REPLY,🙏🙏🙏*

[@pratiiiiiik]

Abj Sir : You can skip the qns Every MF student : Stays*

[@snehakumari4326]

Chahiye samajh aaye ya na aaye adv ke naam pe sab karenge 😂

[@rashmiverma9677]

Sir RBD kab shuru Hoga aprrox (date)

[@namanarora2005]

30:00 Answer for T at 37° came =1.98 By adding psedo + mg component+T in string and finally equating all to the equation of Circular motion of 2m wrt to m. Is it correct?

[@krishnarawat4459]

BINA MASS KE TENSION KI VALUE NIKAL DI LEGEND

[@MayankSingh-pw3jv]

Out of so many answers no two are matching 😀😀 I am also getting different: 6130mg/9747

[@priyam6496]

Bhai tera toh pura hi glt h....mg se chota aa rha ans tera mine is 3.02 mg

[@RoseGautam-sw6ly]

54:40 sir samajh nhi aaya why to neglect mg. Mg is too 4N(0.4×10) and vertical impulse is also 4N-sec. Can someone explain it methamatically or logically?

[@Indian14754]

Think mathematically Normal force=(4kgm/s) /(0.1 sec) =40 newton (much more than 40) whereas gravitational force is 4 newton

[@Indian14754]

Force ko compare karo

[@satyammishra2316]

Sir Prabal batch se selection hote hai ya nhi kyunki maine compitishun pe Jayda tar bacho ka selection Praveen,Pratham se hi dekhe hai aur inme aap bhi padhate ho mgr Prabal pe aap nhi padhate tabhi Mai aap se hi padhta hu you tube pe bass questions unke lagata hu mtlb jee mains level ke.

[@rajabarik54]

T=(46/3)mg answer of h.w question

[@TheCreator67]

Bhai isme agr hme bead ki disp find krni h toh hum m1x1 =m2x2 kr skte h ?

[@TheCreator67]

@technobladev2164..

[@rabiulawaljihad1463]

I have just watched the 1st question by namo kaul sir.and i could not understand a thing.

[@risingpheonix7732]

From where to get DPPs

[@neerjasharma120]

Sir what is the link of your telegram channel please tell sir .I just wanted to verify my answer of that 37 degree question

[@yashsaxena2835]

If work done by internal forces is zero then total mechanical energy must be conserved in case of a man walking on a plank. But when the man starts walking on plank both man an plank have kinetic energy ( net system has kinetic energy) even though initial kinetic energy is zero. ( potential energy for system is constant.)

[@JEE2025-mo9lm]

Internal forces are zero as whole for the system individually definitely there is chamge in mechanical energy of the body

[@Satyam_100]

for someone looking for solution : refrence : kzbin.info/www/bejne/iXrMfGOur9qkY6s (explanation in short) : 1. net gain in kinetic energy(man & plank as system)= energy produced by internal mechanism in man's body. 2. force of static friction is equal for both.(3rd law of motion), and displacement of point of contact of man's foot & plank is equal (as static friction holds,therfore no relative motion) . now we can calculate work done by friction on plank directly through work energy theorem but we cannot calculate it by applying work energy theorem on man as work done by internal mechanism of man's body is unknown. to calculate work done by internal forces , apply work energy theorem on man : velocity of man wrt ground& displacement of point of contact of man & plank are opposite (if man goes right , then plank goes left) hence static friction does negative work on man as a system. calculate work done by static on plank now W(int mechanism)+work done by static friction(-ve) = change in kinetic energy of man.

[@AastikTiwari-em2yg]

21:42 it would be 4v square I guess sir has written wrong

[@harsh9023]

my answer is coming ; T = 5770mg/3363 or T = 1.7 mg (approx)

[@priyam6496]

1.46 ans sure

[@rekhamishra2000]

Im gettin 21/5 mg

[@Satyam_100]

for someone searching this doubt : (a)how is total work done by tension on the system (mass + bead) = 0 OR is total work done by an internal force on a system always zero? may refer this video : kzbin.info/www/bejne/fYeWd32OYs9siNU (b) can we apply work energy theorem in non inertial frame ? may refer this video : kzbin.info/www/bejne/iInEinipr5xgf6Msi=hoLw2lDwMhNXOMWe

[@sst940]

t=690/361 mg??

[@harshverdhanverma6092]

How did you solve please tell

[@aaaaaaaaryan]

30:35 bhai ye kya he....i mean 5 secs was fine...itna der !?

[@FAUXVIKINGIITB]

The first question came in my minor test for jee mains in numericals section

[@Indian14754]

Which test series😢😢

[@VkKingdom-j7u]

Bhai itna tough questions jee mains test me kaise de sakte hai yar coaching wale

[@himanshu_27tt]

Dear competeter ,how many lectures do you watch daily ??

[@ujjwalkumar8629]

In 2 or 3 days I complete one chapter depending upon the length of chapter. What about you?

[@UCENipobithiBagchi]

I lec everyday of ABJ sir Chem ka swad anusaar😁 Maths th full 2 hr for each chap

[@hilop1414]

Kisika hua??

[@authoryajat]

42mg/5 is this right sir please tell please please please

[@ajitgupta3045]

yes same

[@ruhanizm]

I think you forgot the psuedo force

[@krishnaupadhyay1057]

CAN M/C be used in 2nd last q because fext=0

[@archit1939]

is their material and homework from resonance study material ..because i have that...or their own material

[@mnjy]

Koi b kro

[@Predator-pv7of]

Is material available for free??

[@rhymehasan1537]

Impulse 31:50

[@amanpatel1784]

Sir mai droper hu aur aapki lecture se padhta hu Sir please tell me kb tk syllebus complete ho jayega usi ke according mai apna routine bnata please reply sir

[@vkasciencechannel1047]

Hw. Answer 2.368 mg am I correct?

[@D7XTE7]

Absolutely

[@killermansh1715]

i just came to this channel in classes ki kuch fixed dates timing hoti hai kya ?

[@sumitdatta8818]

Ans of 30:28??

[@SamDeshmukh-lc3cl]

0:00 to 30:00 first question

[@gap900]

127mg/99 ?? How do i check now if its correct?😶

[@debashis_]

Hw ans : 1.91 mg Steps in short : v² = (6gLsin theta )/ 1+2 cos² theta Than putting the value of v² in the eq T + 2Tcos² theta - 2mg sin theta = 2m v² / L T = 2mg sin theta {1+ 6/(1+2 cos² theta)} / 1+ 2 cos² theta Putting value of theta = 37° We get T= 690/361 = 1.91

[@parvgoyal3805]

momentum consevation sai velocity sirf horizontal wali balance hogi par ground mai 2m kai par vertical mai bhi velocity hogi to jab total meachical tmec lagaingai to velocity net jo hai voo laingai,,,

[@sahilrawat-we3de]

Bhai Mera vo v^2 Wale eqn mein( 1- 2 cos^2 theta ) ara hai..can u recheck it..baaki aage to sahi hai mene bhi same wahi Kiya...

[@not_mukul]

how did you apply Linear Momentum Conservation and TMEC???

[@dakshgiri3367]

Yar mera to 1.89mg aa rha hai

[@harshverdhanverma6092]

Bhai Teri eqn me 2Tcos²theta kidhar se aaya

[@adarshsinghal8465]

sir...... for an accelarated frame of reference ,if we know the velocity at a particular moment....cant we simply ignore the pseudo force for that particular moment....27:18

[@aayushjoshi3681]

Naah bro, we can't ignore it. Consider this situation: if at a particular instant,car A has velocity 40 m/s and car B has velocity 30 m/s in opposite direction, can't we take velocity of car A wrt car B = 40 m/s at this instant as we know velocities of both cars ? Your query is analogous to this but we clearly can see that velocity of car A wrt car B= 70 m/s. So, I'm not offending you but your question doesn't make sense. Try to relate to this example. If you still have doubt, then comment again.😇

[@susmitashit487]

Sir main lecture dekhe ne phele like 👍 deta hu kiuki muje pata hai lecture amazing hoga

[@abhinavjha376]

Answer of hw question is 1.66mg

[@SSEpsilon]

Thank you sir

[@Vivek-ko7xm]

Sir q1 is given in Allen with ring rather than bead I have already solved it😁

[@priyam6496]

Bead ring me kya difference h?

[@anshhayy]

Fiitjee too

[@arunabhadhal3662]

Thank You So Much Sir

[@nishgoel5876]

4mg ans as told by sir

[@Just_someone_out_here]

Where?

[@milikumari5474]

Great work sir.. thanks a lot😌

[@samarthsai9530]

SIr 37 degree wale ka answer plz bata dein taki check kr lu. Thank you

[@harshpandey8170]

Thank you sir ji 😊

[@iliyanparin_iitism]

Q1 T=500mg/969 ? (p and b are particle and bead) I got vpb= 5vb/2, vp=7vb/2 by m/c in horz direction. by TMEC vb = root(8gl/85) vpb=root( 10gl/17) Then on particle, by pseudo force component, T+2macos37 =2m (vpb)^2 / l putting a=Tcos37/m, got T = 500mg/969. Confirm if correct, correct if wrong.

[@Just_someone_out_here]

Can someone post the answer with solution in the comment sectio ?

[@vipulsharma7655]

@mohit tyagi sir..... I miss your classes sir....

[@Rohit_Cs]

Sheet kaha h?

[@messi0510]

0:05 40:10 46:00 1:02:40 1:08:40 1:17:10

[@himanshumech133]

THANKS

[@woo6777]

Can someone tell me how is momentum at 1:23:25 conserved? I didn't understand the explanation given by sir, please help.

[@krishnaupadhyay1057]

along string fext is 0

[@Samarthkhokhar]

sir where can i find the sheet you were talking about in lectures

[@MohitTyagi]

The sheet you're referring to is available in Competishun's paid batches or DLP study material. For more information, feel free to contact us at 8888000021. - Team Competishun

[@swatisharma5230]

Is the answer 34mg/7

[@dbdoxbxb]

Do anyone know the correct answer of the homework question? I am getting 2.7mg

[@Theadityasharma7]

Yuppp i got 2.59 mg

[@suvankarchowdhury4714]

10.23mg is the answer of HW question

[@SHIVAM_KAMAR]

18:23 ❤😊😊

[@yuvrajbijarniyafscompetish6332]

Thanks sir

[@gigachad9597]

1:10:00 wala question came in JEE ADVANCED 2021

[@k.ashvanth1735]

came here to say exactly this, but you did the job for me

[@theUnmeshraj]

@@k.ashvanth1735 Paper 1 or 2 ?

[@Krishna.1829]

Damn. isse pehel lect 4 mei projectile wala bhi advanced 2021 mei aya tha

[@poonamrastogi7708]

Sir yeh bata dijiye ki 1st q mein vmax of bead vertical mein kyo hoga please sir

[@DeepakKumar-gd1wg]

When the system is vertical there is maximum loss in potential energy which is converted to kinetic energy of the particles. To clarify this you can calculate the speed of the particles at any general angle theta by applying energy conservation. Also you can clearly see that the bead has zero initial and final speed...and the system is following symmetric pattern so the maximum velocity must be at the time when the system has completed half process...that is when it is vertical. Because the bead again decelarates after this.

[@shafin3365]

I didn't understand why people don't like to recommend this Channel. Huge love from Bangladesh 🇧🇩, sir. Hatts off

[@cosmic.pluviophile]

competition, my friend. thats why :)

[@shafin3365]

@@cosmic.pluviophile I've understood. In India, JEE ADVANCED is super competitive I know, same case for our BUET. But problem is, at least students should see the beauties of Physics. They should see what's going on this eternal Entity. Probably they may not get chance to IIT but one day, by being curious, they will discover something that getting chance to IIT and UPSC is nothing in front of that getting Noble Prize for that 👍👍

[@itz_robby-e6j]

Ans 1.9 mg hai kya 37 degree pe

[@harshverdhanverma6092]

Bhai kaise aaya thoda bata do

[@captainamerica1834]

Sir what is your telegram channel group name

[@dj.2958]

Sir where to get.. Com sheeet

[@Yash12409]

How to find acceleration of bead