i am enjoying your lectures more than life

Thank you so much Prof. Lewin and MIT. THIS is what the internet was supposed to be!

I'm 54, loved physics all of my life, but I was bad at it in School and at the University when it came to tests. And the teachers where by far not of Your class. So I ended up with a job in IT. :-D Now I'm learning again by watching Your lectures, only because I love Physics. It's just amazing how You can explain even complex stuff in a way, that someone not gets lost. Thank You!

One of the funniest things about these videos is watching the purple circles under the students' eyes get deeper as the semester wears on.

The center of mass reference frame is such a nifty little trick to solving these collision problems. If your thinking is stuck in the box of only using the lab reference frame, you end up having to solve a quadratic formula which could get you lost. But the COM reference frame allows you to avoid that quadratic formula altogether, and in my opinion, shows a better perspective of what is happening in the collision.

I don't know that I'm regularly watching your videos which you have been posted 5 years ago. And I really get feel in physics ( Means reading 1 line from textbook and understanding 5 lines) Thank you

He is the only person which can make us love physics and proceed physics as our carrier

Thank you so much for Prof. Lewin and KZfaq for this wonderful resource!

from momentum conservation v(wall)=2*m(ball)*v(ball)/m(wall) and since m(wall)->infinity v(wall)->0 and his KE->0 and a big thank for you legend

Wow!!! I love your lectures so much!!! It's really amazing and wonderful

I loved physics before but your lectures have made me enjoy physics. Thanks a lot

For the wall/ball (i.e. sleepless night) problem: Since we're assuming the mass of the wall is much much larger than the mass of the ball, we can apply the analysis from earlier in the lecture where m1

Thank you so much sir for such a wonderful work.... i have never thought physics in this way.... again thank you from all the students from india...

Sir, I have recently become a physics major at my college, and I love this subject with all my heart. Despite some difficulty in getting adjusted to the study of this mighty field, I embrace it all with open arms, and most importantly, an open mind. Thank you for contributing to my burgeoning zeal for this discipline, it is truly wonderful. Stay safe!

I'm 15 years old, currently working on my application at MIT, I enjoy studying physics and these lectures are just amazing, there is not a single video where I don't smile because of how perfectly the physics come together. This really inspires me and helps a lot in understanding these topics, because at my age in school we learn the basics, that's why I decided to homeschool, and these videos are the main reason why I will spend the next 2 years of my life trying to get into to MIT. I hope I will succeed, and if not, I will try to make the most of my passion for engineering in my native Poland :D

Love from india mesmerized by ur practical way of teaching

you are great, helped a lot in my preparation for competetive exams

I always liked to picture the extreme case as a train hitting a golf ball. A change of reference frame /almost/ makes it intuitive! From the perspective of someone on the train, the train isn't moving, the golf ball is approaching them at velocity V, and it bounces off the front of the train with velocity -V. That's a delta of 2V. From the perspective of someone on the ground, watching the train barrel toward the golf ball at velocity V, they must observe the same delta 2V! So they should see the golf ball at rest, fly away from the train, at 2V.

I have my exam in coming 10 days . If I passed that, Iam free from physics for life time But I cant stop my self from watching this person's lectures. He is so amazing. I mean How can a person say that he or she dont like physics after watching his videos. He really makes you love and understand physics in a different way. Thankyou professer!!! Iam a big fan of yours from pakistan.

Sir, so to answer your question, the wall has 2mv momentum, and so we can equate it to MV(momentum of wall), so we get V=2mv/M, since m

I think it would be good if we put it like this for people to understand easily for the ball-wall collision question. 1. original momentum of ball = + mv. 2. Final momentum of ball = -mv. 3. Original momentum of wall ~= 0 4. Change in momentum of ball = -2mv for the momentum to be conserved, change in momentum of wall = +2mv.

You've proved to me of very great help that you cannot think of, means very helpful for my studies and i enjoy it professor.

for the question at 23:00 the ball gains 2v only if you look at it in the perspective of the wall(moving at v speed) hitting the ball(which is staying still). in that frame of reference, the wall keeps on moving at V speed to the same direction, while the ball moves at 2V speed to the direction the wall was moving in. If you look at it from the perspective of the ball hitting the wall, the ball's velocity is reversed, but the wall's velocity doesn't change (because of the infinity mass). So the kinetic energy for the wall stays 0, and the kinetic energy for the balls stays the same (only direction reversed).

Excellent lecture. Thanks to great Physicst of world 🙏🙏🙏🙏

Regarding the last problem. In this case, since it is a completely elastic collision, both momentum and kinetic energy are conserved. The final velocity (Vf) has to be the same as the initial velocity (Vi). If we consider that each ball has a mass m, in general terms we have: ymVi=xmVf, where y is the number of balls moving before the collision and x is the number of balls moving after the collision. Velocities and m's cancel out and we get y=x. In other words, since their masses are the same, the amount of balls that move after the collision will always be the same as the amount of balls that were moving before the collision.

these lectures will really make you fall in love with physics

Congratulations for the set-up of the experiment!!

07:50 That's intuitive if you know that all inertial systems are equivalent. You have v_cm=v_1. And in the CM system (where V_CM=0), you really have V_1=0 and V_2= - v_1 (I use capital letters for the new system.) And then V_2' = -V_2 = +v_1 as object 2 bounces off object 1 as if object 1 is a wall. Going back to the original system you get v_2' = V_2' + v_cm = 2v_1. So all it takes is to consider that object 1 is a huge train on which you sit and object two is a small railroad cart you approach. But you sit on the moving train and see it from the train! That's a beautiful idea, in my opinion!

j'ai beaucoup ses cours good explanation!

Regarding the tennis ball-wall problem, the wall can have a finite p if only v is 0 (m is infinite) and KE=(mv).v.1/2, ie, a finite no. mv multiplied by 0. so the end result is 0 for KE. Is this correct??

Regarding the last problem of the Newton's cradle, I guess the balls are not really touching, so at each collision we simply apply the case of a ball colliding with a stationary ball of the same weight. So if we start with one ball moving then this ball collides with the second (and stops), the second travels a very little distance and collides with the third and so on to the last. If we have more than one moving ball at the start (for example, as you showed, three balls) then the third ball (the central one) collides with the fourth and stops, in a supershort time the second arrives from behind and the central starts again, now the central hits again the fourth that has just hit the fifth and stopped almost instantaneously, etc...

based on hours of algebra and what I've read online, it seems conservation of KE and momentum are not enough to solve that Newton's cradle question, because there are actually infinite solutions that would conserve both. It seems the physics for why it happens in this particular way is much more advanced. Is this correct?

The ball is pushing twice the wall! The first time, the ball is hitting the wall, and it is compressed, and the two bodies are accelerating in the same direction. The momentum mv= (m+M) v’ so the wall gets a first impulse Mv’= mv- mv’ with v’ almost 0, that mean Mv’=mv. Then the ball is decompressed by the elastic forces to his normal shape, the ball is pushing again the wall, and the two bodies are accelerating in opposite directions! Which mean the wall gets a second impulse Mv’’= mv. The ball is pushing twice the wall, first time because is coming with the speed v, the second time because is departing with the speed -v, so therefore wall gets a double impulse from the ball, in the same direction p=MV=2mv. The impulse of wall is p=2mv, but the kinetic energy of the wall is E=pV/2, with V almost zero, therefore E=0. The speed of the ball is the same, before and after hitting the wall, and this can be proved, but we can not prove the momentum of the wall, is based just on the theory. Today, based on this theory, the momentum of the wall is p=2mv. But I'm not so sure about tomorrow...

You are really legend of Physics

My take on the tennis ball problem :) The total momentum of the system before and after the collision is zero. Indeed if we calculate the velocity of the center of mass: (mv+M0)/(m+M) =mv/(m+M) = v(CM) we notice that if the mass of the wall is infinite/big, then v(CM)=0 Hence, because of the wall/enormous mass our frame reference is incredibly near to the center of mass of the system. Hence it doesn't matter if the collision is elastic, inelastic or superelastic...the total momentum is always zero! By not recognizing that we are in the center of the mass can lead to the paradoxical (yet correct) result that the wall has a momentum = 2mv.

I am a student of mathematics but I enjoy and feel your tasty lecture. Thanks Prof. WL

hey walter if these two objects are moving towards each other meaning there positions change, meaning that the sum of the products of the mass and the position vector should also change, shouldn't that result in a changing center of mass location?

your lecture are so good that we loves to be there

Thanks for the wonderful explanation, Professor. I was thinking about the problem with the balls and I believe that 5 balls have to be in motion after the collision due to the conservation of kinetic energy. If fewer than 5 balls were in motion, then due to the conservation of momentum they would have to be moving faster and because Kin. En. is proportional to the square of the velocity and momentum is proportional only to the magnitude of velocity, Kinetic Energy would increase after the collision. But if the collision is completely elastic, that cannot happen. Thus the only possible outcome is having the same number of balls in motion before and after collision with the same velocity. Is that correct?

Loving these lectures

RE tennis ball/wall/momentum: could this also be explained by using v_2 ' = ( 2m / (m+M) ) v_1 ~ 0 ?

Hello professor lewin. In the last few minutes you mentioned about 8 balls. Is it really possible to solve the problem mathematically ?

The question regarding the tennis ball bouncing off the wall and the wall not moving even though the ball has had a momentum change Is the answer related to the inertia of the wall? the ball does provide a change in momentum and hence a force on the wall but because of the wall having a really high inertia, it does not move

Related to your question about the collision between the ball and the wall. Is it due to the fact that moment, mv, where m = x as x goes to infinity can be different from 0 (v = 2/x for instance) but (1/2)*m*v² would still go to 0 with those same values? Also, can it simply be that the earth to which the wall is attached applies an external force on the wall, and so it's the larger system (ball, wall, earth) that has its momentum conserved? Thanks in advance.

Mass of wall (M) is infinite M➡∞ 2mv/M=Vₘ therefore Vₘ➡0 and KE➡0

Good evening sir, I'm attempting the challenge you gave at the end... could I get a few hints? If there are k identical balls of mass m, with n balls from that group initially moving with speed U, then: Total momentum= nmU Let the velocities of all balls after the collissions be V1, V2, V3... Vk By Cons of momentum; nU= V1+V2+... +Vk and by energy conservation, nU^2=V1^2+V2^2+...Vk^2 So now we have k unknowns with only 2 equations... we still need k-2 more equations to fully solve! I suppose another piece of information comes from the fact that Vk>Vk-1>...>V2>V1, since obviously the balls cannot travel through each other. I am stuck!

Dear Sir Lewin, I've been thinking about it deeply and the reason why the wall gets momentum 2*m1*v1 after being hit by the ball does really not have a physical meaning. Instead, its a consequence of the mathematical expression you mentioned and thus: v2' = 2*m1/(m1+m2)*v1, hence: p2' (wall) = m2*v2' = 2*m1*m2/(m1+m2)*v1. And if I now calculate the limit for m2 going to infinity: lim(m2->inf.) p2' I get exactly the 2*m1*v1!!! So mathematically it throws your result but in reality because the speed (and thus also the KE) of the wall = 0 anyway it's meaningless, ie. the wall doesn't move an inch what perfectly matches the idea that its mass is infinite. If it was finite (which is the real scenario) it would get a teeny wheeny bit of speed which, of course, is unnoticeable. Provided the collision is perfectly elastic. Do you agree??

Relating to the balls on a string problem... could it be because the number of balls create a pulse (like a wave) with a wavelength equal to 2n (where n is length of the number of balls pushed), and since the balls were only given a positive velocity (not oscillating), only the positive amplitude of the wavelength is represented. The positive pulse then travels through the balls when they strike each other until it reaches the end, where it knows it has to create momentum as the preexisting momentum is gone, and that this momentum must affect the number of balls which could not get rid of the pulse, equal the length of the pulse (n) which is also the total length of the balls?

for the ball and the wall question: after the collision the ball has a momentum -m1v. Since momentum is conserved, m1v = -m1v + 2m1v, meaning the wall has gained a momentum of 2m1v, with m1 being mass of the ball, v being the velocity of the ball. The wall has momentum because: P = mv; 2m1v = m2v. m2 being the mass of the wall. 2m1v / m2 is the velocity of the wall, which, because m2 --> inf, v --> 0. Because of its huge mass, it doesn't have visible speed, but it has momentum. The wall has no KE because: KE = p * 1/2 * v, there is momentum, but v --> 0, KE is essentially 0. Is that correct?

Sir, honestly speaking, you are the most favorite physics teacher in my life. Your lectures helped me understand calculus-based physics much easier (Though I am 13 years old). It is very very very true that Lectures of Prof. Walter Lewin makes us love physics. :):):) Regards Mayukh Chatterjee India

I would like to point out that the 2v case of the ping pong ball when hit by a billiard ball at 11:32 is, though not intuitive, but quite obvious. Suppose a ball is lying on a train platform. The train, when collides with it, continues to go with the same velocity. Now, from the frame of the train (both are const. Velocity inertial frames) I would see the ball coming to me with same Velocity as the train's ground Velocity. So the ball must bounce back from the train's front in the forward direction with same Velocity in the train's frame. Hence in the ground frame, it would go with twice as speed. Thanks Glad if you correct where was I wrong of I was

24:08 well i think that the wall will have no kenetic energy because if the ball elastically collided with the wall we technically assumed that the mass of the wall is much larger that that of the ball and the formula for the Kenetic energy is KE=Mv^2/2 or p^2/2M or (2mv)^2/2M this means that if we devided the mass of the ball to the mass of the wall which is way lager we will get zero!

The answer to the tennis ball question could be given easily as follows: (First things to be taken into consideration mass of tennis is negligible in comparison with the ball so as per the equations you had given v1'=v1 and v2'=0) Since v2 and v2' are going to be zero before and after collision we can say that KE of wall will be zero since KE=1/2 mv squared and velocity is zero so it is clear that kinetic energy will be zero I don't know this is right or wrong but the result Is indeed correct.

48:20 if we have to prove it analytically we just use the approach of relative velocity and try to solve the collision problem by assuming the moving 5 balls to be at rest and find the velocity of those resting 3 balls with respect to the moving 5 balls, now the problem becomes equivalent to the previous simple case... solve the equations and get the answer. And the second approach i would suggest here is to look the situation from the centre of mass frame...

Thank you so much prof l love this lecture and help me to understand physics

Sir, in the tennis problem when you say that ke is 0 it isn't exactly 0,right? mv+0 = -mv + MV MV=2mv. KE cannot be considered exact zero it is a product of a finite and a very small quantity, it is just infinitesimally small. If we agree that it has it has gained a velocity then there should no reason why it should not gain a kinetic energy.

For the last question. Let's say we have any number of balls. I pulled and dropped from one side "x" number of balls with velocity v. After the elastic collision "y" number of balls moves with Φv. (1/2)xmv^2=(1/2)ym(Φ^2)(v^2) so x=yΦ^2 (1) xmv=ymΦv so x=yΦ (2) combining (1) and (2) yΦ=yΦ^2 and Φ=1. Which means x=y.

@Professor Lewin..... From the equations v'1=(m1-m2)*v1/(m1+m2)....&......v'2=2m1*v1/(m1+m2).........as m2>>m1....we have v'1=(-v1);. ... and...v'2=0...... Then the tennis ball's momentum after bouncing back is (-m1v1)...and the Wall's momentum is 0......... isn't it?.... I'm not clear how wall has momentum of 2m1v1......I know total momentum must be same before and after collision...but from the equations I'm unable to prove that. .. could you please help....

By using the time how can you explain the elastic collision

Prof. Lewin, may I request to please correct me if I am wrong. Wikipedia meaning of explosion: An explosion is a rapid increase in volume and release of energy in an extreme manner The applications of explosion in the system the principle of Thermodynamics fail. The Internal Energy is multiple times lower than the work done by the system or The WORK OUTPUT is multiple times higher than the WORK INPUT I. A 15 ball billiard pool to be strike by white mother ball. F = 30lbs (White Mother Ball); D = 3ft (Distance from the white mother ball to the first ball to strike); W = Work, F = force; D = Distance The mother ball to strike the 15 ball is the work input The first collision is explosion followed by increase in force and travelled distance by individual ball followed again a series of collisions. The mother ball transferred the force to the group of 15ball. The momentum is conserved to the group of 15 while the explosion increases to 15 individual ball. The kinetic energy was absorbed by the group of 15ball. The explosion increases rapidly multiplied to 15ball individual. SOLVE FOR WORK INPUT W = F X D; W = 30lbs x 3ft = 90Ft.lbs SOLVE FOR WORK OUTPUT W = F X D; W = 30lbs x 3ft = 90ft.lbs X 15ball = 1,350 ft-lbs. II. For more heat energy the ball to replace with steel ball and steel side board. After strike are explosions of 15 balls. The explosion increases the group energy into individual 15. The work output higher than work input followed by series of collisions that produces more heat more work output, more sound energy more work output, more travelling distance more work output. III. To get even more energy output: Put additional two group of 6balls in triangular shape to be placed somewhere at the back of 15balls at least 4inches apart from the sidewall and back wall. A total of 15balls + 6balls + 6balls = 21balls. The explosion creates collisions after collisions increasing the energy output. The Internal Energy in I, II, III are the same however, after explosion the work done increasing 15 times the Internal Energy. That principle can use to make free energy machine. Thank you. Abel Urbina Abel Urbina Free Energy Super Machine

you are far better than physics wallah and all they just show formulas to memorise to get marks and score, and doesn't care about concepts to feel

Hello Sir! What are glancing collisions? Is any collision in 2-D or 3-D called glancing collision? Or is there something special about glancing collisions. Im not clear with Google's answers. Help please!

U did infinity large mass . Now though the wall has a momentum but as the mass is very high it has verry low velocity almost comparable to zero

Hello Dr.Walter Lewin and thank you for your open meterial. At 26:35 you say that the momentum is 0 before and after the collisions.I thought we only took one equation for the momentum,having to do with the 'before' and 'after' the collision.I do not understand how can we take momentum only for the 'after' part.There is no collision in that part. And last,why when we are at COM we say that we experience objects with 0 velocity? I appreciate your help.

Hello professor good evening how you ,! How it is going . I hope it will better !! I have doubt about ICOR please help me !! From India 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳

As with every lecture there is always something that sets me thinking new thoughts.

Are elastic/inelastic collisions related to the elasticity of the object ? So can an elastic object have an inelastic collision and vice versa ?

So the wall thing V'2 = 2m1/m1+m2 × V1 Now because m2 = ∞ V'2 will equal approximately 0 meaning KE = 0 And we know that it still has momentum because even if v'2 = 0.0000000001, it multipled by infinity wouldn't be 0, it will probably be some other giant number or just infinity. I think I'm not sure

Thank you very much Mr. Lewin. I studied twenty years ago and I watch your show some evenings. It is fantastic and MIT is fantastic too. I am studied in the Czech republic unfortinetely. Your last question is collision balls. My answer: Balls do not collision in the same time, but first ball move last ball, second ball move second last ball, third ball move thirt last ball.

07:58 I can’t believe even such an explanation can bring such expressions on the face of students.

Professor, Which Physics books you used to read in your time ?

Is momentum of the system is conserved in all types of collisions? And should the conservation of momentum be applied along the line perpendicular to the line where deformation of the bodies takes place ?

This is a difficult chapter but a very interesting one! You got me on the last billiard ball experiment...I got two wrong

8 years later still gem ❤

The wall has gained momentum with respect to the ball (i.e. we are in the frame of wall), Now the velocity of wall is zero and since energy is absolute (K.E. is scalar) the K.E. after the collision is zero. Is it correct?

The wall has momentum but not significant kinetic energy because its enormous mass causes the change in velocity (due to the impact) to be extremely small, leading to negligible kinetic energy. The momentum transfer occurs, but the wall's kinetic energy remains practically zero due to its large mass and the energy distribution into other forms.

Momentum is scaled velocity vector. If m is very high then P will be high (velocity vector gets scaled by mass even if it is very low). Kinetic energy is a number. If velocity vector is low , then it's magnitude is also low . Then |v|^2 is also low. So KE~=0. Thus wall has p=2mv but KE~=0. Is this correct?

Why you multiply the t2' with 2 by 3 & t1' by 1/3

in completely inelastic collision you told that energy lost is in the form of heat but the sound produced during collision is also a form of energy. where does that come from?

Dear sir, There is a question which is iching my taughts since many days, that why speed of sound is constant??? Even though sound propagate by means of elastic collisions between molecules, speed of sound should depend upon the speed of molecules near the source of sound. Yes I have heard it already that molecule speed is directly proportion with the increase in pressure so the bulks module remains constant and hence the formula holds. But why then same phenomena is not observed in collision of two identical elastic balls, were the velocity of second ball depends upon the velocity of the first ball and hence the velocity of propagation of this collision also depended on the velocity of first.

in the last problame the balls may be transfering momentum and energy one to another in quantized way...

I think it’s: the wall can have non-zero momentum but zero kinetic energy because in the limit as the mass of the wall approaches infinity, the velocity of the wall approaches zero, and the momentum is linear in both m and v, so their multiplication can be a nonzero constant. However, kinetic energy scales with v^2, so as mass approaches infinity and velocity approaches zero, the velocity term dominates and kinetic energy becomes zero. Then for the 8 ball newton’s cradle, I think the intuition for it is that while momentum could be conserved with many different combinations of output balls and velocities, the only solution that also conserves kinetic energy is one where the output balls have the same mass (and velocity) as the input balls.

According to the pendulum problem at the end of the video when 5 balls r pushed the initial momentum is 5mv and according to the law of conservation of momentum 5 balls should come out the is is because the 3 stationary balls absorbs the kinetic energy from 3 of the 5 balls the there is still two balls which still possesses the same kinetic energy hence we get 5 balls moving after collision. Sir is this correct???

For the tennis ball hitting the wall; As P=Mv, v=P/M And due to M=infinity, P/M must be 0? Similarly I think this can be shown using F=ma, since the change is momentum is equal to the product of net force and time. Either way, v=0, hence K.E.=0J Also a question, if we imagine that the wall is moving towards the tennis ball instead, then we get a similar (perhaps the same, just from a different viewpoint) result, and this is like the m2=0 scenario, in which v1'=v1, so the velocity of the wall cannot change, and neither does kinetic energy of the wall. Is this also a valid solution? Thank you for reading this

Good mornig Dr.Lewin. Is there any way to send you docs or pics in order to comment more easily doubts and suggestions? Thank you, Sir.

I have a doubt in a collision question A block A of mass 2m is placed on another block of mass 4m which in turn is placed on a fixed table. The two blocks have the same length (not height) 4d. The coefficient of friction (both static and kinetic) between block B and table is μ. There is no friction between the two blocks. A small object of mass m moving horizontally along a line through the center of mass of the block B (parallel to the ground) and perpendicular to the face with a speed v collides elastically with the block B at a height d above the table. What is the minimum value of v required to make the block A to topple?

We love you!

Sir, In that question of yours if the mass is same (m), and velocity is also same (i.e v), them how come the momentum became 2mv for the tennis ball??

what book do you use through the course?

Professor please help me solve the first problem you gave, I asked my teacher he said you cannot conserve the momentum in this specific case

*8.01x Lecture 16* -------- 0:00 2 equations of super elastic collision/ elastic collision/ inelastic collision 4:27 elastic collision + play with the result (use your Intuitition) 9:48 experiment with the pendulum / ping-pong ball and billiard ball 12:23 elastic collisions (+ Air track eg) 23:00 “sleepless question” 24:30 frame of centre of mass (u) 32:22 energy lost in the inelastic collision (quiet amazing) (Lost energy) - special cases 39:50 air track experiment (completely inelastic) 46:38 last question: who can tell me the answer? ---------- *See more here:* zyzx.haust.edu.cn/moocresource/data/20080421/U/01220/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/Video-Segment-Index-for-L-16.htm ---------- *Thanks to Prof. Walter Lewin, Education is free and Physics is amazing!*

So interesting! Thank you!

Thanks a lot....ur lectures help me during my preparation for MBBS entrance test

Thank you Professor Lewin. I'd like to try to respond to your question at 23:50, what is the momentum imparted to the wall when the tennis ball collides with it.I thought of three possible approaches:1. You said at the beginning of the lecture that if there is no external force, then the momentum must be conserved. I would suggest that there is an external force present, namely the wall is anchored to the ground. The earth will exert opposing force to resist translation of the wall, so momentum really isn't conserved.2. There is no reason to assume the extreme case that the collision is perfectly elastic. Most likely there is a Q, and so some energy is expended in the form of heat.3. If we look at the wall and the earth together, and grant the extreme case that the collision is perfectly elastic, suppose the ball retained 98% of the momentum after the collision, then m2v2 would be only 2%, a small number. Since m2 (the earth) is almost infinitely larger than m1, v2 would have to be infinitesimally small, perhaps imperceptible. You explained at 8:00 than if m2 >> m1, m1 approaches 0, then v1' = v1 and v2 = 0.But if we ignore this and we have to impart a tiny by of momentum to be imparted into the earth, my guess is the tiny bit of momentum, with no kinetic energy, is manifested as minute waves radiating from the point of impact, like a drumstick hitting a drum, dissipating with the square of the radius.Now I can sleep.

Sir if we consider the the system to be earth, the wall attached to it, and the ball so when the collision happens the wall and earth gain momentum but since their combined mass is very large compared to the ball they don't move much so, momentum is still conserved. Is that a correct way to think about this situation? Hope you're having a great time ❤️ and thanks a lot for these valuable lectures ❤️❤️❤️

23:52 Sir you asked us to show mathematically that the wall may have zero Kinetic Energy but it still has a momentum 2mv PLEASE LOOK INTO MY SOLUTION 2/v * K. E = momentum ( here we consider the absolute value of v) 2/v * K. E = 2mv ( for the wall ) K. E = mv*v 1/2 mv*v = mv*v Which is only possible if mv*v is zero so K. E is zero

the example which you showed at 10:20 is with the billiard balls and i was trying to replicate the same experiment with different kind of balls and the first ball didn't stop after it hit the second ball.. what is the meaning of this.

hi proff at 8:49 i think v2 prime should be (2)x(m1)x(v1) cuz on the top there is nothing that we can compare m1 with in the bootom the is m2 that is the reason we should neglect m1 in the bottom and not on the top ....... plz help me get a clarity at this

at 24:45 lect 16 . Answer : The wall is no part of the system , the reaction force of the wall is external at system "tennis ball" . Thank for your precious lessons Prof.W.Lewin

how do we actually derive v2'=(2m1/(m1+m2))*v1?