Rocket are fascinating - they allow humanity to achieve things almost unimaginable even 100 years ago. Thank you Professor for describing their basic principles so elegantly!

Who needs sitcoms when you have Prof. Lewin on KZfaq? Even my wife, completely unrelated to the realm of physics, enjoys these lectures and has learnt a LOT!

I love how you end with the Oberth effect. Always try to burn at periapsis!

Proof of 4:47 done, Sir! h=L*(1-cos(theta)) Using the first 2 terms of the taylor's expansion of cos(theta) h=L*(theta^2)/2 (almost) x=L*sin(theta). Using the first term of the taylor's expansion of sin(theta), x=L*theta (hence X is almost equal to the arc). Hence theta=x/L. Hence h is almost equal to x^2/2L.

Professor Lewin you are a legend , I listen to your lectures all the time. Thank you for your work in physics, I hope that one day I can be as great as you.

For the basketball-tennis ball problem I think the tennis ball will go up 9 times higher than the original h? The tennis ball and the basketball will have the same velocity right before the moment that the basketball hits the ground. Both the tennis ball and the basketball experience an elastic collision. The basketball hits the ground with a (velocity v) which is the same as the velocity of the tennis ball at that point(same time duration, same acceleration, and both start from rest). The basketball will have a final velocity -v(elastic); the same idea applies, when the tennis ball and the basketball collide, the tennis ball will have a velocity of -v(elastic); since (m tennis)

After filling lakhs of fees, real knowledge is found on KZfaq for free.❤️🎉🎉🎉🎉

Sir you are the best thing it happend to my career. Formal Greetings from an Argentinan student

Hello, Prof Walter I think you answered the question of last lecture which was about tennis ball against wall here in that lecture at 15:00 ! the momentum of the ball changed by 2mv .

This is a great lesson. There is a great teacher.

Is there a continuation on orbital mechanics? always really wanted to learn it!

There is a subtlety in the derivation of the rocket equation. It seems puzzling how u(dm/dt) can be the thrust acting on the rocket when the actual velocity of the escaping gas (the mass dm) is v - u (upwards taken positive). But this is explained by noticing that at the initial instant (just before escape), the mass dm was moving up with the rocket with velocity v, and therefore the CHANGE in its momentum due to the ejection is u(dm) since "u" is the change in its velocity. And hence this CHANGE in momentum, divided by the time dt, is precisely the force it applies on the rocket.

The leaving style, this is how a physicist leaves an impulsive impression❤

I don't know why I am watching and keenly following these lectures as I am looking to get into medicine, but I sure am enjoying the series. I did Mathematics and Applied Mathematics in high school so I am easily following, but there sure are a lot of "aha!" moments where the loose concepts of high school are fully and theoretically cemented.

I wish I had a chance to meet Prof Lewin in person. He is my inspiration

Wow sir.. Excellent rocket with a fire extinguisher...

You are right, this is the utmost of professorship.

There’s a Yale prof doing E&M and when I tried to watch him I thought ppl pay top dollar tuition to be put to sleep. By contrast Prof Lewin is always energetic and enthusiastic as well as informative and entertaining. I’ve been binge watching him and Prof van Biezen, Parth G, Dr. Matt Anderson, Dr. Matt O’Dowd, Kathy Loves Physics, Physics with Elliot, Khan Academy India English, and others. Can’t get enough physics.

Hi sir, I love your physics lectures. that's awesome. (hope this isn't new type of comment as lecturers like u often get these views but nevertheless i would like to thank you for your sincere efforts). I am from India. I had my physics lecturer same as you who has made me love physics. Till now I don't remember a single formula in phy. since formulas are required only for problems while physics is not a problem. :) ... i use to see your lectures for reference for any concept . as it contains visual proof for most of the concepts its very easy for me to remember the concept and apply it immediately. i have gained a lot of knowledge from your lectures. But my wish is to attend your lecture atleast once in person (live). hope that happens soon. once again thank you sir. :)

Again, Love you SIR hope you live forever

Here's my take on 23:27. The ratio is h(final)=6*h(initial). (9 times is way too much for my taste, Sir!). I took into account the following coefficients: a) i re-calculated the speed of the basketball because it reached only about 0.75 of the initial height when thrown alone (in other words, i considered it a partially elastic collision) b) i did consider the different masses by imposing the constrain mass(basketball)= 12*mass(tennis ball) c) i moved the frame in the (real?!) center of mass, then back to the initial frame My estimate is that our dear Sir dropped the balls from 1.75 meters. The speed of both the balls as they reached the floor was about 5.8 m/s The speed of the basketball after the partially elastic collision was about 5 m/s The speed of the tennis ball after the elastic collision was about 14.1 m/s and it reached a total height of about 10,4 meters (10.2 m + 0.2 m

My take on the tennis ball. If we admit that the basketball changes its direction from downward to upward before the tennis ball does, reaching the floor, impacting it, being deformed and giving a cushion to the tennis ball which is still going down, then we have: Let v be the speed just before impact related to the initial height. Basketball goes up at a v speed. Tennis ball has a relative velocity to its floor, the basketball, of 2v downwards (v_rel=v-v_basketball; v_basketball=-v) just before impact. Tennis ball hits the basketball and changes direction elastically, to approx 2v upwards with respect to the basketball, hence being his absolute (relative to the lab) speed 3v (v_basketball+v_rel). From there one can easily get the final height.

Great lecture, thank you a lot

Sir i am in class 7 and watching your lectures and I am understanding it very easily since I have completed my jee syllabus. Thanks to you sir for this great content

Great lecture, but I have 2 questions. 1) How can I can calculate the u term before the launch ? 2) Where can I find the assignments of your lectures ? Thanks a lot.

Very good lecture Sir. Thanks and Regards 🙏🙏🙏🙏🙏🙏🙏

Famous professor. A book I was reading mentioned his lectures. ❤

Hi Sir. Lecture was super good. I have a doubt. Say we drop a ball from a particular height. If it were a perfect elastic collision, then would the ball be jumping perpetually?

At 46:45 when you write equation for gain in KE, isn’t the final mass lower because of the fuel that’s burned and lost? Therefore should you not have lower m for the final KE?

21:45 Regarding the tennis ball and the basket ball. Consider the point at which the two balls change direction and they are at their maximum squashiness. Assuming perfect elasticity all the energy that started off as height potential energy became kinetic energy and is now elastic potential energy. As the two balls unsquash themselves the potential energy becomes kinetic energy in the two balls. From the start of the unsquashing let the time that it takes the tennis ball to separate from the basket ball be tt and let the time it takes the basket ball to separate from the floor be tb. Now these times are unknown but I think they are critical to finding out the height reached by the tennis ball. I expect tt is much less than tb. If we assume uniform transferring of energy then all the potential energy drops to zero during the time tb but it gets converted to kinetic energy in the tennis ball during time tt (all the forces on the basket ball cancel out) and then gets converted to kinetic energy of the basket ball during time (tb-tt). So, if the total energy is E, then as the balls leave the ground the kinetic energy in the tennis ball is (tt/tb).E and the kinetic energy in the basket ball is ((tb-tt)/tb).E. Let the masses of the two balls be M and m. Let the height dropped from be l. So E is (M+m)gl. Let the new height of the tennis ball be h. We have (tt/tb).E = mgh, (tt/tb).(M+m)gl = mgh, h = l.tt.(M+m)/tb.m. Let’s now make up some numbers. Let’s say tb = 8ms, tt = 2ms, M = 500g, m = 50g, l = 2m. tt/tb = ¼. h = 2m x 550/(4 x 50) = 5.5m. Around the height of the lecture theatre.

The arrangement in the thumbnail was my question in jee mains mock test.

Hiiiii Prof Walter Lewin! Your lectures are sooo great! I just have a small question about the rocket equation you talked about in 8.01 Lect 17: If we consider the gravity of the rocket and its fuel when it is launched vertically from the earth, isn’t this gravity force an external force on the system? If so, why does the conservation of momentum still hold? Thanks a lot!!

in the ke demonstration at the end where is the mass change of the rocket

Tnx sir💗💗💗 I literally addicted for you

Good afternoon, I am going through the assignment 5 solutions and I am a little confused at parts of the solution to 5.6 with the binary star system. I understand that the system is centred about the centre of mass however I don’t understand the derivation of m2r2 - m1r1, why is there a minus between them? Finally, how do you related the period omega1 = omega2 = omega, with m2r2 = m1r1? I hope you are able to help me or please direct me to a resource that would help. Thanks.

In the assignments, pages of which book have been referred to?

Thank you 🙏🙏🙏 professor

Sir may you upload those assignments you are talking about?

Professor Lewin, Do you have any recommended sources for learning more about Taylor Series Expansions? Thanks, Coby

Thanks you professor.

when will you be online according to your nation timing

Answer to the tennis ball and basket ball dropped from some height: Its Ellastic collision. lets assume both are falling at velocity v0. Basket ball hits the floor first and bounce back with same velocity vo. When basketball is about go up, at that time, tennis ball is falling at velocity v0. so relative velocity of tennis ball becomes 2v0 w.r.t Basketball which is going up at velocity v0. So tennis ball is moving up at 2v0 + v0 = 3v0. Correct ?

Sir, if I throw some solid ball 10kg/sec with a velocity 3m/sec and it bounces from a surface with the same velocity within 1sec ,then the force experienced by the surface will be 10kg/sec ×6m/(sec)×(sec)=60 N/sec. Am I right.

I'm starting to think WL's puns are intentional haha

SO CLASSSIC

Shouldnt the velocity vector of v-u be directed downwards since it was spewed out from the rocket at 34:20?

can I ask u any question,how can I contact u online sir

I was not able to get the minus sign at 39:28 when i did the integration. So I checked the lecture notes mentioned in description. There, i found 2 things. One, instead of ignoring gravity during the super-elastic collision & conserving momentum (@ 36:43), they have equated the change in momentum with time to the external force acting on the system - gravity. This doesn't explain the minus sign. This just shows a new perspective to derive the same equations. The other thing i found is, they mentioned modulus dm/dt --> | dm/dt | And i realised, since rocket is losing mass, dm/dt would be negative in sign. So, i took the new mass of the rocket after gas release as m+dm instead of m-dm. And the gas's mass as -dm instead of dm and ran the same equations and this answered the minus sign. Hope this helps others and i am right !

Dear Professor Lewin, In the case of the basketball and the tennis ball, the tennis ball goes up nine times higher than its original height. Now my question is: "Is the collision of the basketball and the tennis ball elastic or superelastic?" I mean I know the kinetic energy of the tennis ball increases. So does the kinetic energy of the basketball decrease by the same amount after the impact? I thought if it does, then it must be elastic because the kinetic energy of the system would be conserved and if the kinetic energy of the basketball does not change after the impact (supposedly because the mass of the tennis ball is negligible) then it must be superelastic. So which is it?

What an Exit 😂👍

What is this course called?

oh my dear sir ..... Im an ex MIT student and I was really lucky to have you as my professor ... im Harsh Sekhar Aggarwal from india , btw you u remember me? Sir it really hurts me now to not see you anymore ..... i really miss those lectures and your guidance ... only if there was some way that i could meet you one day :') .... btw im now a software engineer at google and currently residing in chicago

44:50 But if the same amount of fuel is burned where does this extra energy come from?

Hi, I was told that the way the Rocket Equation was derived at 35:56 was actually not so accurate, because the speed of the exhaust at that particular time t + delta t is v + delta v and not v. So there's no need to omit the delta m * delta v term. Is this correct physically?

first the ball has a momentum +mv (in the right direction), then the momentum of the ball changes to -mv (in the left direction). the change in momentum becomes -2mv, thus according to conservation of momentum the wall must have a momentum +2mv. is that right sir?

I have a question related to difference in change in kinetic energy at 45:20. If we have two observers one on a train which is moving with 10m/s and one at rest on the platform. Both view a same event at a distance where an object of mass m is accelerated. Suppose if object's velocity is 10m/s from platform frame and is accelerated by an external agent to 20 m/s, then the change in K.E. or work done will be 300m/2 J. But from train frame object was at rest and then moving with the velocity of 10 m/s after being accelerated. So the work done comes out to be 100m/2 J. I get why kinetic energy should depend on inertial reference frames but work done depending on inertial reference frames seems counter-intuitive. On the internet I found answers like energy does depend on reference frames but momentum-energy four vector does not. Is that the only explanation?

I love your lectures sir...these lectures are assets for students like me all around the globe...I have a question sir...do you follow spacex and how they are changing the space industry..?...what do you think about reuseablity of rockets.?

The rocket equation was developed by the Russian Tsiokolvsky, and is known as the Tsiokolvsky equation. This man's life is just fantastic and I think he deserved to be quoted in his excellent lecture.

sir when bullet is absorbed in block, heat energy is produced...right?? my question is.... does that make any changes in momentum before and after ?? Does the momentum 100% conserved??

Dear Professor Lewin. At 46:44 minutes into the lecture, you wrote "m" for the mass of the rocket, but shouldn't there be two different masses, initial m and final m as the rocket is burning fuel?

Sir, I will go to class 11. Sir your videos are very simple to understand. Sir I will continue my studies through this channel. 🙏

I think i have the solution for the tennis ball problem: Assuming the collisions are completely elastic and there is no air drag, First the basketball will collide with the ground and will bounce up with the same velocity it had before the collision, Second the tennis ball will have the same velocity the basketball has but in opposite direction and another collision will take place such that the tennis ball will now have a velocity of (2(m1-m2)/(m1+m2)-1)*V where V is the original velocity of the tennis ball, and it will be negative because it will be traveling in the opposite direction, assuming the basketball is 10 times heavier in mass compared to the tennis ball and the ratio of the heights using the energy conservation will be H/h = (V2/V1)^2 and plugging in the numbers we will get roughly 6.95(the exact value is 841/121) meaning it will travel 6.95 times higher.

There's that dotting the board technique. So suave

Sir. In this lecture I think there is contradiction between 17:45-18:00 and 26:45-27:00

48:09 thats why we love lewin

Hi Professor, Thank you very much for this fantastic lecture！I have a question. You said that "when you burn a certain amount of fuel for a given amount of time, you obtain a fixed change in velocity, but the change of kinetic energy is not fixed. " And you gave an example which showed that the kinetic energy increases more when the initial velocity is higher. My quesiton is that the increase of kinetic energy comes from the chemical energy of fuel, while the given amount of fuel for a given amount of time generate fixed energy, and the rocket with higher initial velocity will elevate more, thun more gravitational potential energy. According to E = K + U, E is fixed, U is higher, then K will be smaller when inital speed is higher. This conclusion is not consistent with that in your calculation. I am very curious why these two conclustions are opposite. Would you please tell me the problem in my derivation? Thank you very much!

I can't seem to download the notes and assignments on mobile, did anyone have any luck downloading them on PC?

(23:30) sir would you please say how can we calculate the velocity of the tennis ball considering them as solids?

Hope I'm not mistaken, but perhaps at 4:43, the approximation for h is (l * x^2)/2?

How can we Know the acceleration of the rocket at every moment?

What a gem!

Hello Prof. Lewin, I have some doubts about this lecture. 1) at 37:59 you wrote the rocket equation after some derivation. My question is that why didnt you just conclude it at 30:10 !? Because it is just F(thrust) = ma = dm/dt * u (in the case of no gravity) 2) at 46:52 why there is only one mass in the second expression of the change of KE? I think 2 masses must appear in the equation because the rocket has an initial mass which is not the same as the final one! I really appreciate that you will devote some of your precious time to answer these two questions Thank you.

where should be an object placed infront of a concave mirror to form image on itself

27:42 that v seems as printed one

at 26:58 If I have a superball and a tomato of the same mass in my hand, and I let it go, they will reach the bathroom with the same force, according to Newton third law, the bathroom should have exerted the same force to the object and it should show the two object have the same weight, which contradict to what you have said.

The tennis ball and basketball problem: The basketball and tennis ball will approximately reach the speed, when the basketball hits the ground, it bounces back with the same speed V(upwards), so when the tennis ball hits the basketball with the speed V(downwards), thus the relative speed is 2V, so I(tennis)=4mv, the tennis ball bounces back with the speed of 2V. We studied before ½gv²=h, we can calculate that the hight is about 4 times higher than before. Is this correct, professor?

I think that at 25:11 you must divide nmv by dt?

Lecture 17 - 25:30 how do we know that (delta(p)/delta(t))=F_avg still holds for changing mass? I think it was the beginning of Lecture 15 where you derived force as the time derivative of momentum, but this assumed constant mass. Also, what if the velocity of the tomatoes is not constant, eg gravitational acceleration, does F = (dm/dt)*v still hold?

why is there a --ve sign before "u" ?

when i calculated the integral for rocket's initial and final speed it comes out to be positive ln(Mfinal/Minitial) but you stated it negative is there another thing that we need to take care of during integration??

hello prof, I am one of the huge fans of you .You have been consistently saying much detailed notes will be put in web, may we know where could we access to that notes exam questions.

respected sir, i am going through yuor lectures, you really make people love physics (especially me). sir, i wanted to know the solution for the tennis ball and the wall problem in which you told that the wall has a momentum of 2mv.can you please explain it.

prof Walter do you recommend using Ohanian 2nd now? i am studying from your lectures because i am preparing for GRE PHYSICS. thnx

sir at 46:40 you wrote change to be .5 m ( 1100^2 - 100^2) but mass during intial velocity and final velocity are also different so in place of m wouldn"t it be m initial and m final?

Sir, i have a question ,if u is velocity of ejected fuel with respect to rocket and v is velocity of rocket with respect to the earth ,then the velocity of fuel would be v-u or v+u . Because in books like introduction to mechanics by kleppnner and kolenkow it is given v+u.... Now im confused a lot...😓

hi Professor at 46.41 What is this mass m？Isn't the mass of a rocket constantly changing?

At the last of the video, you emphasized the fact that we consider vertical launch from Earth.......... but If we do a little bit inclined launch from Earth rather than vertical launch then I think we will consider the vertical component of the thrust force.........

Sir at the end you told us to show analytically how Newton's cradle works PLEASE CHECK MY SOLUTION Since it is nearly an elastic collision m1v1 = m2v2 (momentum is conserved ) (momentum before and after collision respectively) 1/2 m*v1*v1 = 1/2 m*v2*v2 (kinetic energy is conserved) m1v1=m2v2 v1/v2=m2/m1...............(1) 1/2 m1*v1*v1 = 1/2 m2*v2*v2 using equation (1) we get here that m1= m2 And putting this in equation (1) We again get that v1 = v2 Thus, these conclusions completely predict the behavior of Newton's cradle.

Hi Sir, I absolutely love your lectures. I have one question: How can we justify the difference between kinetic energy differences resulted from burning the same amount of fuel in the same time? Where is the remainder of the chemical energy going when we start with a zero velocity? Can we say that it stays in the exhaust? In other words, if burning fuel can make such a huge shift in kinetic energy when we start with a huge velocity, it must mean that the chemicals have enough internal energy in them to do so, but if we start with zero velocity they will do much less work. Where is the remainder of their chemical energy going? I appreciate if you can shed more light.

Hello, Mr. Lewin. I love your lectures. At 39:22 I went through the integral and looked at the notes, but I don't get why u is negative.

at 46:58, I think you have done some calculation mistake.......... It should be 210,000 and not 200,000

Could you not just take the average mass instead off all the calculus?

WOW!

Is there any book written by Mr. Walter.

Amazing lecture. I have three questions that are killing me, though. Question 1: At 00:04:40 where you make the small angle approximation, I guess I measured h = x2(2l) - not divided by it. Am I wrong dear professor? Question 2: At 00:26:49, you pointed out that "If they weren't tomatoes, if they were super balls which would bounce up, then the momentum change would be double, and so the bathroom scale would indicate 40 Newtons." But I guess earlier in the lecture you mentioned that in this case there is also a change in the impact time. So if the impact time doubles as well as the momentum, wouldn't the force effectively still be 20 Newtons? Question 3: I derived the rocket equation integral (I also checked the PDF), but I still can't figure out why there is a minus sign before u! I mean we have dm/m which is (lnm) evaluated between final and initial masss, isn't it? It would be great if you would be so kind as to explain these to me. Thanks a million for the difference you made in the world through your lectures.

Beste meneer Lewin, thanks for these lectures and answering all these questions! About the ballistic pendulum: if 99.94% of the kinetic energy is converted to heat inside the block, how can we still apply the work energy theorem to get v' out of x? You mention in the lecture that at point x all the kinetic energy is converted into potential energy, but that must mean then: all the kinetic energy that was left after 99.94% of it was converted into heat. But still, the result of the calculation is awfully close to the measurement in Lecture 2, so clearly my reasoning is wrong. But where? Thanks! (ps genoten van uw optredens in De Wereld Draait Door!)

Sir why we do not consider in rocket equation that ...there is net gravitational force on the system (rocket+exhaust) and thus momentum will not be conserved ??

(25:19) change in momentum of 'n' tomatoes is 'nmv'............... so this should be equal to "delta p" and not "(delta p)/(delta t)"..........

sir, i think that the mass lost should be considered at 46:59 while you are calculating the change of kinetic energy.

sir,what are the topics in maths that will be usefull for physics