8.01x - Lect 17 - Impulse, Rockets

[@JoshuaBautista1409]

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

[@obayev]

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!

[@lecturesbywalterlewin.they9259]

Very welcome

[@Zonnymaka]

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.

[@lecturesbywalterlewin.they9259]

:)

[@kurtu5]

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

[@mrtomato7668]

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.

[@Zonnymaka]

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

[@ayn4671]

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)

[@positivegradient]

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.

[@sheriefatalla7391]

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 .

[@johnlysons3176]

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.

[@tsigilis]

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.

[@lecturesbywalterlewin.they9259]

Αλέξανδρος Τσιγγίλης 1) what is the number of the lecture also tell me how many minutes into the lecture. 2) the assignments are posted as pdf files. Look below the videos. I also posted the exams and the solutions of homework and exams + Lecture notes. Not all lectures have Lecture Notes.

[@tsigilis]

Lecture 17 39:25. At the equation Vf - Vi = -u*ln(Mf/Mi) - gt

[@lecturesbywalterlewin.they9259]

Αλέξανδρος Τσιγγίλης u is the speed of the gas (relative to the rocket) that propels the rocket. I will ask my close friend and colleague Jeffrey Hoffman what a typical value for u is. He has flown 5 times on the Shuttle including the flight that repaired the Hubble ST.

[@lecturesbywalterlewin.they9259]

Lectures by Walter Lewin. They will make you ♥ Physics. Here is Jeffrey's answer: Hello Walter! "Isp of solid rockets is ~250 sec, so exhaust velocity is 250*9.8=2450 m/s Iip of the cryogenic liquid rockets is ~450 sec, so exhaust velocity is 450*9.8=4410 m/s

[@shrinivasabhat4628]

@@lecturesbywalterlewin.they9259 Wow you really are great

[@prasadpawar7027]

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?

[@lecturesbywalterlewin.they9259]

*KE in frame of ref 1 is not the same as KE in ref frame 2 if the two frames move relative to each other*

[@prasadpawar7027]

@@lecturesbywalterlewin.they9259 So would the two observers also disagree on the work done by the external agent?

[@ShivamKumar-of2tn]

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

[@lecturesbywalterlewin.they9259]

Keep it up

[@vaanivijay6552]

How is ur jee preparation going on now? How did u do in 10th grade?

[@balasubramanianramasamy1229]

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. :)

[@marksille]

Balasubramanian Ramasamy ok

[@emiliocoria3757]

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

[@tysonwu6312]

47:00 Here Prof. Lewin seems to say that you may gain a lot more kinetic energy from a burn if you have a higher initial velocity. My question is that say your in space and there aren't any reference points and your going 1km/s, are you not effectively going 0 m/s would you still have a larger kinetic energy gain? I think not... after all the fuel is also moving at 1km/s so when you expel it, it actually moves at 1-0.something km/s but not negative like the problem assumes.

[@VickysTuition]

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 !

[@lecturesbywalterlewin.they9259]

I cannot add to the clarity of this lecture. Watch it again.

[@VickysTuition]

Okay. I will go through it again 😅

[@alexandredemasure]

Hi, I was also very confused with the minus sign. It was not clear if u is negative or not. It seems there is as many method as there are people explaining it... So, just to be clear, the vectoriel notation is *v_fuel* = *v_rocket* + *u* (with u negative in the direction of movement of the rocket). But Mr Lewin used the scalar notation, therefore the minus sign is here. Also we can see two kind of dm here. The dm_fuel and the dm_rocket which are opposite in sign. The mass of the rocket at a time t+dt is m+dm_rocket (with dm_rocket negative) or m-dm_fuel (with dm_fuel positive) and that's really not so intuitive at first. Especially that we have to keep only the dm_rocket and not the dm_fuel to integrate the equation containing the mass of the rocket. (We could also keep track of the mass of the fuel instead and therefore keep dm_fuel ! But we have to choose and that's not intuitive which is which. ) I have been stuck two days with this concept. And now, I think I start to get it. I post this to help people who had the same trouble. You can also find a different way of explaining it on EdX "MITx: 8.01.2x Mechanics: Momentum and Energy" - Lesson 19 Hope it will help somebody Greetings from France

[@themanavthakur]

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

[@yash29210]

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

[@lecturesbywalterlewin.they9259]

yup 210,000 !

[@akilvanam3391]

@@lecturesbywalterlewin.they9259 Well to be more specific shouldn't the mass change for initial and final? I believe it was taken same for both initial and final

[@godevil072]

@@akilvanam3391 just make another comment he usually doesnt reply in others comment..... Man of physics he is. .

[@padraiggluck2980]

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.

[@juanmanuelespanabolacuenta4048]

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

[@esaujunior6770]

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

[@etrabajos]

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?

[@lecturesbywalterlewin.they9259]

What I have is fine as the mass of the fuel in any rocket burn to change the orbit is much much smaller than the mass of the satellite/rocket.

[@xavierpoes4985]

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.

[@mranonymous2729]

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.

[@williamleung8588]

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.

[@lecturesbywalterlewin.they9259]

>>> they will reach the bathroom with the same force, according to Newton third law>>> *NOOOOOOOO* you do not understand forces at impact on the floor and you do not understand Newton's 3rd.

[@vinbhaskaran]

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?

[@yash29210]

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

[@yash29210]

you said that 'nm' is number of kilograms per second of tomatoes that I throw on the floor...........but this is only the case when 'n' - which you have defined as number of tomatoes that you throw - this is only the case when 'n' is the NUMBER OF TOMATOES THAT YOU THROW PER SECOND.........

[@lecturesbywalterlewin.they9259]

It's obvious in the context that n is 1/sec (sorry, I should have said that) Thus nmv=delta(p)/delta(t).

[@mohammedalsaqqa1546]

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.

[@lecturesbywalterlewin.they9259]

I watched at the times that you mentioned. I cannot add to the clarity of the lecture. I suggest you watch it again.

[@rafaqatmir01]

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

[@JohnSmith-ol5kh]

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?

[@lecturesbywalterlewin.they9259]

F_t = (dm/dt)*v_t this holds at any time t. Thus it may be different at time t1 than at time t2.

[@JohnSmith-ol5kh]

@@lecturesbywalterlewin.they9259 that makes sense, but how do we know that (dm/dt)*v is the result of a force? The units work out the same, but im not convinced that it represents a force just like m*(dv/dt) does.

[@stonecold799]

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

[@yan4339]

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

[@iaexo]

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?

[@lecturesbywalterlewin.they9259]

use google

[@lollolzi2996]

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

[@efeguleroglu]

v-u is the speed in frame of reference of a human standing on earth. In the frame of reference of shuttle it is v-u-v=-u and that comes to your question. In frame of reference of shuttle it is downwards. But in the frame reference of human it is downwards only if u>v

[@ahmadislam2364]

Why at 34:26 velocity of the exhaust is not (v+delta(v))-u? Isn't u relative to rocket always constant?

[@lecturesbywalterlewin.they9259]

the speed of the gas relative to the rocket is constant

[@ahmadislam2364]

Lectures by Walter Lewin. They will make you ♥ Physics. Isn't the rocket moving with v+delta(v)? So shouldn't the exhaust speed be v+delta(v)-u?

[@lecturesbywalterlewin.they9259]

What I have is correct I cannot add to the clarity of this lecture. If you need more help. use google

[@ahmadislam2364]

Lectures by Walter Lewin. They will make you ♥ Physics. Thank you, sir.

[@AIphy9]

@Trex you are right but Mr. Lewin has omitted delta(v) because we can. Re-write the momentum equation and you will notice infinitesimally small delta(m) when multiplied with infinitesimally small delta(v) can be ignored for all practical purposes. therefore exhaust momentum P as seen from inertial frame = delta(m)[v + delta(v) -u] = delta(m)*v - delta(m)*u.

[@sudiptosarkergarbo3674]

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

[@lecturesbywalterlewin.they9259]

question unclear. watch my solution

[@sudiptosarkergarbo3674]

@@lecturesbywalterlewin.they9259 sir where i can find your solution?

[@randhirkapoor2955]

Topic of impulse will start at 12:08....before that it's an experiment using the concept of work energy theorem and conservation of energy...thank me later for this..

[@arjunvg4719]

42:28 sir, i think there is a mistake here..'u' cannot be a postive number because it is the velocity of the hot gas relative to the rocket which is always negative since the rocket pushes the gas always down...and so the expression would turn to be a negative value still...beacause there is -u in the expression... this is something i noticed weird and so i solved the differential equation myself and the inconsistency i noticed with the above equation is that there shouldn't be a minus sign near u....i.e, the eq uation should be as follows...Vf-Vi=u ln[Mf/Mi] - gt..pls correct this sir.

[@lecturesbywalterlewin.they9259]

no mistake in my lecture

[@tech_science_tutos4155]

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

[@lecturesbywalterlewin.they9259]

slip of the pen

[@prakharbhalla9461]

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?

[@lecturesbywalterlewin.they9259]

If we know the mass loss dm of the rocket due to the rocket burn then you can take it into account (after the burn the mass is then (m-dm) but maybe in my case dm is negligibly small compared to the mass m of the rocket. I do not remember. Please tell me. I don't want to watch the lecture again. *In any case the ratio of the KE change in the 2 cases is 200,000 which is independent of the mass loss as (m-dm) is the same in both cases.

[@prakharbhalla9461]

@@lecturesbywalterlewin.they9259 considering dm to be very small as compared to m then it will be ok. It was not mentioned in the lecture. thanks sir

[@asjathahmath5769]

Tnx sir💗💗💗 I literally addicted for you

[@lecturesbywalterlewin.they9259]

Glad to hear that

[@noneeded152]

You are right, this is the utmost of professorship.

[@ГнусныйШкольнек]

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

[@Jason-ke4jf]

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.

[@lecturesbywalterlewin.they9259]

brush up on your math

[@Jason-ke4jf]

@@lecturesbywalterlewin.they9259 Okay, thank you sir.

[@Jason-ke4jf]

@@lecturesbywalterlewin.they9259 Hi sir, I just want to let you know i brushed up on my math and realized that it's negative because the final mass is smaller that the initial mass. Thanks!

[@VickysTuition]

In the lecture notes, they mention a modulus --> | dm/dt | Since rocket is losing mass, dm/dt is negative, dm is negative. So, if we need to account for this, then we need to take the masses as m+dm and -dm instead of m-dm & dm. This will correct the equations.

[@maheshwarsambari6178]

when will you be online according to your nation timing

[@lecturesbywalterlewin.they9259]

post your questions here. I will answer them within 24 hr.

[@d4rya38]

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.

[@harshasn406]

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 ?

[@lecturesbywalterlewin.they9259]

+Harsha s n looks good

[@kahoung9901]

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

[@lecturesbywalterlewin.they9259]

I watch from 45:30 - 47:30. What I did is correct. I cannot add to the clarity of that part.

[@kahoung9901]

but it should be a mass decrease due to the burn of fuel. i supposed that delta K should be "1/2*(m-delta m) * v' ^2 - 1/2 * m* v ^2, shouldn't it ?

[@lecturesbywalterlewin.they9259]

watch my lecture again. I demonstrated something that is well known in Physics namely that momentum change is NOT equivalent to energy change. The 2 differ as much as apples and coconuts. The change in KE for a given change in momentum depends strongly on the speed before the change; the higher that speed was, the larger will be the increase in KE. *END OF STORY.* Feel free to take the small % decrease in mass of the rocket into account. Your conclusion will not change.

[@Zonnymaka]

I'm a stubborn guy Sir :) so i beg you pardon but the difference in KE (of the rocket only!) is equal to (10^5/2)*[11*m(final)-10*m(initial)] hence if m(final)/m(initial)=10/11 the change in KE is zero. If the rocket (after the thrust) lost more than 9,1% of his initial mass, the change in KE would be negative!

[@AIphy9]

@zonnymaka Although you are right about mass as rockets follow Mf = Mi * e^(Vi-Vf/u), Mr. Lewin is just trying to elaborate the non-intuitive fact about momentum vs KE. Just plot two graphs linear (p=MV) and parabola (KE = 1/2 MV^2) with V along X-axis; and notice the difference along Y-axis for same change in velocity. Again, you are correct about the decreasing mass of rocket.

[@tajpa100]

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.

[@lecturesbywalterlewin.they9259]

:)

[@papalegba6759]

@@lecturesbywalterlewin.they9259 the rocket equation violates newton's 1st law. it's pseudoscience.

[@mornemorkel1483]

27:42 that v seems as printed one

[@1_adityasingh]

His handwriting is the best

[@hamidsoleimani2136]

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.

[@lecturesbywalterlewin.they9259]

1. simple math 2. Impact time is unrelated to momentum change 3. I cannot change Newton's Laws, Use google.

[@hamidsoleimani2136]

@@lecturesbywalterlewin.they9259 1. That's true, sir. I'm sorry I just meant "l" appears upstairs. 2. Thanks a lot. 3. I was just wondering about the minus sign, but thanks anyway.

[@Ahmad-pq6ft]

Again, Love you SIR hope you live forever

[@johnnymendez6094]

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

[@lecturesbywalterlewin.they9259]

which book was that?

[@Suseenthar1]

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?

[@glkglkglkglk9193]

Of course ,if there is no damping by air resistance

[@sandeepmishra3275]

Sir may you upload those assignments you are talking about?

[@prasadpawar7027]

They are in description of every video.

[@erenpol4t]

I am a highschooler, so I dont know what "" stands for as in . Can you please help me out with that.

[@lecturesbywalterlewin.they9259]

it may mean "average" value but I would have to know when I wrote it.

[@erenpol4t]

@@lecturesbywalterlewin.they9259 ohh yes, thank you sir that's it.

[@hamidsoleimani2136]

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?

[@TheRuffusMD]

certainly the rocket exhaust , pushing against the air , adds to the overall thrust...in fact that is the only propulsion you get from a jet engine . thus how do you add this additional trust to the equation ? when the rocket is in space , in a vacuum , this type of thrust against the air is absent .. do astronauts experience the difference in the amount of thrust experienced ?

[@lecturesbywalterlewin.they9259]

>>>>certainly the rocket exhaust , pushing against the air , adds to the overall thrust..>>> *this is incorrect* watch my 8.01 lecture on rockets.

[@TheRuffusMD]

thanxs for your feed back... this is something i had been thinking about for some time...after your comment i went and watched some videos...i have the book you used and i read in that too ..i love your class..i am reading the book and watching your videos .. @@lecturesbywalterlewin.they9259

[@papalegba6759]

@@lecturesbywalterlewin.they9259 now you're saying w=PV is incorrect.

[@andersonexatas]

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

[@subhankarpaira8662]

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.

[@mickeyanderson7963]

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?

[@lecturesbywalterlewin.they9259]

not correct

[@saya-tv2rg]

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

[@saya-tv2rg]

46:41

[@lecturesbywalterlewin.they9259]

I cover that in one of my 8.01 lectures about rockets

[@margodphd]

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

[@lecturesbywalterlewin.they9259]

kzbin.info/www/bejne/ioK1Z4mhjt90m7s

[@maheshwarsambari6178]

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

[@lecturesbywalterlewin.they9259]

I prefer you ask your questions here. Make them brief; refer to the lecture number and also to how many minutes into the lecture.

[@sheriefatalla7391]

Hello prof Walter Happy to write you again :) in min 10:01 when the pendulum swings and give its velocity to be measured in centimeters (x direction)! Did we take in our consideration the friction that this ruler could make?or did we take also in our consideration the collision between those parts (block, ruler) because may be if the ruler is too light, it bounces more and give high reading in x direction(for example 10 cm ).thnx

[@lecturesbywalterlewin.they9259]

The ruler is not connected with any part of the demo. It does not move. Thus no bounce and no friction

[@harshaggarwal6632]

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 ??

[@lecturesbywalterlewin.they9259]

www.grc.nasa.gov/WWW/K-12/rocket/rktpow.html

[@harshaggarwal6632]

Thank you sir

[@absharma534]

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

[@soumyadey7387]

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

[@lecturesbywalterlewin.they9259]

If you come to Cambridge MA, USA I'll be happy to meet again with you. Which course did you take from me and in which year?

[@surendrakverma555]

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

[@tudoriacob9632]

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

[@turtle8558]

The mass changes overtime, if you take the average mass you might not have enough thrust to actually start the rocket, or perhaps the rocket might stop midway

[@philipmilosevski2099]

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.

[@satyamjyotisamal2371]

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

[@lecturesbywalterlewin.they9259]

All the best

[@satyamjyotisamal2371]

@@lecturesbywalterlewin.they9259 Thank you sir 🙏

[@КириллИржевский]

Hello professor. I don't know why, but links on assignments are not working since 17 lecture 8.01. Can you please fix it?

[@lecturesbywalterlewin.they9259]

use playlist "8.01 Homework, Exams, Solutions and Notes"

[@КириллИржевский]

@@lecturesbywalterlewin.they9259 Thank you, sir!

[@qingfangzhang7856]

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!

[@lecturesbywalterlewin.they9259]

KE and Momentum are very different. A 10 sec rocket burn changes the momentum of the spacecraft by a fixed amount. But the change in KE can be HUGE or SMALL. I have 2 mases (1 kg and 10 kg). They both move with a speed of 1 m/sec. I change the momentum of each by 10 SI units. The change in momnetum is the same but not the change in KE. simple physics

[@faridhasanov6094]

Sir , I couldn't find the solution for problem of tennis ball and basketball . Can u briefly explain how yo calculate that ?

[@lecturesbywalterlewin.they9259]

+Фарид Гасанов Treat it as an elastic collision between the 2 balls. If m1

[@faridhasanov6094]

+Lectures by Walter Lewin. They will make you ♥ Physics. I got it. Thank you!

[@vanoscrap6296]

If the ratio between the two masses is 10, my calculations give a ratio of 17 between the speed of the tennis ball before and after impact. Do you get the same result?

[@commentor6439]

17 times the speed would probably dent the roof of MIT! The ratio is indeed 9. Realize that the ratio between the two masses is in this case negligible. The key here is to find that the velocity of the tennis ball is thrice its initial velocity as seen in our frame of reference. => Height = (3v)^2/2g. Substitute sqrt(2gh) for the velocity and you will have your relation.

[@fifil6251]

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!!

[@haroonahmad9354]

There is negligible gravitational pull coming from that rocket

[@xrisku]

if the ballistic pendulum moves so little, can one not stand closer or use a less dense material?

[@lecturesbywalterlewin.they9259]

question unclear

[@xrisku]

Lectures by Walter Lewin. They will make you ♥ Physics. Sorry, I forgot the time mark. At about 4:50 +, in the beginning, you state the ballistic pendulum may only move < 1 mm upon having a bullet fired into it, and that is nearly impossible to measure. you then rearrange the equation to incorporate x. My question is: by replacing either the block that your shooting into inside the pendulum (making it less dense and therefore more likely to be be affected by the bullet's velocity/transferred momemtum) or by placing the gun or rifle closer to the ballistic pendulum, would the ballistic pendulum then not then be able to record an accurate reading? I suppose that is not as standardized as a better equation. p.s. I nearly had a heart attack when you walked up in front of the rifle after you had inserted the bolt, put a bullet into it and cocked it. scary.

[@lecturesbywalterlewin.they9259]

momentum is conserved. Thus mv = (M+m)v' The smaller M the larger v'. 0.5*(m+M)*v'2 = (m+M)gh . We measure horizontal displacement which is related to h. I cannot add to the clarity of the lecture.

[@xrisku]

thank you for the reply. the demo is very good.

[@AnkushKumar-vi2dk]

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

[@CaptainCalculus]

Wouldn't you just apply Triolkovsky's rocket equation at 31:23? Sorry for the endless questions. oh ok you get there in the end...

[@harshitaharsh7671]

Dear sir,what is the purpose to introduce impulse as there is no specific use of this quantity why can't we use change in momentum directly to calculate avg force?

[@lecturesbywalterlewin.they9259]

>>> there is no specific use of this quantity >>> ha ha ha are you kidding?

[@harshitaharsh7671]

avg force can directly be calculated by delta p/delta t then why we calculate first impulse then equate it to delta p then give a new formula of avg force - F=I/delta t ,why?

[@zombiesalad2722]

Impulse is itself change in momentum.

[@mrpotatohed4]

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

[@schattopadhyay1901]

Sir, if rockets gain more energy when they have a non-zero initial velocity, by burning the same amount of fuel....then where does the extra energy come from, given that the only source of energy is the chemical energy released from the combustion of the fuel??

[@schattopadhyay1901]

Yes. I can see that, momentum is proportional to V, where as KE is proportional to V^2. But my question is, when that happens...how does the extra energy come from the same amount of fuel?

[@lecturesbywalterlewin.they9259]

That is the wrong way to think. If you change the momentum by a fixed amount the KE can change little or a lot depending on the speed of the object. Momentum is NOT energy. If you hit an object with a hammer (impulse), KE is in general not conserved, it's an inelastic collision.

[@yedmavus]

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

[@CobyHerford]

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

[@lecturesbywalterlewin.they9259]

used google

[@arnoldstallone938]

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.?

[@lecturesbywalterlewin.they9259]

SpaceX believes a fully and rapidly reusable rocket is the pivotal breakthrough needed to substantially reduce the cost of space access.

[@lecturesbywalterlewin.they9259]

en.wikipedia.org/wiki/Reusable_launch_system

[@manassrivastava7448]

Sir you discuss a problem where you made a basketball and tennis ball simultaneously fall there tennis ball was above basketball the tennis ball rise so high sir I think the reason behind it must be that when they collided with floor and basketball ball respectively they get twice there momentum I observed the situation in video and found that basketball ball transferred it's momentum to tennis ball . you discussed a similar situation where are a Ping Pong ball when collided with a heavier ball in pendulum then Ping Pong ball got twice of its original velocity sir is my reasoning correct

[@lecturesbywalterlewin.they9259]

>>>they get twice there momentum>> that's not correct >>> Ping Pong ball got twice of its original velocity>>> not correct

[@manassrivastava7448]

Sir I remember you discussed a situation where you discussed a situation with the help of a pendulum demonstration where ping pong ball got twice the velocity of heavier ball. I think In situation where you discussed that why tennis ball reached such a high velocity when made to fall from height above a basket ball I observed in the video that after bouncing back to the floor the basket ball collided with tennis ball then according to law of conservation of momentum the tennis ball due to smaller mass got higher velocity sir now am I right

[@mohamedraafat3715]

Yes you treated it correctly but you used wrong equation because the tennis ball is moving while the other collide with it

[@swarnatejathota9318]

sir suppose i tied a mass to a string fastened one end. then i released the mass from a height. what will be the tension developed in the string?? is this related to impulse?

[@lecturesbywalterlewin.they9259]

if we ignore air drag, everything in free fall is weightless. Thus the tension is zero.

[@swarnatejathota9318]

at the time when the ball reaches the end of the string the ball pulls the string. is that in this case 2mg i.e m(g+g)

[@lecturesbywalterlewin.they9259]

I misread. I now understand that you do not drop the string with the ball (mass m); the string is attached to the ceiling and the mass is at the ceiling. If you drop the mass from the ceiling the string will also drop (free fall). Thus both will be weightless; zero tension in the string. Then there comes a time near the very end of the falling mass, that the string will slow down the mass and ultimately stop the mass. Then the tension in the string will become very briefly even larger than mg when the string is fully stressed and extended (assuming that the string is massless). Shortly after that the tension will become mg (assuming the string is massless). The transition from zero tension to larger than mg to mg occurs in a very short amount of time depending on the stiffness of the string.

[@swarnatejathota9318]

sir what do you mean by "even larger than mg" how can i find it??

[@lecturesbywalterlewin.they9259]

Then the tension in the string will become very briefly larger than mg

[@omozaas]

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??

[@lecturesbywalterlewin.they9259]

When a bullet gets stuck in a piece of wood, momentum is conserved. BEFORE the coll bullet mass m and speed v , block mass M, speed after the coll of bullet together with the block V (assuming that the block can move freely) , then mv=(M+m)*V. This is an inelastic collision. KE is converted into heat.

[@theblackhole1]

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

[@sreyassuresh1886]

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?

[@lecturesbywalterlewin.they9259]

+Sreyas Suresh yes that is correct. But after the collision the wall has near zero KE. Thus the wall has momentum but no KE.

[@crazyplayer1000]

I am not sure the signs are correct in the rocket equation? At some point you write mdv=udm, that is: dv=u(dm/m). Integrating both members: vf-vi=u*[ln(mf)-ln(mi)]=u*ln(mf/mi)]. This is obviously wrong, as it would lead to a decrease in speed!!! And in fact without showing it the calcs, you write vf-vi=-u*[ln(mf)-ln(mi)]. Where is the mistake? I do not have access to the online demonstration you mentioned.

[@lecturesbywalterlewin.they9259]

>>>I am not sure the signs are correct in the rocket equation?>>> *they are correct - use google*

[@crazyplayer1000]

I beg to differ. The signs at the end are obviously correct I do not disagree with the final formula, but with all respect, I dont need to use Google to integrate: and from the blackboard I see: $dv=u(dm/m)$. As I said, the result is $vf-vi=u*ln(mf/mi)$ without the minus sign that you rightly apply at the end to make the thrust consistent with the increase of speed. But this is an adjustment made not supported by the integration. On the other hand, if, as you state, $mdv=u*dm$, that can only imply that dv>0 only if dm>0, i.e. the mass increases in the rocket. Which is clearly a fallacy in how the equation was derived until that point.

[@crazyplayer1000]

Another thing, again with due respect: at the age of 50, I am from the generation that still used paper books to study. Not google. If I look up things on google, I might be inclined to believe that the Earth is flat, or that gravity doesn't exist. I prefer texts like the Resnick-Halliday (although a bit to simplistic for my taste), or more mathematically based texts (which you probably don't know as they are from Italian authors). Google is not a reliable source of information.

[@crazyplayer1000]

No, really, you can''t get away with a smile.....it's driving me crazy :-) Where is the stray sign in your explanation??? :-)

[@lecturesbywalterlewin.they9259]

Please do not go crazy. Read my lecture notes (by Tipler) which are attached to my lecture as pdf file. freepdfhosting.com/a3a29b78f4.pdf.

[@manassrivastava7448]

Sir whenever we hit a door with a door with force F and impulse "I" the door simply open but if we hit a door with same force but smaller impulse It stats vibrating . Sir what can be reason behind it.

[@lecturesbywalterlewin.they9259]

send me a video. If I give my doors a small impulse they do not vibrate.

[@manassrivastava7448]

Sir actually what I did was. I first applied some force to close the door from outside by my feet. After i opened it again and kicked it. It started vibrating I was not able to get why this happened

[@manassrivastava7448]

Lectures by Walter Lewin. They will make you ♥ Physics. sir I redid the same exp but again the door vibrated

[@manassrivastava7448]

Sir where should I send video

[@lecturesbywalterlewin.they9259]

you can send videos here: kzbin.info/www/bejne/npuXf62Lap6qoM0

[@NaderHGhanbari]

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.

[@lecturesbywalterlewin.they9259]

question unclear.

[@NaderHGhanbari]

46:53 Case 1: Rocket burning fuel starting with v0=0 Case 2: Same rocket burning same fuel starting with v0=1000 We burn the same amount of fuel in the same time in Case 1 and Case 2, but the increase in kinetic energy in Case 2 is much higher. How can we justify this? The same chemical should result in the same amount of energy and I'm assuming all the chemical energy is converted to kinetic energy so what causes the difference here? Is this assumption wrong? Could it be that the difference between the two is because of the difference in generated heat in two cases?

[@lecturesbywalterlewin.they9259]

Energy is conserved. If you add up the energy increase of the rocket and the energy of the hot exhaust and its speed, that is the same as the chemical energy of the fuel. But The gain in KE of the rocket is exclusively due to the impulse on the rocket during burning. IMPULSE IS NOT energy!

[@NaderHGhanbari]

Thank you very much Professor Lewin for the clarification. By the way, given this fact, a paradox comes to my mind. Let's say a rocket is going with a huge velocity vr (near infinity) and we have a finite amount of fuel burning in a finite amount of time. According to the above calculation, the fuel can still increase the velocity of the rocket by vc (a small constant velocity), we see that the change in kinetic energy is 1/2*m[(vr+vc)^2 - vr^2] which we can approximate by m*vc*vr which is near infinity (given that vr is infinity, and vc and m are finite amounts). Can we conclude that a rocket with a finite mass can not possibly have a nearly infinite velocity? Is this somewhat related to the fact that the classical mechanics does not work for very very large velocities?

[@lecturesbywalterlewin.they9259]

NO paradox at all. When we deal with very high speeds we need special relativity, Newtonian physics does not apply.

[@archanaagrawal4642]

Si,In ballastic pedulum illustration shouldn't 1/2(m1+m2)(v')^2=-(m+M)gh

[@archanaagrawal4642]

Oh i did w=ki-kf

[@randomvideosbyarandomguy2709]

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...😓

[@lecturesbywalterlewin.they9259]

watch my lecture on rockets and/or use google. Always trust Kleppner and Kolenkow.

[@randomvideosbyarandomguy2709]

Lectures by Walter Lewin. They will make you ♥ Physics. Ok sir

[@colonyresident7151]

Dear Sir, I have a doubt on the Ballistic Pendulum method which was used to measure the speed and energy of a bullet. My question is: Why we don't measure the Angle θ, while we measure the x and h which are able to be calculated by l and θ easily. Besides, the measurement of x and h has some uncertainty greater than the protractor which can be synchronized at the pivot of the Pendulum.

[@lecturesbywalterlewin.they9259]

use google "Ballistic Pendulum"

[@colonyresident7151]

Yes, I did google it. Most of the topics pop up was about the principle and calculation of the Ballistic Pendulum. There are some topics in which the people measured the Angle θ to calculate x and h, instead of directly read the x or h from their scales. My concern was the exact reason of the parameter of the Robins' formula adopted was X and h, why not the angle of θ. In my imagination to do the Ballistic Pendulum, the angle θ can be achieved without any friction while to get travel distance of the huge mass, there was some friction. Mr. Robin was such a smart person, I am not sure if my method would be better than his. Would you mind to share the story behind it if you have any clue?

[@lecturesbywalterlewin.they9259]

>>>Why we don't measure the Angle θ>>> All we have to measure is x and that can be done with an accuracy of about 1 mm which is about 2%. To measure h would be lunatic. It's only about 1 mm. The error in h would be near 100% thus the error in theta would be near 100%. *To measure ONLY x is THE ONLY sane way of measuring v. End of story.*

[@colonyresident7151]

Thank you, Professor.

[@sheriefatalla7391]

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

[@lecturesbywalterlewin.they9259]

I recommend Ohanian because I used it in the lectures on this channel. But it's not better than most other college physics books.

[@sheriefatalla7391]

thanks for your reply Prof Walter, my goal is to do well in GRE PHYSICS so is it enough to study 8.01x, 8.02x, 8.03x or should i study something else beside those courses? I graduated from faculty of pharmacy but i really fascinated with light and want to pursue my education with quantum photonics that's why i am preparing for GRE PHYSICS to be able to enroll in photonics research program. so any ADVICE from you Prof Walter for that Goal?thnx

[@lecturesbywalterlewin.they9259]

You can find out on the web which topics are being covered on the GRE. You definitely should buy a book that prepares you for the GRE.

[@kutbuddinrangwala1039]

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

[@teamdatamanagementdrive6292]

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??

[@VickysTuition]

Yes, this is because we assume dm is positive. But since the rocket is losing mass dm/dt is negative & dm is negative. So, instead of m-dm & dm. Use m+dm and -dm for the new mass of the rocket and released gas. Your equations will get corrected. In the lecture notes, they mention this as "modulus" dm/dt -> | dm/dt |