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!
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
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
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
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!
@@lecturesbywalterlewin.they9259 Thank you for your reply. This spring I will be enrolled in my second semester of calculus-based physics at my local Maryland college. Your electricity and magnetism lectures will surely help motivate this mighty course. I started as a History major as I began college, subsequently became enamored with mathematics and then soon after developed a fondness for physics. My embrace for this subject is predicated on a deep admiration for the higher-level reasoning skills that define this discipline. I know you frequently declare that physics is not about equations but concepts. This is why the subject is such a worthy endeavor. The humanities and arts matter, but physics is such a powerful body of knowledge. I believe I have made the right choice for my future studies. A prominent modern mathematician once said that the difference between someone who embraces math and everyone else, is that the math-minded individual is not afraid of not knowing the answer, since their logic and reasoning will enable them to navigate the path to a solution. I feel the same way about physics, with the path to a deeper understanding deriving from a keen conceptual insight. Thank you for your time and I hope you and your loved ones remain happy and healthy!
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.
carultch thanks alot for you important contributions in answering questions of my viewers. I have been reading them with pleasure. Your responses will help many. *THANKS* \\/\//////@lter Lewin
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.
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...
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).
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??
I don't understand this answer. The mass of the wall is large, but not infinite. The impuls imparted on it is not zero, and therefore its speed after the impact cannot be 0. What happens when you take a tennis ball machine gun and bombard the wall with tennis balls? It will still not move. There must be some impuls from the other side that pushes it back. Where does that impuls come from? Assuming that it is a free standing wall, I bet the force that pushes it back is gravity. The imparted speed will cause the wall to start to topple over a tiny bit, that is, its front side is lifted a bit off the ground. Then gravity pull that back, and cause the reverse impuls on the wall. That will in its turn impart a tiny impuls on the Earth itself, which will cause the Earth to move a bit faster. So the real answer should be: because the wall is attached to the Earth, it will not move wrt. the Earth, and the Earth will take over the impuls.
@@renedekker9806 The impulse comes from elecreostatic forces between the tennis ball and the wall during the collision, gravity has nothing to do with it.
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 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.
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.
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
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!
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.
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??
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?
Hi. if KE of wall is zero because it’s v is close to zero, then momentum of wall should also be zero because it’s v is zero. So I don’t think your answer is correct.
@@AJ-ss3jy no it is correct because it isn't exactly 0. It is close to zero and the mass of the wall is large enough. The energy lost is in the form of potential energy of the wall.
We can also think about this in this way: The law of conservation of momentum does not hold for the problem as there is an external force of the ground on the wall.
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...
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
You are always free to choose the boundaries of your system. If you choose the wall to be out of your system, then it´s easy to understand what you have explained. The problem that Walter presents is when you choose the wall to be part of your system, then it looks as if the laws of physics are breaking apart.
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
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?
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.
+luvinthe jazz I appreciate your comments. The question is, can an object which has a mass which in the limiting case goes to infinity have momentum but no KE? In outer space bounce a tennis ball (mass m, speed v) off a wall with mass, M, of 10^10 kg and with speed,V=0. The collision is elastic. Clearly after the coll, the tennis ball will have to a veryyyyyyyyyy high degree of accuracy the same speed as before the coll but the direction will be reversed. Before the coll M has speed V=0.. After the coll M has speed V'. In the limiting case that M=>infinity V' must be zero. If not, the kinetic energy of the wall would be infinitely high and that violates the conservation of mechanical energy (in this case the conservation of KE). Yet the momentum of the wall must be +2mv as momentum is conserved. How can the wall have momentum but no KE?
+Lectures by Walter Lewin. They will make you ♥ Physics. Maybe this will help you. www.physicsforums.com/threads/why-momentum-must-be-in-the-wall.287325/
+Lectures by Walter Lewin. They will make you ♥ Physics. I Got it!!! So the change in momentum is +2mv. So if v=0, then 2mv=0. As Foghorn Leghorn said, two nothins is nothin.
+Walter Lewin I think both KE and P are not zero, but v tends to zero so KE really tends to zero because the quadratic of v, talking always in an ideal closed-system case. You made m1 tends to zero to deduce m2 Ke is zero, but that's not completely true. Thanks for all your material.
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.
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.
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.
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!
For the problem of the wall and the ball. Consider instead of a wall, a huge cement block sitting on the ground (acts like wall). When the ball hits the block, it exerts a momentaneous force on the block. This force does transmit kinetic energy to the block, but at the same time, the ground exerts a static friction force on the block, taking away that energy, hence, the block doesn't move. Momentum isnt conserved as there is a net external force acting on the system (the friction force) as the collision takes place. The ball bounces back due to action-reaction principle (the block exerts an equal but opposite force on the ball, that accelerates it, changing the direction of its velocity). In other words, there is a net impulse over tge system that changes its momentum. We can replace the block with a wall and the friction f. with forces due to the configuration of the building. If the block was sitting on a frictionless surface or the wall was inserted in a peculiar system of frictionless rails attached to the ceiling and to the ground, then there wouldnt be a net external force and momentum would be conserved. In this case we would see indeed that both the wall and the block would move (with the corresponding velocities arising from the equations).Would this reasoning be correct sir?
@@lecturesbywalterlewin.they9259 Ive read a comment below suggesting you can consider the wall to be the earth. I dont understand this, as the ball itself would get attracted/accelerated towards the object it is going to colide with (due to gravity force). Wouldnt this make it impossible to apply conservation of momentum? In other words, in order to apply the M>>>m approximation dont you have to make sure the difference of masses is not big enough so that the gravitational interaction is no longer negligible in the first place?
@@lecturesbywalterlewin.they9259 what formula do we use to determine the momentum of the wall? Mass is huge (infinite) but velocity is 0, so momentum would still be zero.
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.
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...
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
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?
I call what is happening to billiard balls when they collide "Momentum Transfer". Here is why. If the balls have the same mass and there is no friction or external force between them, the moment and kinetic energy is conserved. Now, if a ball with mass 'm'--b1--collide with 2 balls at rest, its momentum is transferred to the first ball in contact--b2--and again, b2's moment is transferred to the third ball--b3--. So, mv1=mv3 and b3 takes off. If we use mathematics, v1´=0 and v2´=v1(before it transfers momentum to b3). When v2 transfers, v2´=0 and v3´=v1. As an example, if 3 balls collide with 2 balls, the two balls in the 3 balls system push 2 balls at rest with 2mv and momentum is transferred ,so, the two balls takes off and 2 balls momentarily at rest and the remaining ball collide with them and as in previous argument, another 1 takes off. This all happened in a flash so, we see 3 balls take off. Professor Lewin,please correct me if I'm wrong.
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.
+Pedro Candeias KE=P^2/2M. In this case, P of the wall is 2mv (m is the mass of the tennis ball, v is its speed). Since M=>infinity, KE is ZERO, but P is not zero.
Regarding brain teaser at end of lec: Let us assume k1 balls are moving with velocity v1 initially and k2 balls are moving with velocity v2 finally. K1mv1=k2 mv2(conservation of momentum) Hence K1v1=k2v2........... (1) Since collisions are elastic KE is conserved. Hence, K1(1/2m(v1)^2)=K2(1/2m(v2) ^2) So, k1(v1) ^2=k2(v2) ^2.......... (2) By solving equation (1) and (2) we get k1=k2.hence same no. Of balls should be ejected which initially hit. Hence, proved. IS IT CORRECT?
You assumed that k1 balls stopped after the collision, why? if you have let's say 2 balls with masses m1 m2 and velocities v1 v2 initially and u1 u2 finally even if v2=0 that doesn't mean that u1 will be 0 only if m1=m2 that will be true I think the same apply in your situation with k1 k2 so you assumed that u1=0 so it's like you assuming that k1=k2 and that is why you end up to this result
imagine the wall is the earth, mu=m(-u)+Mv p of the earth= Mv = 2mu since M>>m velocity of the earth v~=0 therefore, velocity of the tennis ball bouncing back =-u and approximately no change in KE for the ball is this correct?
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 balls momentum after bouncing back is (-m1v1)...and 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
sir you say V=0 then you are saying P not equals zero. P=MV. so if V=0 then P=0. but here V is only approximately zero, I think that's why you are telling there is momentum present in the wall. then it is obvious that it has a velocity of very very small magnitude and indeed it possesses a very very small kinetic energy. Amazing sir truly, to think the wall has a very small velocity
About the problem with the tennis ball and the wall: We know that the momentum of the wall is: p(wall) = m(wall) * v(wall) Then, by the equation for v1' and v2' for elastic collisions, we can substitute v(wall), and we get: p(wall) = (2*m(wall)*m(ball)*v(ball)) / (m(wall) + m(ball)) We know that m(wall) >> m(ball), so as we take the lim m(wall) -> infinity we get that: p(wall) = 2*m(ball)*v(ball) * lim (m(wall) -> inf) ( m(wall) / (m(wall) + m(ball)) ) the right side limit is of the indetermined form inf/inf, so we apply L'Hopital's Rule and we get that the limit is exactly 1, so: p(wall) = 2*m(ball)*v(ball) = 2mv Is this correct?
ofcoz p of the wal is +2mv after the collision, but the KE of the wall after the collision (if its mass M is thousands of tons) *is veri close to zero & can be made as small as you want to by increasing M.*
for the last problem...lets change the frame of reference....lets consider frame is on 5th ball and then we will see 3 balls are approaching to us and what we will see is 3 balls get detached from behind and balls which were approaching gets attached to 5th ball
it's a multiple collision problem. It's easy to show that the end result is consistent with the conserv of momentum and KE. But that is not the complete solution. use google
+Shivam Agarwal Momentum can be HIGH and KE effectively zero for very massive objects. I suggest you figure this out on your own. If you don't succeed in 2 weeks, contact me again.
+Lectures by Walter Lewin. They will make you ♥ Physics. Is it because even if the have velocities very very low velocities, they have a high momentum because of the high amount of mass they posses ?
+Lectures by Walter Lewin. They will make you ♥ Physics. Well if m2>>m1 and the momentum imparted to m2 is 2mv, then the speed of m2 will be given by 2mv/m2. But, because m2 is very very high, the velocity will approach 0. Thus KE which is given by 1/2m2v^2 will be about zero. Is that correct ?
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
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
Lectures by Walter Lewin. They will make you ♥ Physics. Well, taking a different approach, surely if the wall gained any velocity after the collision then the energy is the system would be infinitely great which would violate the conservation of energy. However, why does it not work such that P is finite and M is infinite so P/M=0, and despite the wall having gained momentum, it cannot have gained and velocity?
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.
@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....
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.
Sir as you said I watched this 💕 amazing lec but I have still doubt 🙏 plz plz sir clear it🥺 My teacher said *KE is conserve "in" elastic collision but not "during" elastic collision* Is he correct ?
@@lecturesbywalterlewin.they9259 Sir he is talking about elastic not inelastic 🙏plzzz sir clear it 🙏 He said that we cannot use term "during" bcoz @ exact time of collision KE is not conserved as it may be converted into potential energy for that Instant so the kE is conserved during elastic collision is wrong 🙏🙏🙏Plz 🥺 sir confirm that is he correct or we can use both "in" & "during"
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
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?
Do you think parallel universes exist and dreams are glimpses into them? As I listened to your lecture 16 using iPad on my bed, I fell into sleep. I had a dream in which I sat in your MIT auditorium watching your alternative lecturing momentum and elastic collision. Instead of using ping pong and billiard balls to demonstrate the three cases, your alternative was using different sized globes. Everything looks realistic.
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.
I have a really stupid question I guess; Why use V1, V2, etc.; why not use V, V2, etc. (isn't the first V assumed to be V1, so using V1 would be redundant?)? Same for Time, Mass, etc. Also, what is Prime and its value? I'm not a math guy, but I love watching these lectures; for some reason this seems to make more sense to me than algebra - algebra makes me hate math so I never bothered trying to move past it (is it really necessary to understand algebra in order to under stand stuff like this?). Thanks for your time!
V1 is not redundant. It specifically refers to the velocity of the object we call object 1. V would simply mean velocity in general. As for the prime notation, it could either mean Newton's notation of derivative function notation (such as f'(x), pronounced "f prime of x"), or it could mean a way to label a second instance of a variable you've already used. In this particular lecture's example, the prime notation indicates the velocity of the center of mass in the lab reference frame.
Regarding the problem at the last of the lecture (48:01)- Let the mass of each ball be 'm' and their velocity is 'v'...... then the initial momentum of the system= 5mv Now since it is an elastic collision so some balls after the collision will stay put and some will move with the same velocity 'v'... Let the number of balls which after collision stay put be 'x'....... then the number of balls which will move with velocity 'v' after the collision will be (8-x)........ So the momentum after the collision will be (8-x)mv......... Since momentum of system is conserved so 5mv = (8-x)mv......... which gives x=3 AND SO 3 BALLS WILL STAY AT REST AND OTHER 5 BALLS WILL MOVE WITH VELOCITY 'v'......... Is it correct?
Nice try. It's easy to show that the "surprising"outcome of any combination of balls that you swing is consistent with the conservation of KE and with the conservation of momentum. But that is only a "consistency test". But that is not a proof. The actual physics is much more complicated as it is a multiple-collision problem. My advice is to live with your "consistency test" and not bother about the "proof". en.wikipedia.org/wiki/Newton%27s_cradle
Thank you sir for the lecture.... I am not able to find a way to prove the happenings at the end of the lecture. Why is it so that when we hit 5 balls on 3 balls, 3 balls remain and 5 move on?
Very roughly speaking,Because as we can see all balls are identicle and if we consider energy transfer by five balls is 5 times 1/Mv2 then the transfer of energy to three balls is 3 times 1/Mv2 and the system has still excess 2 times 1/Mv2 and that will transfer to last two ball of five ball system and then system will came to equilibrium .....
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!
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?
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 ?
Momentum is conserved as long as there is no net impulse from external forces. This will happen if: 1. There are no external forces 2. External forces add up to zero 3. The collision happens so quickly, that any cumulative effect from external forces is zero, during the instant the collision happens.
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?
I would like to leave a question about the ball-wall collision. If the ball with mass m collides with an initial velocity v and comes out with a velocity v in the other direction , than there must be a momentum transfer to the wall of +2mv, because of the conservation of momentum of the system. I understand that K=P/2M, where M is the mass of the wall, and can be thought as a limit case where M->infinity, threfore K->0, even though P is not zero. However, where is the momentum in the wall? I mean, physically speaking, I can not understand how can something have momentum and no velocity.
At the collision the tennis ball transfers a momentum p=2mv to the wall. m is the mass and v the speed of the ball. The KE of the wall is p^2/2M Here M is the mass of the wall. p is finite (+2mv) If M ==> infinity, KE ==> 0 but p remains +2mv. There is another way to look at this. The momentum of the wall is 2mv=MV (V is the velocity of the wall after the collision). Thus MV is fixed M=> infinity then V must go to zero to make sure that MV is finite. KE of the wall is 0.5MV^2=0.5pV. p is finite but V=>0. Thus KE=> zero.
let momentum of wall be M by conservation of momentum mv=- mv+M M=2mv let kinetic energy of wall be K as the collision is perfectly elastic kinetic energy is conserved so, 1/2mv^2=1/2m(-v)^2+K K=0 i think that must be answer tell me if i m right or wrong
+Safder Shakil Momentum is conserved. When the tennis ball approaches the wall, its momentum is mv, after the bounce it is --mv. The difference is 2mv.
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?
Suppose a ball is thrown upwards in a room and it hits the ceiling and comes back down. We observe that the velocity of the ball increases. Conservation of momentum does not take place in this case, does it? External force is indeed applied initially.
If the the ball hits the ceiling with a "decent speed" of about 1 m/s and if the collision takes place in a very short time (0.1 msec), then to a very high degree of accuracy the magnitude of the momentum of the ball just before and just after the collision are the same (assuming that it is an elastic collision). The reason is that the change in velocity due to gravity during the 0.1 msec collision is only 10*10(-4) = 10^(-3) m/sec.
Good question. The reason we commonly get away with using these terms interchangeably, is that Earth's gravitational field is close enough to uniform in most situations, that there is no significant difference between these points. But if you did have a non-uniform gravitational field in the problem in question, there would be a difference. Center of mass is what you calculate when you take a "weighted average" of the positions of all the masses that comprise an object or system of objects. By "weighted average" I mean that you multiply mass*position for each object, add it up, and then divide by total mass. This calculation is gravity-agnostic, as it ignores any gravitational fields that apply to the objects in question. It exclusively studies the inertial masses of the objects. Center of gravity is what you get, when instead of using mass alone to take the weighted average, you account for the gravitational force when taking a weighted average of the positions of the point masses in the system. Center of gravity is the point at which gravity appears to act, as if it were a single concentrated force. It is the point where you could apply a single point support force, and not require any torque to balance it. If you apply a point support force at exactly the center of gravity, the object will be in an equilibrium of neutral stability. Above, and it will have a stable equilibrium; below, and it will have an unstable equilibrium.
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!
Thanks for sharing!
i am enjoying your lectures more than life
thanks for your kind words
Lectures by Walter Lewin. They will make you ♥ Physics. Is that really you replying?
Yes Mohamed I am the REAL Walter Lewin and I run this channel.
Where are the answers of questions you asked ?
@@lecturesbywalterlewin.they9259 omg i've just learned about that now, so embarassing, my previous replies to you were awkward haha
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
I'm doing the same thing as you, good luck! :)
@@23s83 same to you man ;)
He is the only person which can make us love physics and proceed physics as our carrier
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
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
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
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!
You are most welcome
@@lecturesbywalterlewin.they9259 Thank you for your reply. This spring I will be enrolled in my second semester of calculus-based physics at my local Maryland college. Your electricity and magnetism lectures will surely help motivate this mighty course. I started as a History major as I began college, subsequently became enamored with mathematics and then soon after developed a fondness for physics. My embrace for this subject is predicated on a deep admiration for the higher-level reasoning skills that define this discipline. I know you frequently declare that physics is not about equations but concepts. This is why the subject is such a worthy endeavor. The humanities and arts matter, but physics is such a powerful body of knowledge. I believe I have made the right choice for my future studies. A prominent modern mathematician once said that the difference between someone who embraces math and everyone else, is that the math-minded individual is not afraid of not knowing the answer, since their logic and reasoning will enable them to navigate the path to a solution. I feel the same way about physics, with the path to a deeper understanding deriving from a keen conceptual insight. Thank you for your time and I hope you and your loved ones remain happy and healthy!
One of the funniest things about these videos is watching the purple circles under the students' eyes get deeper as the semester wears on.
Thank you so much for Prof. Lewin and UA-cam for this wonderful resource!
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.
carultch thanks alot for you important contributions in answering questions of my viewers. I have been reading them with pleasure. Your responses will help many. *THANKS* \\/\//////@lter Lewin
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.
How did your exam go?
I loved physics before but your lectures have made me enjoy physics. Thanks a lot
these lectures will really make you fall in love with physics
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...
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).
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??
Yes, that is correct
@@lecturesbywalterlewin.they9259 So it is a problem which only occurs in a limit of infinity.
I don't understand this answer. The mass of the wall is large, but not infinite. The impuls imparted on it is not zero, and therefore its speed after the impact cannot be 0. What happens when you take a tennis ball machine gun and bombard the wall with tennis balls? It will still not move. There must be some impuls from the other side that pushes it back. Where does that impuls come from?
Assuming that it is a free standing wall, I bet the force that pushes it back is gravity. The imparted speed will cause the wall to start to topple over a tiny bit, that is, its front side is lifted a bit off the ground. Then gravity pull that back, and cause the reverse impuls on the wall. That will in its turn impart a tiny impuls on the Earth itself, which will cause the Earth to move a bit faster.
So the real answer should be: because the wall is attached to the Earth, it will not move wrt. the Earth, and the Earth will take over the impuls.
Sir what does it mean in physical terms that a body has some finite momentum but zero velocity ?
@@renedekker9806 The impulse comes from elecreostatic forces between the tennis ball and the wall during the collision, gravity has nothing to do with it.
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.
Wow Great!
Congratulations for the set-up of the experiment!!
The art of scientific experiment design!!
its much more fun when things don't work surely
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.
+Harsha s n correct
Harsha s n
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
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.
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
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!
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.
j'ai beaucoup ses cours good explanation!
So many things got cleared in my head...thanks a ton ❤️
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.
Glad to hear that!
@@lecturesbywalterlewin.they9259 Thank you sir.
Wow!!! I love your lectures so much!!! It's really amazing and wonderful
I am a student of mathematics but I enjoy and feel your tasty lecture.
Thanks Prof. WL
It's my pleasure
As with every lecture there is always something that sets me thinking new thoughts.
maybe that's one of my secrets
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??
Loving these lectures
your lecture are so good that we loves to be there
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?
Hi. if KE of wall is zero because it’s v is close to zero, then momentum of wall should also be zero because it’s v is zero. So I don’t think your answer is correct.
@@AJ-ss3jy no it is correct because it isn't exactly 0. It is close to zero and the mass of the wall is large enough. The energy lost is in the form of potential energy of the wall.
We can also think about this in this way: The law of conservation of momentum does not hold for the problem as there is an external force of the ground on the wall.
@@mohammedbinalimaqqavi6599 momentum is conserved in the horizontal direction when the balls are at their lowest points and when they collide
and that is *KEY*
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
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...
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
You are always free to choose the boundaries of your system. If you choose the wall to be out of your system, then it´s easy to understand what you have explained. The problem that Walter presents is when you choose the wall to be part of your system, then it looks as if the laws of physics are breaking apart.
you are great, helped a lot in my preparation for competetive exams
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
You are really legend of Physics
Excellent lecture. Thanks to great Physicst of world 🙏🙏🙏🙏
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...
8 years later still gem ❤
Mass of wall (M) is infinite M➡∞
2mv/M=Vₘ therefore Vₘ➡0 and KE➡0
Thanks a lot....ur lectures help me during my preparation for MBBS entrance test
Glad to hear that
Sir u r the best physics lecturer of 22nd century
:)
Magneto Xmen Are there flying cars in the 22nd Century?
Love from india mesmerized by ur practical way of teaching
07:58
I can’t believe even such an explanation can bring such expressions on the face of students.
XD
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?
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.
+luvinthe jazz I appreciate your comments. The question is, can an object which has a mass which in the limiting case goes to infinity have momentum but no KE? In outer space bounce a tennis ball (mass m, speed v) off a wall with mass, M, of 10^10 kg and with speed,V=0. The collision is elastic. Clearly after the coll, the tennis ball will have to a veryyyyyyyyyy high degree of accuracy the same speed as before the coll but the direction will be reversed. Before the coll M has speed V=0.. After the coll M has speed V'. In the limiting case that M=>infinity V' must be zero. If not, the kinetic energy of the wall would be infinitely high and that violates the conservation of mechanical energy (in this case the conservation of KE). Yet the momentum of the wall must be +2mv as momentum is conserved. How can the wall have momentum but no KE?
+Lectures by Walter Lewin. They will make you ♥ Physics. Maybe this will help you. www.physicsforums.com/threads/why-momentum-must-be-in-the-wall.287325/
+Lectures by Walter Lewin. They will make you ♥ Physics. I Got it!!! So the change in momentum is +2mv. So if v=0, then 2mv=0. As Foghorn Leghorn said, two nothins is nothin.
+luvinthe jazz 2mv is NOT zero. m is the mass of the tennis ball, v is its speed. After the collision the wall has momentum +2mv but its KE is zero.
+Walter Lewin I think both KE and P are not zero, but v tends to zero so KE really tends to zero because the quadratic of v, talking always in an ideal closed-system case. You made m1 tends to zero to deduce m2 Ke is zero, but that's not completely true. Thanks for all your material.
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.
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.
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.
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!
For the problem of the wall and the ball. Consider instead of a wall, a huge cement block sitting on the ground (acts like wall). When the ball hits the block, it exerts a momentaneous force on the block. This force does transmit kinetic energy to the block, but at the same time, the ground exerts a static friction force on the block, taking away that energy, hence, the block doesn't move. Momentum isnt conserved as there is a net external force acting on the system (the friction force) as the collision takes place. The ball bounces back due to action-reaction principle (the block exerts an equal but opposite force on the ball, that accelerates it, changing the direction of its velocity). In other words, there is a net impulse over tge system that changes its momentum. We can replace the block with a wall and the friction f. with forces due to the configuration of the building. If the block was sitting on a frictionless surface or the wall was inserted in a peculiar system of frictionless rails attached to the ceiling and to the ground, then there wouldnt be a net external force and momentum would be conserved. In this case we would see indeed that both the wall and the block would move (with the corresponding velocities arising from the equations).Would this reasoning be correct sir?
you have not yet solved the wall - tennis ball problem. In the case of very high mass of the wall, the wall can have a HUGE momentum but ZERO KE.
@@lecturesbywalterlewin.they9259 Ive read a comment below suggesting you can consider the wall to be the earth. I dont understand this, as the ball itself would get attracted/accelerated towards the object it is going to colide with (due to gravity force). Wouldnt this make it impossible to apply conservation of momentum? In other words, in order to apply the M>>>m approximation dont you have to make sure the difference of masses is not big enough so that the gravitational interaction is no longer negligible in the first place?
@@lecturesbywalterlewin.they9259 what formula do we use to determine the momentum of the wall? Mass is huge (infinite) but velocity is 0, so momentum would still be zero.
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.
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...
Sir, i am a student and I'm preparing for NEET and your lacture help me in understanding pysics❤❤❤❤
Thank you so much prof l love this lecture and help me to understand physics
You are most welcome
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
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?
Are elastic/inelastic collisions related to the elasticity of the object ? So can an elastic object have an inelastic collision and vice versa ?
depends on the circumstances of the collision
I call what is happening to billiard balls when they collide "Momentum Transfer".
Here is why. If the balls have the same mass and there is no friction or external force between them, the moment and kinetic energy is conserved. Now, if a ball with mass 'm'--b1--collide with 2 balls at rest, its momentum is transferred to the first ball in contact--b2--and again, b2's moment is transferred to the third ball--b3--. So, mv1=mv3 and b3 takes off.
If we use mathematics, v1´=0 and v2´=v1(before it transfers momentum to b3). When v2 transfers, v2´=0 and v3´=v1. As an example, if 3 balls collide with 2 balls, the two balls in the 3 balls system push 2 balls at rest with 2mv and momentum is transferred ,so, the two balls takes off and 2 balls momentarily at rest and the remaining ball collide with them and as in previous argument, another 1 takes off. This all happened in a flash so, we see 3 balls take off. Professor Lewin,please correct me if I'm wrong.
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.
+Pedro Candeias KE=P^2/2M. In this case, P of the wall is 2mv (m is the mass of the tennis ball, v is its speed). Since M=>infinity, KE is ZERO, but P is not zero.
Regarding brain teaser at end of lec:
Let us assume k1 balls are moving with velocity v1 initially and k2 balls are moving with velocity v2 finally.
K1mv1=k2 mv2(conservation of momentum)
Hence K1v1=k2v2........... (1)
Since collisions are elastic KE is conserved.
Hence, K1(1/2m(v1)^2)=K2(1/2m(v2) ^2)
So, k1(v1) ^2=k2(v2) ^2.......... (2)
By solving equation (1) and (2) we get k1=k2.hence same no. Of balls should be ejected which initially hit.
Hence, proved.
IS IT CORRECT?
You assumed that k1 balls stopped after the collision, why? if you have let's say 2 balls with masses m1 m2 and velocities v1 v2 initially and u1 u2 finally even if v2=0 that doesn't mean that u1 will be 0 only if m1=m2 that will be true I think the same apply in your situation with k1 k2 so you assumed that u1=0 so it's like you assuming that k1=k2 and that is why you end up to this result
imagine the wall is the earth,
mu=m(-u)+Mv
p of the earth= Mv = 2mu
since M>>m
velocity of the earth v~=0
therefore, velocity of the tennis ball bouncing back =-u and approximately no change in KE for the ball
is this correct?
correct. KE=(1/2)M*V^2 M is infinity V is speed of Earth after colllision
KE=(1/2)P*V. P is 2mv, but V=0 thus KE of M is zero, but P is not zero.
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 balls momentum after bouncing back is (-m1v1)...and 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
sir you say V=0 then you are saying P not equals zero. P=MV. so if V=0 then P=0. but here V is only approximately zero, I think that's why you are telling there is momentum present in the wall. then it is obvious that it has a velocity of very very small magnitude and indeed it possesses a very very small kinetic energy. Amazing sir truly, to think the wall has a very small velocity
>>>P=MV. so if V=0 then P=0.>>>
that is not true of M=> infinity
what will be P if M>=infinity. will it be a definite number? I cant get it sir
This is a difficult chapter but a very interesting one! You got me on the last billiard ball experiment...I got two wrong
About the problem with the tennis ball and the wall:
We know that the momentum of the wall is: p(wall) = m(wall) * v(wall)
Then, by the equation for v1' and v2' for elastic collisions, we can substitute v(wall), and we get:
p(wall) = (2*m(wall)*m(ball)*v(ball)) / (m(wall) + m(ball))
We know that m(wall) >> m(ball), so as we take the lim m(wall) -> infinity we get that:
p(wall) = 2*m(ball)*v(ball) * lim (m(wall) -> inf) ( m(wall) / (m(wall) + m(ball)) )
the right side limit is of the indetermined form inf/inf, so we apply L'Hopital's Rule and we get that the limit is exactly 1, so:
p(wall) = 2*m(ball)*v(ball) = 2mv
Is this correct?
ofcoz p of the wal is +2mv after the collision, but the KE of the wall after the collision (if its mass M is thousands of tons) *is veri close to zero & can be made as small as you want to by increasing M.*
@@lecturesbywalterlewin.they9259 yes, i notice it when I made the limit as m2 -> inf of the KE for the wall, and it goes to 0
for the last problem...lets change the frame of reference....lets consider frame is on 5th ball and then we will see 3 balls are approaching to us and what we will see is 3 balls get detached from behind and balls which were approaching gets attached to 5th ball
it's a multiple collision problem. It's easy to show that the end result is consistent with the conserv of momentum and KE. But that is not the complete solution. use google
For your question, you said that KE is zero since the wall has no velocity, then how can it have a momentum without a velocity ?
+Shivam Agarwal Momentum can be HIGH and KE effectively zero for very massive objects. I suggest you figure this out on your own. If you don't succeed in 2 weeks, contact me again.
+Lectures by Walter Lewin. They will make you ♥ Physics. Is it because even if the have velocities very very low velocities, they have a high momentum because of the high amount of mass they posses ?
+Shivam Agarwal you are getting close. You can express all this in one equation which will tell the whole story. Can you do that?
+Lectures by Walter Lewin. They will make you ♥ Physics. Well if m2>>m1 and the momentum imparted to m2 is 2mv, then the speed of m2 will be given by 2mv/m2. But, because m2 is very very high, the velocity will approach 0. Thus KE which is given by 1/2m2v^2 will be about zero. Is that correct ?
+Shivam Agarwal yes that is correct.
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
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
If P is not zero, how can KE be zero?
Lectures by Walter Lewin. They will make you ♥ Physics.
Well, taking a different approach, surely if the wall gained any velocity after the collision then the energy is the system would be infinitely great which would violate the conservation of energy.
However, why does it not work such that P is finite and M is infinite so P/M=0, and despite the wall having gained momentum, it cannot have gained and velocity?
ok
thank you so much sir... your videos help me a lot in understanding physics.
:)
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.
en.wikipedia.org/wiki/Speed_of_sound
@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....
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.
Sir as you said
I watched this 💕 amazing lec but
I have still doubt 🙏 plz plz sir clear it🥺
My teacher said *KE is conserve "in" elastic collision but not "during" elastic collision*
Is he correct ?
but not during *inelastic* collisions
@@lecturesbywalterlewin.they9259 Sir he is talking about elastic not inelastic 🙏plzzz sir clear it 🙏
He said that we cannot use term "during" bcoz @ exact time of collision KE is not conserved as it may be converted into potential energy for that Instant so the kE is conserved during elastic collision is wrong
🙏🙏🙏Plz 🥺 sir confirm that is he correct or we can use both "in" & "during"
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
how do we actually derive v2'=(2m1/(m1+m2))*v1?
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?
use google. this cradle is discussed extensively on line
Do you think parallel universes exist and dreams are glimpses into them?
As I listened to your lecture 16 using iPad on my bed, I fell into sleep. I had a dream in which I sat in your MIT auditorium watching your alternative lecturing momentum and elastic collision. Instead of using ping pong and billiard balls to demonstrate the three cases, your alternative was using different sized globes. Everything looks realistic.
>>>Do you think parallel universes exist>>>
use google
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.
I have a really stupid question I guess; Why use V1, V2, etc.; why not use V, V2, etc. (isn't the first V assumed to be V1, so using V1 would be redundant?)? Same for Time, Mass, etc. Also, what is Prime and its value? I'm not a math guy, but I love watching these lectures; for some reason this seems to make more sense to me than algebra - algebra makes me hate math so I never bothered trying to move past it (is it really necessary to understand algebra in order to under stand stuff like this?). Thanks for your time!
V1 is not redundant. It specifically refers to the velocity of the object we call object 1. V would simply mean velocity in general.
As for the prime notation, it could either mean Newton's notation of derivative function notation (such as f'(x), pronounced "f prime of x"), or it could mean a way to label a second instance of a variable you've already used. In this particular lecture's example, the prime notation indicates the velocity of the center of mass in the lab reference frame.
Understanding physics without sufficient knowledge in algebra is like using a fork to eat a soup
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 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳
Regarding the problem at the last of the lecture (48:01)-
Let the mass of each ball be 'm' and their velocity is 'v'......
then the initial momentum of the system= 5mv
Now since it is an elastic collision so some balls after the collision will stay put and some will move with the same velocity 'v'...
Let the number of balls which after collision stay put be 'x'.......
then the number of balls which will move with velocity 'v' after the collision will be (8-x)........
So the momentum after the collision will be (8-x)mv.........
Since momentum of system is conserved so 5mv = (8-x)mv.........
which gives x=3 AND SO 3 BALLS WILL STAY AT REST AND OTHER 5 BALLS WILL MOVE WITH VELOCITY 'v'.........
Is it correct?
Nice try. It's easy to show that the "surprising"outcome of any combination of balls that you swing is consistent with the conservation of KE and with the conservation of momentum. But that is only a "consistency test". But that is not a proof. The actual physics is much more complicated as it is a multiple-collision problem. My advice is to live with your "consistency test" and not bother about the "proof".
en.wikipedia.org/wiki/Newton%27s_cradle
Thank you sir for the lecture.... I am not able to find a way to prove the happenings at the end of the lecture. Why is it so that when we hit 5 balls on 3 balls, 3 balls remain and 5 move on?
This is called Newton's cradle. Searched the web for solutions.
Very roughly speaking,Because as we can see all balls are identicle and if we consider energy transfer by five balls is 5 times 1/Mv2 then the transfer of energy to three balls is 3 times 1/Mv2 and the system has still excess 2 times 1/Mv2 and that will transfer to last two ball of five ball system and then system will came to equilibrium .....
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!
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?
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 ?
Momentum is conserved as long as there is no net impulse from external forces.
This will happen if:
1. There are no external forces
2. External forces add up to zero
3. The collision happens so quickly, that any cumulative effect from external forces is zero, during the instant the collision happens.
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?
I do not solve problems for viewers. I teach physics. Watch my lectures and you will find your answer.
@@lecturesbywalterlewin.they9259 okay...Thank you sir
I would like to leave a question about the ball-wall collision. If the ball with mass m collides with an initial velocity v and comes out with a velocity v in the other direction , than there must be a momentum transfer to the wall of +2mv, because of the conservation of momentum of the system. I understand that K=P/2M, where M is the mass of the wall, and can be thought as a limit case where M->infinity, threfore K->0, even though P is not zero. However, where is the momentum in the wall? I mean, physically speaking, I can not understand how can something have momentum and no velocity.
>> I understand that K=P/2M> That is not correct. K=p^2/2m.
Other than that your reasoning is OK
Thank you professor, I didn't notice that! But my question remains on how can you have momentum without velocity, that seems weird.
At the collision the tennis ball transfers a momentum p=2mv to the wall. m is the mass and v the speed of the ball. The KE of the wall is p^2/2M Here M is the mass of the wall. p is finite (+2mv) If M ==> infinity, KE ==> 0 but p remains +2mv. There is another way to look at this. The momentum of the wall is 2mv=MV (V is the velocity of the wall after the collision). Thus MV is fixed M=> infinity then V must go to zero to make sure that MV is finite. KE of the wall is 0.5MV^2=0.5pV. p is finite but V=>0. Thus KE=> zero.
let momentum of wall be M
by conservation of momentum
mv=- mv+M
M=2mv
let kinetic energy of wall be K
as the collision is perfectly elastic kinetic energy is conserved so,
1/2mv^2=1/2m(-v)^2+K
K=0
i think that must be answer tell me if i m right or wrong
I assume you mean momentum of the wall after impact is MV (here V is the velocity after the collision).
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??
+Safder Shakil Momentum is conserved. When the tennis ball approaches the wall, its momentum is mv, after the bounce it is --mv. The difference is 2mv.
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?
Suppose a ball is thrown upwards in a room and it hits the ceiling and comes back down. We observe that the velocity of the ball increases. Conservation of momentum does not take place in this case, does it? External force is indeed applied initially.
If the the ball hits the ceiling with a "decent speed" of about 1 m/s and if the collision takes place in a very short time (0.1 msec), then to a very high degree of accuracy the magnitude of the momentum of the ball just before and just after the collision are the same (assuming that it is an elastic collision). The reason is that the change in velocity due to gravity during the 0.1 msec collision is only 10*10(-4) = 10^(-3) m/sec.
Lectures by Walter Lewin. They will make you ♥ Physics. But in real cases, velocity increases by a rather large and considerable value?
So interesting! Thank you!
My pleasure!
Can anyone tell me what is the difference between Center of Mass And Center of Gravity ?
Good question. The reason we commonly get away with using these terms interchangeably, is that Earth's gravitational field is close enough to uniform in most situations, that there is no significant difference between these points. But if you did have a non-uniform gravitational field in the problem in question, there would be a difference.
Center of mass is what you calculate when you take a "weighted average" of the positions of all the masses that comprise an object or system of objects. By "weighted average" I mean that you multiply mass*position for each object, add it up, and then divide by total mass. This calculation is gravity-agnostic, as it ignores any gravitational fields that apply to the objects in question. It exclusively studies the inertial masses of the objects.
Center of gravity is what you get, when instead of using mass alone to take the weighted average, you account for the gravitational force when taking a weighted average of the positions of the point masses in the system. Center of gravity is the point at which gravity appears to act, as if it were a single concentrated force. It is the point where you could apply a single point support force, and not require any torque to balance it. If you apply a point support force at exactly the center of gravity, the object will be in an equilibrium of neutral stability. Above, and it will have a stable equilibrium; below, and it will have an unstable equilibrium.
Sir please guess me some books for physics to do in engineering
ask an engineer
Who liked it
ask an engineer
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