For anyone wondering where the formula v1 + v1' = v2 + v2' comes from, this is it: In a collision between two objects, net momentum is conserved, therefore: p1 + p2 = p1' + p2' This can be expanded to: (m1 x v1) + (m2 x v2) = (m1 x v1') + (m2 x v2') Next, we subtract (m1 x v1') from both sides of the equation: (m1 x v1) + (m2 x v2) - (m1 x v1') = (m2 x v2') Then, we subtract (m2 x v2) from both sides of the equation: (m1 x v1) - (m1 x v1') = (m2 x v2') - (m2 x v2) Next, we group similar terms: m1(v1 - v1') = m2(v2' - v2) Finally, we can make m1 the subject of the equation by dividing both sides by (v1 - v1'): m1 = m2(v2' - v2) / (v1 - v1') Put a mental pin on this equation, we'll come back to it later. An elastic collision is defined as a collision where net kinetic energy is conserved, therefore: Ek1 + Ek2 = Ek1' + Ek2'. This can be expanded to: (1/2 x m1 x v1^2) + (1/2 x m2 x v2^2) = (1/2 x m1 x v1'^2) + (1/2 x m2 x v2'^2) Next, we multiply both sides of the equation by two to get: (m1 x v1^2) + (m2 x v2^2) = (m1 x v1'^2) + (m2 x v2'^2). Then, we subtract (m1 x v1'^2) from both sides of the equation to get: (m1 x v1^2) + (m2 x v2^2) - (m1 x v1'^2) = (m2 x v2'^2) Then, we subtract (m2 x v2^2) from both sides to get: (m1 x v1^2) - (m1 x v1'^2) = (m2 x v2'^2) - (m2 x v2^2) Next, we group similar terms: m1(v1^2 - v1'^2) = m2(v2'^2 - v2^2) We can now go back to the pinned equation and, using substitution, replace m1 with (m2(v2' - v2) / (v1 - v1')): (m2(v2' - v2) / (v1 - v1')) x (v1^2 - v1'^2) = m2(v2'^2 - v2^2) Now get rid of the extra brackets: m2(v2' - v2) / (v1 - v1') x (v1^2 - v1'^2) = m2(v2'^2 - v2^2) Now divide both sides by m2: (v2' - v2) / (v1 - v1') x (v1^2 - v1'^2) = (v2'^2 - v2^2) Now divide both sides by (v1^2 - v1'^2): (v2' - v2) / (v1 - v1') = (v2'^2 - v2^2) / (v1^2 - v1'^2) Now expand the squares: (v2' - v2) / (v1 - v1') = (v2' - v2)(v2' + v2) / (v1 - v1')(v1 + v1') Now multiply both sides by (v1 - v1'): (v2' - v2) = (v2' - v2)(v2' + v2) / (v1 + v1') Now divide both sides by (v2' - v2): 1 = (v2' + v2) / (v1 + v1') Finally, multiply both sides by (v1 + v1'): (v1 + v1') = (v2' + v2) Get rid of the brackets: v1 + v1' = v2' + v2 This can be rearranged to: v1 + v1' = v2 + v2' Thx for listening to my TED Talk
In summary, the formula comes from using substitution to combine the formulas for conservation of momentum and conservation of kinetic energy, followed by simplification. It's therefore important to note that the formula v1 + v1' = v2 + v2' only applies to elastic collisions because this is the only kind of collision in which kinetic energy is conserved; it does NOT apply to inelastic collisions.
Thank you so much for your videos! I take college physics and my professor is not good at all. I spend an hour in his class and learn way more watching your 15 minutes videos. You're awesome!!
This explanation saved my life and my career as a physics student. you wouldn't believe how confusing my textbook is at explaining this. thank you so much!
I am a college student and last week, we just solved a collision problem that can't be solve with just one equation like your example. The solution our teacher taught us takes about 10 mins to use and we find it so difficult. After watching this video now, I am so amazed that those 10 mins can be shortened into 2 mins by this method that you taught (use velocities instead of KE). Thank you so much man.
this is SO much easier than the method my professor is having us do. like the algebra is so much easier and it's way less time-consuming. thank you!!!!
Thank you! I could not find a video about this topic no matter how hard I searched, but it seems like you have come to my rescue once again. Thank you so much!
Can you explain how you got the v1 + v1' = v2 + v2' from the kinetic conservation equation? I started from them all being 1/2mv^2 and I can't get rid of the m. Its really confusing how you could just simplify it that way.
I’m currently in an 1 & 2 AP physics class for the fact that I won’t have to take a science next year if I pass. Thanks for helping me get through the struggle!
my professor did the same problem in very difficult way which I couldn't understand. But, after watching your video it feels like a piece of cake. Thank you very much
This is a great use of algebra, and I will definitely be using this from now on. However, it doesn't seem to account for the units kg and p, which seem to be used as multipliers and place holders. Dividing kgms^-1 by ms^-1 wouldn't produce a velocity, it would produce a mass. Is there any way to use this method while also accounting for those units? Or am I making an error?
i didnt not know about the equation u1+v1=u2+v2 i think my teacher forgot to explain it or i wasnt focusing either way thank you. you learn a new thing every day
So if I'm doing this in two dimensions, where the velocities are vectors, would I just do this twice? Once for the x velocities and once for the y velocities?
Correct me if I am wrong, but this equation v1f = v2i + v2f - v1i; only holds true when the mass of moving object is bigger than the object which is at rest. The generic one that works in all situations is v1'= [ (m1-m2) / (m1+m2)*v1 ] + [ 2*m2 / (m1+m2)*v2 ] v2'= [ 2*m2 / m1+m2*v1 ] + [ (m2-m1) / (m1+m2)*v2 ]
@@Manu-sk7qx if you're in college physics is more application based and explains concepts using if you take general physics (which is what is usually offered in high school if you don't take ap physics) it won't explain concepts w/ calculus since that's more geared towards engineers and physicists; it will be more concept-based
Thanks for the explanation. You made the explanation easy to follow. I just have a question - if the initial angles of trajectory of the balls are different (i.e. the paths of travel converge to the point of impact) and they are both in motion, traveling at different velocities towards each other, how would this be calculated. I would imagine that you would calculate the x and y components separately. But I'm really not sure of the process...
Please how do I become a member of this channel to get access to the membership videos. I can't find the join button. Please help me I need to access a video in the membership section. Thank you.
if anyone is interested, you should end up with these 2 equations: v2f = ((m1*v1i) - (m1*v2i) + P) / (m1 + m2); v1f = v2i + v2f - v1i; where P=total momentum before/after as it is assumed to be conserved.
I have a concern, the first equation had a momentum which means it's Kgm/s and the second one was velocity m/s , how can you add/subtract the two. they are not simple constants right?
I don't get one moment in collisions. Let's take the example from the video. 4kg ball moves towards the 2kg ball and touches it. 4kg ball exerts force on the second ball for some amount of time and the second also exerts force on the first (Newton's third law). During this, second ball will accelerate while the first is deaccelerating. At some moment they both will reach equal velocity. If they have equal velocities, they no longer exert force on each other. So, they will continue their motion with equal velocities. I would really appreciate if some kind person could help me to understand this or maybe could give me a link to a page where it is explained well.
in the first example specifically the algebra part the constant in the first equation is MV while on the second equation is just V why did you use systems?
Can someone help me how to find one missing variable using the m1v1' + m2v2' equation Question: a 3kg object is moving at east direction at 5.0m/s. It strikes a 6.0 kg object moving at east 2.0m/s.Objects have one dimensional collision in the east-west direction. The velocity of the 3.0kg object after collision is?
You are my god sir 😭😭😭 I invented a new formula because of you sir thanks a lot ....we are proud to have you ..thank you for explaining everything with proofs very nicely ! U are Einstein and I'm minkowski..sir thanks again
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For anyone wondering where the formula v1 + v1' = v2 + v2' comes from, this is it:
In a collision between two objects, net momentum is conserved, therefore:
p1 + p2 = p1' + p2'
This can be expanded to:
(m1 x v1) + (m2 x v2) = (m1 x v1') + (m2 x v2')
Next, we subtract (m1 x v1') from both sides of the equation:
(m1 x v1) + (m2 x v2) - (m1 x v1') = (m2 x v2')
Then, we subtract (m2 x v2) from both sides of the equation:
(m1 x v1) - (m1 x v1') = (m2 x v2') - (m2 x v2)
Next, we group similar terms:
m1(v1 - v1') = m2(v2' - v2)
Finally, we can make m1 the subject of the equation by dividing both sides by (v1 - v1'):
m1 = m2(v2' - v2) / (v1 - v1')
Put a mental pin on this equation, we'll come back to it later.
An elastic collision is defined as a collision where net kinetic energy is conserved, therefore:
Ek1 + Ek2 = Ek1' + Ek2'.
This can be expanded to:
(1/2 x m1 x v1^2) + (1/2 x m2 x v2^2) = (1/2 x m1 x v1'^2) + (1/2 x m2 x v2'^2)
Next, we multiply both sides of the equation by two to get:
(m1 x v1^2) + (m2 x v2^2) = (m1 x v1'^2) + (m2 x v2'^2).
Then, we subtract (m1 x v1'^2) from both sides of the equation to get:
(m1 x v1^2) + (m2 x v2^2) - (m1 x v1'^2) = (m2 x v2'^2)
Then, we subtract (m2 x v2^2) from both sides to get:
(m1 x v1^2) - (m1 x v1'^2) = (m2 x v2'^2) - (m2 x v2^2)
Next, we group similar terms:
m1(v1^2 - v1'^2) = m2(v2'^2 - v2^2)
We can now go back to the pinned equation and, using substitution, replace m1 with (m2(v2' - v2) / (v1 - v1')):
(m2(v2' - v2) / (v1 - v1')) x (v1^2 - v1'^2) = m2(v2'^2 - v2^2)
Now get rid of the extra brackets:
m2(v2' - v2) / (v1 - v1') x (v1^2 - v1'^2) = m2(v2'^2 - v2^2)
Now divide both sides by m2:
(v2' - v2) / (v1 - v1') x (v1^2 - v1'^2) = (v2'^2 - v2^2)
Now divide both sides by (v1^2 - v1'^2):
(v2' - v2) / (v1 - v1') = (v2'^2 - v2^2) / (v1^2 - v1'^2)
Now expand the squares:
(v2' - v2) / (v1 - v1') = (v2' - v2)(v2' + v2) / (v1 - v1')(v1 + v1')
Now multiply both sides by (v1 - v1'):
(v2' - v2) = (v2' - v2)(v2' + v2) / (v1 + v1')
Now divide both sides by (v2' - v2):
1 = (v2' + v2) / (v1 + v1')
Finally, multiply both sides by (v1 + v1'):
(v1 + v1') = (v2' + v2)
Get rid of the brackets:
v1 + v1' = v2' + v2
This can be rearranged to:
v1 + v1' = v2 + v2'
Thx for listening to my TED Talk
😳
😵😵
yo, thanks bro
In summary, the formula comes from using substitution to combine the formulas for conservation of momentum and conservation of kinetic energy, followed by simplification. It's therefore important to note that the formula v1 + v1' = v2 + v2' only applies to elastic collisions because this is the only kind of collision in which kinetic energy is conserved; it does NOT apply to inelastic collisions.
I applaud your dedication. Thankyou!
We should really not skip ads as payback for the goodness this man has given.
True! he's supporting us so, we should support him too
not during the ap exam 1 day away OoO
Agree
I literally skipped ads then when reading this comment, refreshed and watched the ads lmao
@@lovelylovelyabunchlovely5053 t
Thank you so much for your videos! I take college physics and my professor is not good at all. I spend an hour in his class and learn way more watching your 15 minutes videos. You're awesome!!
your videos are getting me through my degree lol ur a better teacher than many of my professors
God bless him
😂😂
This explanation saved my life and my career as a physics student. you wouldn't believe how confusing my textbook is at explaining this. thank you so much!
Jasmine Young fr I’m so happy
Textbooks have to make everything more complicated than it should be. I don't care about the specifics, I just want to finish this course lol
I am a college student and last week, we just solved a collision problem that can't be solve with just one equation like your example. The solution our teacher taught us takes about 10 mins to use and we find it so difficult. After watching this video now, I am so amazed that those 10 mins can be shortened into 2 mins by this method that you taught (use velocities instead of KE). Thank you so much man.
College???? I'm in 9th grade bro and this is our lesson
Thank you Jesus! Finally, something that makes sense. I prayed over this and I was brought to this video. God is so good
just thought this
Don t involve god in this
All the time
Amen to that
@@mrhatman675 ok anti christ god is great
yea failing my test tomorrow
Same :(
Same
Same
same
same
You the only UA-camr that makes me understand physics
Watching from Zambia 🇿🇲
The textbook does no justice, thank you soooo much!!!
I hope people never stop making this kind of videos
this man is saving my semester, thanks
this is SO much easier than the method my professor is having us do. like the algebra is so much easier and it's way less time-consuming. thank you!!!!
Thank you! I could not find a video about this topic no matter how hard I searched, but it seems like you have come to my rescue once again. Thank you so much!
Can you explain how you got the v1 + v1' = v2 + v2' from the kinetic conservation equation? I started from them all being 1/2mv^2 and I can't get rid of the m. Its really confusing how you could just simplify it that way.
If they all have an m , then the m's cancel out.
But KE is proportional to v^2
Take the kinetic energy equation and divide by the linear momentum equation. Only V's will remain.
thank you!!
@Nedgy Thank you so much. You have added the real value to this stream of discussion.
thank you for explaining all this clearly- it helped me a lot
I love you, it's 1 in the morning and i'm better now
Simple formulas of velocities of balls after collision are V1'=(m1-m2/m1+m2)/v1..And for velocity of 2nd ball after collision is v2'=(2m1v1)/m1+m2..
Easiest way for finding velocities after collision...
You are the Best of all at explaining concepts for one to understand. May God Bless in abundance
For an easy short cut use this method:
Vf1=(m1-m2/m1+m2)*V1
Vf2=(2m1/m1+m2)*V1
I’m currently in an 1 & 2 AP physics class for the fact that I won’t have to take a science next year if I pass. Thanks for helping me get through the struggle!
I should just take the L
same here. im failing science 20 sitting at 23%.
I'll join you as well
i wouldn't give up so soon, college is 1 million times worse... :(
@@tenkshadw7370 I am in college, I ended up getting an 80 on it so I'm happy
LMAO exactly once I said after he started doing the elimination part I was like tf is this shit I-
honestly i think i love this man 😊😯
Dude your a frckin life saver
I thought elastic collisions need to use the vf= (m1-2)/(m1+m2)*vo Formulas?
that's also applicable
my professor did the same problem in very difficult way which I couldn't understand. But, after watching your video it feels like a piece of cake. Thank you very much
Will the eq 2 be the same if the equation
e=-(Vaf-Vbf)/Vai-Vbi is used?
My prof would absolutely kill me if I used this method, but this is awesome, thank you.
Thankyou 😭❤️, I used to get so puzzled by having two unknown velocities
I wish I would have found you while I was doing my high school physics. I used Khan Academy, but I find how you teach better.
I think that your channel name must be 'The Organic Science Tutor'
tq so much!! I've always stuck at the perfectly elastic collision parts
Bro this is the only video that helped me thanks alot
You really a big help. Sending you a lot of thanks 😘
Hi. How would save in similar problem in which the first ball was accelerating (or decelerating)?
Your a star teacher...💖
By the time he got to the end I almost forgot the question! Many thanks though, my text book never covered what to do if one of the v's was missing.
This is a great use of algebra, and I will definitely be using this from now on. However, it doesn't seem to account for the units kg and p, which seem to be used as multipliers and place holders. Dividing kgms^-1 by ms^-1 wouldn't produce a velocity, it would produce a mass. Is there any way to use this method while also accounting for those units? Or am I making an error?
My tutor love from Zambia
i didnt not know about the equation u1+v1=u2+v2 i think my teacher forgot to explain it or i wasnt focusing either way thank you. you learn a new thing every day
So if I'm doing this in two dimensions, where the velocities are vectors, would I just do this twice? Once for the x velocities and once for the y velocities?
I just started watching your videos recently and I really like how you explain stuff. Keep up the good work 🔥
Thanks a lot for such an amazing video and examples. God bless you, after hours of searching this video made my assignment questions really easy.
You, my good sir, are a gift from god
thank you so much you put the avengers to shame bc its actually you saving the world !
You're the better 🎉🎉
thank you so much for this video🌸🌸
Correct me if I am wrong, but this equation v1f = v2i + v2f - v1i; only holds true when the mass of moving object is bigger than the object which is at rest.
The generic one that works in all situations is
v1'= [ (m1-m2) / (m1+m2)*v1 ] + [ 2*m2 / (m1+m2)*v2 ]
v2'= [ 2*m2 / m1+m2*v1 ] + [ (m2-m1) / (m1+m2)*v2 ]
Thanks so much. I have finally conquered this question 🥰
A "thank you" is not at all enough for your help.....u really helped me
5:08 can I use the substitution method?
Thanks for making my studies easy😘God bless you
You know what’s crazy, I really had a 100% in the class before this unit 😂😢😭
Thanks a lot for your videos they have been of help to me
Where does the "4" multiplier in (v1'+v2'=5 )(4) comes from?
thanks man🙏, really appreciate this
Am impressed with your solving
what a guy!! Thank you
Awesome!
6:01 Like Newton Law Gravitation.
G= 6.67 * 10^-11
Thank you so much for these videos, they have been helped me a lot!
do u have a video where u have done with the kinetic energy formula?
Thanks mate. May God bless you.
Thanks a lot! This has been beautifully helpful.
I watch all ur videos and i like them all
Thank you so much! I'm failing my college physics class and this is a huge help since I'm afraid to talk to people lol
This is taught in college?!?!? I'm in 10th grade
@@Manu-sk7qx if you're in college physics is more application based and explains concepts using
if you take general physics (which is what is usually offered in high school if you don't take ap physics) it won't explain concepts w/ calculus since that's more geared towards engineers and physicists; it will be more concept-based
or u can find the 2 final velocities directly by this formulas:
v1'= [ (m1-m2) / (m1+m2)*v1 ] + [ 2*m2 / (m1+m2)*v2 ]
v2'= [ 2*m2 / m1+m2*v1 ] + [ (m2-m1) / (m1+m2)*v2 ]
Thanks for the explanation. You made the explanation easy to follow. I just have a question - if the initial angles of trajectory of the balls are different (i.e. the paths of travel converge to the point of impact) and they are both in motion, traveling at different velocities towards each other, how would this be calculated. I would imagine that you would calculate the x and y components separately. But I'm really not sure of the process...
Please how do I become a member of this channel to get access to the membership videos. I can't find the join button. Please help me I need to access a video in the membership section. Thank you.
6:07 How do u know which equation u will substitute V2 into?
sir you are a great teacher i want to meet you one day when iam a great student in physics
Thanks so much these videos have helped me so much
Brother you are god of physics even I learner many maths terms from your channel , i hate the books write the content
if anyone is interested, you should end up with these 2 equations:
v2f = ((m1*v1i) - (m1*v2i) + P) / (m1 + m2);
v1f = v2i + v2f - v1i;
where P=total momentum before/after as it is assumed to be conserved.
I really enjoyed the lesson
I have a concern, the first equation had a momentum which means it's Kgm/s and the second one was velocity m/s , how can you add/subtract the two. they are not simple constants right?
You really help me thank u
Really helpful ❣💪
2:03 “now you don’t want to use the equations for *kinetic energy* , it’d be a lot of *work* ”... I see what you did there
please more complex examples. your examples are often too easy compare to my physics course. thanks,
Same for me, but he is just giving a base for people who don't even know the simple ones (it includes me btw).
I don't get one moment in collisions. Let's take the example from the video. 4kg ball moves towards the 2kg ball and touches it. 4kg ball exerts force on the second ball for some amount of time and the second also exerts force on the first (Newton's third law). During this, second ball will accelerate while the first is deaccelerating. At some moment they both will reach equal velocity. If they have equal velocities, they no longer exert force on each other. So, they will continue their motion with equal velocities.
I would really appreciate if some kind person could help me to understand this or maybe could give me a link to a page where it is explained well.
can someone explain to me in what class/grade do u learn these in the US?
Good explanation teacher😉
in the first example
specifically the algebra part the constant in the first equation is MV
while on the second equation is just V
why did you use systems?
Thank you once again.
you're better than my physics teacher
Can someone help me how to find one missing variable using the
m1v1' + m2v2' equation
Question: a 3kg object is moving at east direction at 5.0m/s. It strikes a 6.0 kg object moving at east 2.0m/s.Objects have one dimensional collision in the east-west direction. The velocity of the 3.0kg object after collision is?
Thank you 😍
thank you !!!
I love you man!
man thank you so much
My teacher discussing this topic*
My braincells: sleeps*/
This dude in UA-cam
My braincells: my time has come~
You are my god sir 😭😭😭 I invented a new formula because of you sir thanks a lot ....we are proud to have you ..thank you for explaining everything with proofs very nicely ! U are Einstein and I'm minkowski..sir thanks again
Nice 👌
what do i do when the moving ball at the beginning is lighter than the ball at rest?
dear God this guy is fantastic
My textbook list a different equation, v1+v2=v1'+v2'. Do you know why that is? Please help.
This velocities of ball before collision and velocities of balls after collision and formula is v1+v2=-v1'+v2'..
Has anyone come up with equations for V1' = and v2' =?
Thank you!!
Insha Insha nzn
Thank you tutor