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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

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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.

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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...

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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

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Thankyou 😭❤️, I used to get so puzzled by having two unknown velocities

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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.

tq so much!! I've always stuck at the perfectly elastic collision parts

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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.

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Hi. How would save in similar problem in which the first ball was accelerating (or decelerating)?

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I thought elastic collisions need to use the vf= (m1-2)/(m1+m2)*vo Formulas?

My tutor love from Zambia

Will the eq 2 be the same if the equation e=-(Vaf-Vbf)/Vai-Vbi is used?

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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..

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My prof would absolutely kill me if I used this method, but this is awesome, thank you.

For an easy short cut use this method: Vf1=(m1-m2/m1+m2)*V1 Vf2=(2m1/m1+m2)*V1

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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?

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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?

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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?

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do u have a video where u have done with the kinetic energy formula?

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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 ]

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is this for igcse??

Super logic.

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 ]

Awesome! 6:01 Like Newton Law Gravitation. G= 6.67 * 10^-11

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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?

can someone explain to me in what class/grade do u learn these in the US?

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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.

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.

i'm confused, i got the second equation different, mine is: Va(initial) = Va(final) + Vb(final) the rule is: the left hand of the equation is for initials or before the collision, and the right hand is for the final or after the collision. correct me if im wrong tho

5:08 can I use the substitution method?

Thank you tutor

6:07 How do u know which equation u will substitute V2 into?

please more complex examples. your examples are often too easy compare to my physics course. thanks,

Has anyone come up with equations for V1' = and v2' =?

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You can also use subtitution to solve the 2nd equation into the 1st equation you just have to arrange the 2nd equation to solve for one of the velocities.

Super easy 😍

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Where does the "4" multiplier in (v1'+v2'=5 )(4) comes from?

How you get - v1'

what do i do when the moving ball at the beginning is lighter than the ball at rest?

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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.