You remind me of what Prof Paul Dirac once said. He stressed the beauty of mathe and the importance of being able to predict the behaviour of an equation rather than just able to solve it. Your tenacity in working out a generalised form of an equation for intuition certainly satisfy the intellectual starvation of mathe and physics lovers like me. Thank you, please keep up your good work.
Your equation (3') tells you that v1f and v2f are two legs of a right triangle with hypotenuse v1i, which immediately gives you φ + θ = 90°. The extra algebraic manipulation, while not strictly necessary, was still instructive though. Enjoyed your video.
no, equation (3') is about v1f v2f and v1i satisfied an equations, they has nothing to do with the angle to be 90. v1f v2f and v1i are numbers not vectors so they have no direction. Imagine this: after colliding ball 2 somehow goes a wrong directions but v2f still the same, then equation (3') still satisfied. Or second way to understand: You need both momentum and energy equations for this answer, meanwhile equation (3') only for energy. The algebraic manipulation is just getting information from momentum equations and it strictly necessary
This seems to lack transfer of rotation, which is quite annoying since everyone seems to omit this, even though it's a crucial part of programming billiards/pool/snooker collisions.
So I'm studying for a physics exam, and a similar problem came up where it's an elastic collision with two billiard balls. Ball one moves with an initial velocity of 3m/s, and after collision with ball two, ball one travels 30 degrees to the x axis, where ball two travels 14.5 degrees below the x axis. I went through this whole video, as it was the closest thing I could find on how to solve a problem like this, and at the end noticed 30 and 14.5 do not add up to 90. Now I'm even more confused and have no idea how to solve for the final velocities. I've tried every method I can think of, and nothing works. Could you steer me in the right direction? I've spent over an hour on just this problem alone. Any help would be appreciated, thanks.
😂 I tried to answer your question and then realized it was one year ago! I hope you already got the answer. What I’m thinking is that this video is for perfect elastic collision. But your case is not a perfect elastic one, which involves restitution coefficient, 0
Based on the initial condition, you have a direct central impact which implies both ball velocities after the collision will be along this central line (x-axis) after the collision. You are calculating an “oblique impact” which implies you need to know the “off-axis angle” of the incident ball and resolve its components before the collision which you are not specifying.
@@pgille2 the idea is if a = x and b = y then a + b = x + y ( the same concept of adding like this also works with subtraction, division and multiplication)
This doesn’t help me as I am not given either of the angles to start with. I’m given the means to solve for the final velocity of the ball that was hit using energy.
I'm working on a similar problem that reads Two identical billiard balls are on a frictionless, horizontal surface. Ball 1 has an initial velocity of 5.00 m/s in the positive x direction, and ball 2 is initially at rest. The balls collide, and, after the collision, ball 1 is moving at a speed of 2.00 m/s at an angle of ϕ= 30.0° below the positive x axis, as shown in the diagram. What angle θ, in degrees, does the final velocity of ball 2 make with the x axis? What is the speed, in meters per second, of ball 2 after the collision? The final answer for the first part was 17.01 degrees which contradicts the point made at the end of the video
A 20kg body traveling east at 20m/s collides elastically with a 10kg body moving west at 2m/s. After the collision, the lighter body is scattered in a direction of 30°south of East. Calculate the speeds of the two bodies after collision and the direction of the heavier body
5:00 since it’s elastic, can i use the coefficient of restitution formula in instead of conservation of KE? I tried that but somehow i got different answer, not sure why. Someone please tell me, thanks 🙏
Hi! This question is driving me nuts!😥 Pls help! Two identical frictionless balls moving upward strike each other. If ball A is moving to the right with velocity 30fts^1 and making an angle of 30⁰ with the normal axis and ball B is moving to the left at a velocity of 40fts^s making an angle of 60⁰ with the normal axis. Assume e=0.9. Calculate the magnitude of the tangential and the normal velocoty component of ball B
@@amirrafiqi7232 Mathematically more difficult but write down conservation of momentum and kinetic energy and solve for the 2 final velocities. In this case the angle will depend on the masses.
Momentum is a vector quantity.. depending on the direction you take as +ve (in the video, + y is the positive direction), the vector quantity in the opposite direction will be - ve
@@5alidtheog924 Since the angle is formed below the x-axis, and suppose the angle is -30*(which is formed by going clockwise), you can also write that as 330* in standard position(which just means that that you form the triangle by going counter-clockwise). You would still get the same answer because sin(-30*) and sin(330*) are the same which would be negative 1/2. When he put - [sin(-30*)] it becomes -(-1/2) which then becomes 1/2. That is wrong because it now becomes moving in the positive direction when it is actually moving in the negative direction.He put the negative first because it actually is but when you put in the angle, you get 2 negatives when you just want 1 negative. It should be +m1*vf*sin(-30*)
You literally saved me from having a nervous breakdown cause I wasn't able to understand the method of how a question was solved cause the explanation was written In a book by an idiot
You can do this in 2 lines by going from 1/2 mv^2 = 1/2mv^2 + 1/2mv^2, dividing through by 1/2m and getting a pythagorean relationship between the velocities.
can someone do a video showing moving ball hitting moving ball? cant find this anywhere. If a pool ball moving 10 mph hits a pool ball moving towards it at 5 mph, what will happen? cant find answer anywhere.
m1, m2 masses, v1, v2 initial velocities, u1, u2, final velocities, then you have conservation of momentum for both axes, for the momentum is a vector. For the X axis you have m1 * v1x + m2 * v2x = m1 * u1x + m2 * u2x. You can do the same for the Y component. Now, you also have conservation of kinetic energy, if we talk about elastic collision ( there's no deformation, so no loss of kinetic energy ), so you have 1/2 * m1 * v1^2 + 1/2 * m2 * v2^2 = 1/2 * m1 * u1^2 + 1/2 * m2 * u2^2. Using these three equations you can find all the information you need. Don't forget about that vector components can be written using sin and cos. I am actually trying to simulate collision in 2D using JavaScript so I have to relearn all of this, still learning.
You remind me of what Prof Paul Dirac once said. He stressed the beauty of mathe and the importance of being able to predict the behaviour of an equation rather than just able to solve it. Your tenacity in working out a generalised form of an equation for intuition certainly satisfy the intellectual starvation of mathe and physics lovers like me. Thank you, please keep up your good work.
Thank you so much for the nice compliment.
omg thank you you are the best teacher ever and this explanation was so helpful !!! :DDD
this didn't work on my homework and I am shook. It made so much sense and all for it to not work.
That's quite helpful with all my doubts cleared please please make a video on loss of K.E during 1dimension non elastic collision
Your equation (3') tells you that v1f and v2f are two legs of a right triangle with hypotenuse v1i, which immediately gives you φ + θ = 90°. The extra algebraic manipulation, while not strictly necessary, was still instructive though. Enjoyed your video.
no, equation (3') is about v1f v2f and v1i satisfied an equations, they has nothing to do with the angle to be 90. v1f v2f and v1i are numbers not vectors so they have no direction. Imagine this: after colliding ball 2 somehow goes a wrong directions but v2f still the same, then equation (3') still satisfied. Or second way to understand: You need both momentum and energy equations for this answer, meanwhile equation (3') only for energy. The algebraic manipulation is just getting information from momentum equations and it strictly necessary
This seems to lack transfer of rotation, which is quite annoying since everyone seems to omit this, even though it's a crucial part of programming billiards/pool/snooker collisions.
So I'm studying for a physics exam, and a similar problem came up where it's an elastic collision with two billiard balls. Ball one moves with an initial velocity of 3m/s, and after collision with ball two, ball one travels 30 degrees to the x axis, where ball two travels 14.5 degrees below the x axis. I went through this whole video, as it was the closest thing I could find on how to solve a problem like this, and at the end noticed 30 and 14.5 do not add up to 90. Now I'm even more confused and have no idea how to solve for the final velocities. I've tried every method I can think of, and nothing works. Could you steer me in the right direction? I've spent over an hour on just this problem alone. Any help would be appreciated, thanks.
If the masses are different the angle will NOT be 90 degrees.
If masses are the same the angle must be 90.
😂 I tried to answer your question and then realized it was one year ago! I hope you already got the answer. What I’m thinking is that this video is for perfect elastic collision. But your case is not a perfect elastic one, which involves restitution coefficient, 0
Use sine law
Based on the initial condition, you have a direct central impact which implies both ball velocities after the collision will be along this central line (x-axis) after the collision. You are calculating an “oblique impact” which implies you need to know the “off-axis angle” of the incident ball and resolve its components before the collision which you are not specifying.
best ever lecture
Thank you
Outstanding video and lecture!
Thank you so much, you really explained it perfectly
I like the way you teach its very engaging. Thank you
I don't understand why you can add equations 1'' and 2'' together. Can you explain this please? Thanks
This guy is one curious cat! Can he be helped please?
@@pgille2 the idea is if a = x and b = y then a + b = x + y ( the same concept of adding like this also works with subtraction, division and multiplication)
Beautiful physics❤❤❤
Thank you! 🙂
We need more questions on collisions sir
Very helpful and well explained
Thnx a lot Sir.... Love my India 🇮🇳🇮🇳
i want you teach me physics so that i will understand every content of physics
I have several courses on Udemy. Use code PHYSICS999 to get the lowest price.
This doesn’t help me as I am not given either of the angles to start with. I’m given the means to solve for the final velocity of the ball that was hit using energy.
I only watched this cos i dont know how the cue ball acts after it collides with another ball on a snooker table...
I'm working on a similar problem that reads
Two identical billiard balls are on a frictionless, horizontal surface. Ball 1 has an initial velocity of 5.00 m/s in the positive x direction, and ball 2 is initially at rest. The balls collide, and, after the collision, ball 1 is moving at a speed of 2.00 m/s at an angle of ϕ=
30.0° below the positive x axis, as shown in the diagram.
What angle θ, in degrees, does the final velocity of ball 2 make with the x axis?
What is the speed, in meters per second, of ball 2 after the collision?
The final answer for the first part was 17.01 degrees which contradicts the point made at the end of the video
Your problem doesn’t say that the collision is elastic. If it’s not elastic than it’s a different problem
A 20kg body traveling east at 20m/s collides elastically with a 10kg body moving west at 2m/s. After the collision, the lighter body is scattered in a direction of 30°south of East. Calculate the speeds of the two bodies after collision and the direction of the heavier body
This helped a lot. Thank you.
It's the same that you made a cue ball radial shot. What happens in the opposite case?
This is so counter intuitive, what if one of the angles is 1? Then the other is 89? Right? But if it's 0 then it just bounces back?
5:00 since it’s elastic, can i use the coefficient of restitution formula in instead of conservation of KE? I tried that but somehow i got different answer, not sure why. Someone please tell me, thanks 🙏
Hi! This question is driving me nuts!😥 Pls help! Two identical frictionless balls moving upward strike each other. If ball A is moving to the right with velocity 30fts^1 and making an angle of 30⁰ with the normal axis and ball B is moving to the left at a velocity of 40fts^s making an angle of 60⁰ with the normal axis. Assume e=0.9. Calculate the magnitude of the tangential and the normal velocoty component of ball B
What would happen if the masses were different?
You probably already know now but the angles do not add up to 90 degrees
Nice work, really helped
So if the masses are the same the angles are always going to add up to 90 degrees? Regardless of the initial velocity?
Yes!
Unless it’s a head on collision of course.
@@PhysicsNinja what about different mass?
@@amirrafiqi7232 Mathematically more difficult but write down conservation of momentum and kinetic energy and solve for the 2 final velocities. In this case the angle will depend on the masses.
@@amirrafiqi7232 In 2D collisions, you can only have elastic collisions between objects of the same mass.
Thank you very much for the explaination
What if you were to put spin on the ball? How would you do that calculation?
This is amazing!!
What are some examples of center of gravity?
why did you introduce negative sign at 4:27
Momentum is a vector quantity.. depending on the direction you take as +ve (in the video, + y is the positive direction), the vector quantity in the opposite direction will be - ve
@@5alidtheog924 Since the angle is formed below the x-axis, and suppose the angle is -30*(which is formed by going clockwise), you can also write that as 330* in standard position(which just means that that you form the triangle by going counter-clockwise). You would still get the same answer because sin(-30*) and sin(330*) are the same which would be negative 1/2. When he put - [sin(-30*)] it becomes -(-1/2) which then becomes 1/2. That is wrong because it now becomes moving in the positive direction when it is actually moving in the negative direction.He put the negative first because it actually is but when you put in the angle, you get 2 negatives when you just want 1 negative. It should be +m1*vf*sin(-30*)
How to find both angles then?
thanks a lot!!
You literally saved me from having a nervous breakdown cause I wasn't able to understand the method of how a question was solved cause the explanation was written In a book by an idiot
why don't we calculate kinetic energy for both, x and y, directions? pls answer
Kinetic energy is a scalar NOT a vector, it has no direction, it’s just a number.
@@PhysicsNinja thank you a lot
Good afternoon sir 😁
Is this alhazen's billiard problem?
What happens if one ball hits two stationary balls?
deez nuts
Thank you sir so much
My question is, is this equation true for all elastic Collisions???
No, this formula assumes the mass of the balls are equal.
@@PhysicsNinja yes thats what I needed to know thank you
one question i have is should we make phi negative due to the unit circle and stuff or am I overthinking it or?
I took the sign into consideration when I wrote the conservative of momentum.
adding the two equations at 13:08 did you miss writing '2' V1f^2 ......+ '2' V2f^2
Wo wo please sir reply me please then what is oblique collision what is the difference between oblique and elastic collision in 2d
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I would like to see an example that uses the example of dart satellite impacting with didymos b. A fascinating real world example
Bruuuu ##🔥🔥🔥🔥🔥🔥🔥 iam sooo grateful thank you a ton it was so helpful 1 more sub
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You can do this in 2 lines by going from 1/2 mv^2 = 1/2mv^2 + 1/2mv^2, dividing through by 1/2m and getting a pythagorean relationship between the velocities.
This works nicely for the case of the same mass. It doesn’t work for the case of different masses.
you had a^2 + b^2 = c^2 right from the beginning -- that's Pythagorean's theorem -- didn't have to do all that math -- to prove it's 90 degrees
Almost. You won't get the right answer if the angle theta =0 ( a head on collision). The general solution works for all cases.
The lesson was too fast
but what if I want to know how to solve two different masses
Same approach just more math. The masses will not cancel out if they are different.
Can't really call it billiard if you don't account for a spin. ;)
If it is not stated that the collision is elastic what do we do?
I feel like i waisted so much time when i was told the final answer XD, but thankyou anyways :)
*Wasted 😐
thank you sir
44,000th veiw
can someone do a video showing moving ball hitting moving ball? cant find this anywhere. If a pool ball moving 10 mph hits a pool ball moving towards it at 5 mph, what will happen? cant find answer anywhere.
m1, m2 masses, v1, v2 initial velocities, u1, u2, final velocities, then you have conservation of momentum for both axes, for the momentum is a vector. For the X axis you have m1 * v1x + m2 * v2x = m1 * u1x + m2 * u2x. You can do the same for the Y component. Now, you also have conservation of kinetic energy, if we talk about elastic collision ( there's no deformation, so no loss of kinetic energy ), so you have 1/2 * m1 * v1^2 + 1/2 * m2 * v2^2 = 1/2 * m1 * u1^2 + 1/2 * m2 * u2^2. Using these three equations you can find all the information you need. Don't forget about that vector components can be written using sin and cos.
I am actually trying to simulate collision in 2D using JavaScript so I have to relearn all of this, still learning.